CHEM 122 - GENERAL CHEMISTRY II -
Chemical analysis Question Bank
Question 1
Solution: Step 1: Substitute the absorbance value of the unknown solution
into the standard curve equation. A = 0.05C + 0.02 0.15 = 0.05C + 0.02
Step 2: Solve for C (concentration). 0.15 - 0.02 = 0.05C 0.13 = 0.05C C =
0.13 / 0.05 C = 2.6 mg/mL
Therefore, the concentration of the compound in the unknown solution is
2.6 mg/mL.Question 1: A student performs a chemical analysis to
determine the concentration of a certain compound in a solution.
The student uses a spectrophotometer to measure the absorbance of
the solution at a specific wavelength. The standard curve equation
obtained from known concentrations of the compound is A = 0.05C
+ 0.02, where A is the absorbance and C is the concentration in
mg/mL. The absorbance of the unknown solution is measured to be
0.15. Calculate the concentration of the compound in the unknown
solution.
Solution: Step 1: Substitute the absorbance value of the unknown
solution into the standard curve equation. A = 0.05C + 0.02 0.15 =
0.05C + 0.02
Step 2: Solve for C (concentration). 0.15 - 0.02 = 0.05C 0.13 =
0.05C C = 0.13 / 0.05 C = 2.6 mg/mL
Therefore, the concentration of the compound in the unknown
solution is 2.6 mg/mL.
Question 2
If the chemist requires 20 mL of the NaOH solution to neutralize
25 mL of the vinegar sample, what is the concentration of acetic acid
(CH3COOH) in the vinegar sample?
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction. CH3COOH (aq) + NaOH (aq)
CH3COONa (aq)
+ H2O (l)
2. Determine the molarity of NaOH: Volume of NaOH solution
used = 20 mL = 0.02 L Molarity of NaOH solution = 0.1 M
1
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.Question 2: A chemist is analyzing a sample
of vinegar using titration. The chemist uses a solution of sodium
hydroxide (NaOH) with a known concentration of 0.1 M to titrate
the vinegar sample. The balanced chemical equation for the reaction
is: CH3COOH (aq) + NaOH (aq)
CH3COONa (aq) + H2O (l)
If the chemist requires 20 mL of the NaOH solution to neutralize
25 mL of the vinegar sample, what is the concentration of acetic acid
(CH3COOH) in the vinegar sample?
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction. CH3COOH (aq) + NaOH (aq)
CH3COONa (aq)
+ H2O (l)
2. Determine the molarity of NaOH: Volume of NaOH solution
used = 20 mL = 0.02 L Molarity of NaOH solution = 0.1 M
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.
Question 3
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
2
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.Question 3: A sample of iron ore
weighing 4.75 g was dissolved in acid and then titrated with a stan-
dard solution of potassium dichromate (K2Cr2O7). The reaction is
as follows:
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.
3
Question 4
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.Question 4:
Calculate the molarity of a solution that contains 15 grams of sodium
chloride (NaCl) dissolved in 500 mL of water.
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.
Question 5
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample is
0.100 M.Question 5: A student wants to determine the concentration
of acetic acid in a vinegar sample. The student performed a titration
with 0.100 M sodium hydroxide solution. It took 25.0 mL of the
sodium hydroxide solution to reach the equivalence point. Calculate
the concentration of acetic acid in the vinegar sample. The balanced
chemical equation for the reaction is:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample
is 0.100 M.
Question 6
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
5
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.Question
6: A sample of unknown compound is found to have a mass of 3.50
grams. When the compound is burned, it produces 6.30 grams of
carbon dioxide (CO2) and 1.80 grams of water (H2O). Determine the
empirical formula of the compound.
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.
Question 7
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
6
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.Question
7: A 0.250 g sample of a compound containing only nitrogen and oxy-
gen was burned in excess oxygen, yielding 0.415 g of nitrogen dioxide
(NO2). What is the empirical formula of the compound?
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.
Question 8
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
7
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.Question 8: A student performs an experiment to determine the
concentration of acetic acid in vinegar. The student titrates a 25.00
mL sample of vinegar with 0.100 M sodium hydroxide (NaOH) solu-
tion. It took 18.75 mL of the NaOH solution to reach the equivalence
point. Calculate the concentration of acetic acid in the vinegar.
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.
Question 9
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The em-
pirical formula of the compound is H2O, which is water.Question 9:
8
A sample of a compound contains 2.00 g of hydrogen and 16.00 g of
oxygen. Determine the empirical formula of the compound.
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The empir-
ical formula of the compound is H2O, which is water.
Question 10
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.Question
10: A sample of aluminum metal is reacted with excess hydrochloric
acid to produce aluminum chloride and hydrogen gas. If 5.00 grams
of aluminum react with 20.0 grams of hydrochloric acid, what mass
of aluminum chloride is produced? Convert the masses to moles and
determine the limiting reactant.
9
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.
10
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.Question 2: A chemist is analyzing a sample
of vinegar using titration. The chemist uses a solution of sodium
hydroxide (NaOH) with a known concentration of 0.1 M to titrate
the vinegar sample. The balanced chemical equation for the reaction
is: CH3COOH (aq) + NaOH (aq)
CH3COONa (aq) + H2O (l)
If the chemist requires 20 mL of the NaOH solution to neutralize
25 mL of the vinegar sample, what is the concentration of acetic acid
(CH3COOH) in the vinegar sample?
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction. CH3COOH (aq) + NaOH (aq)
CH3COONa (aq)
+ H2O (l)
2. Determine the molarity of NaOH: Volume of NaOH solution
used = 20 mL = 0.02 L Molarity of NaOH solution = 0.1 M
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.
Question 3
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
2
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.Question 3: A sample of iron ore
weighing 4.75 g was dissolved in acid and then titrated with a stan-
dard solution of potassium dichromate (K2Cr2O7). The reaction is
as follows:
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.
3
Question 4
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.Question 4:
Calculate the molarity of a solution that contains 15 grams of sodium
chloride (NaCl) dissolved in 500 mL of water.
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.
Question 5
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample is
0.100 M.Question 5: A student wants to determine the concentration
of acetic acid in a vinegar sample. The student performed a titration
with 0.100 M sodium hydroxide solution. It took 25.0 mL of the
sodium hydroxide solution to reach the equivalence point. Calculate
the concentration of acetic acid in the vinegar sample. The balanced
chemical equation for the reaction is:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample
is 0.100 M.
Question 6
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
5
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.Question
6: A sample of unknown compound is found to have a mass of 3.50
grams. When the compound is burned, it produces 6.30 grams of
carbon dioxide (CO2) and 1.80 grams of water (H2O). Determine the
empirical formula of the compound.
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.
Question 7
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
6
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.Question
7: A 0.250 g sample of a compound containing only nitrogen and oxy-
gen was burned in excess oxygen, yielding 0.415 g of nitrogen dioxide
(NO2). What is the empirical formula of the compound?
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.
Question 8
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
7
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.Question 8: A student performs an experiment to determine the
concentration of acetic acid in vinegar. The student titrates a 25.00
mL sample of vinegar with 0.100 M sodium hydroxide (NaOH) solu-
tion. It took 18.75 mL of the NaOH solution to reach the equivalence
point. Calculate the concentration of acetic acid in the vinegar.
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.
Question 9
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The em-
pirical formula of the compound is H2O, which is water.Question 9:
8
A sample of a compound contains 2.00 g of hydrogen and 16.00 g of
oxygen. Determine the empirical formula of the compound.
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The empir-
ical formula of the compound is H2O, which is water.
Question 10
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.Question
10: A sample of aluminum metal is reacted with excess hydrochloric
acid to produce aluminum chloride and hydrogen gas. If 5.00 grams
of aluminum react with 20.0 grams of hydrochloric acid, what mass
of aluminum chloride is produced? Convert the masses to moles and
determine the limiting reactant.
9
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.
10
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.Question 2: A chemist is analyzing a sample
of vinegar using titration. The chemist uses a solution of sodium
hydroxide (NaOH) with a known concentration of 0.1 M to titrate
the vinegar sample. The balanced chemical equation for the reaction
is: CH3COOH (aq) + NaOH (aq)
CH3COONa (aq) + H2O (l)
If the chemist requires 20 mL of the NaOH solution to neutralize
25 mL of the vinegar sample, what is the concentration of acetic acid
(CH3COOH) in the vinegar sample?
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction. CH3COOH (aq) + NaOH (aq)
CH3COONa (aq)
+ H2O (l)
2. Determine the molarity of NaOH: Volume of NaOH solution
used = 20 mL = 0.02 L Molarity of NaOH solution = 0.1 M
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.
Question 3
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
2
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.Question 3: A sample of iron ore
weighing 4.75 g was dissolved in acid and then titrated with a stan-
dard solution of potassium dichromate (K2Cr2O7). The reaction is
as follows:
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.
3
Question 4
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.Question 4:
Calculate the molarity of a solution that contains 15 grams of sodium
chloride (NaCl) dissolved in 500 mL of water.
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.
Question 5
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample is
0.100 M.Question 5: A student wants to determine the concentration
of acetic acid in a vinegar sample. The student performed a titration
with 0.100 M sodium hydroxide solution. It took 25.0 mL of the
sodium hydroxide solution to reach the equivalence point. Calculate
the concentration of acetic acid in the vinegar sample. The balanced
chemical equation for the reaction is:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample
is 0.100 M.
Question 6
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
5
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.Question
6: A sample of unknown compound is found to have a mass of 3.50
grams. When the compound is burned, it produces 6.30 grams of
carbon dioxide (CO2) and 1.80 grams of water (H2O). Determine the
empirical formula of the compound.
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.
Question 7
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
6
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.Question
7: A 0.250 g sample of a compound containing only nitrogen and oxy-
gen was burned in excess oxygen, yielding 0.415 g of nitrogen dioxide
(NO2). What is the empirical formula of the compound?
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.
Question 8
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
7
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.Question 8: A student performs an experiment to determine the
concentration of acetic acid in vinegar. The student titrates a 25.00
mL sample of vinegar with 0.100 M sodium hydroxide (NaOH) solu-
tion. It took 18.75 mL of the NaOH solution to reach the equivalence
point. Calculate the concentration of acetic acid in the vinegar.
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.
Question 9
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The em-
pirical formula of the compound is H2O, which is water.Question 9:
8
A sample of a compound contains 2.00 g of hydrogen and 16.00 g of
oxygen. Determine the empirical formula of the compound.
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The empir-
ical formula of the compound is H2O, which is water.
Question 10
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.Question
10: A sample of aluminum metal is reacted with excess hydrochloric
acid to produce aluminum chloride and hydrogen gas. If 5.00 grams
of aluminum react with 20.0 grams of hydrochloric acid, what mass
of aluminum chloride is produced? Convert the masses to moles and
determine the limiting reactant.
9
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.
10
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.Question 2: A chemist is analyzing a sample
of vinegar using titration. The chemist uses a solution of sodium
hydroxide (NaOH) with a known concentration of 0.1 M to titrate
the vinegar sample. The balanced chemical equation for the reaction
is: CH3COOH (aq) + NaOH (aq)
CH3COONa (aq) + H2O (l)
If the chemist requires 20 mL of the NaOH solution to neutralize
25 mL of the vinegar sample, what is the concentration of acetic acid
(CH3COOH) in the vinegar sample?
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction. CH3COOH (aq) + NaOH (aq)
CH3COONa (aq)
+ H2O (l)
2. Determine the molarity of NaOH: Volume of NaOH solution
used = 20 mL = 0.02 L Molarity of NaOH solution = 0.1 M
3. Determine the moles of NaOH used in the reaction: Moles of
NaOH = Molarity
Ö
Volume (in L) Moles of NaOH = 0.1 M
Ö
0.02
L = 0.002 moles
4. Use the balanced chemical equation to determine the moles of
acetic acid (CH3COOH) in the vinegar sample: 1 mole of CH3COOH
reacts with 1 mole of NaOH Hence, moles of CH3COOH = moles of
NaOH = 0.002 moles
5. Determine the concentration of acetic acid (CH3COOH) in
the vinegar sample: Volume of vinegar sample = 25 mL = 0.025 L
Concentration of CH3COOH = moles / volume (in L) Concentration
of CH3COOH = 0.002 moles / 0.025 L = 0.08 M
Therefore, the concentration of acetic acid (CH3COOH) in the
vinegar sample is 0.08 M.
Question 3
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
2
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.Question 3: A sample of iron ore
weighing 4.75 g was dissolved in acid and then titrated with a stan-
dard solution of potassium dichromate (K2Cr2O7). The reaction is
as follows:
6 Fe2+(aq)+14H+(aq)+Cr2O72
−(aq)ß6F e3+(aq)+2Cr3+(aq)+7H2O(l)
If it took 37.5 mL of the potassium dichromate solution to reach
the endpoint, what is the concentration of this solution in mol/L?
The molar mass of K2Cr2O7 is 294.2 g/mol.
Step-by-step solution: 1. Write the balanced chemical equation
for the reaction between iron (II) ions and potassium dichromate. 6
Fe2+ (aq) + 14H+(aq) + Cr2O72
−(aq)ß6F e3+ (aq)+2Cr3+ (aq)+7H2O(l)
2. Determine the number of moles of iron in the sample using the
molar mass of iron and the given sample weight. Number of moles of
Fe = 4.75 g / molar mass of Fe
3. Use the stoichiometry of the balanced equation to determine
the number of moles of potassium dichromate consumed during the
reaction. From the balanced equation, 1 mol of K2Cr2O7 reacts with
6 mols of Fe2+NumberofmolesofK2Cr2O7 = (NumberofmolesofF e)/6
4. Calculate the concentration of K2Cr2O7 in the solution using
the volume and number of moles of K2Cr2O7 used during titration.
Concentration of K2Cr2O7 = (Number of moles of K2Cr2O7) / Vol-
ume of K2Cr2O7 solution used in titration
5. Convert the concentration to mol/L by dividing the concentra-
tion by the volume of the solution in liters.
Now solve the calculations to find the concentration of the potas-
sium dichromate solution in mol/L.
3
Question 4
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.Question 4:
Calculate the molarity of a solution that contains 15 grams of sodium
chloride (NaCl) dissolved in 500 mL of water.
Step 1: Convert the mass of sodium chloride to moles. Given:
Mass of NaCl = 15 grams Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
Number of moles of NaCl = 15 g / 58.44 g/mol Number of moles of
NaCl 0.256 moles
Step 2: Convert the volume of the solution to liters. Given: Vol-
ume of solution = 500 mL 1 L = 1000 mL
Volume of solution = 500 mL / 1000 mL/L Volume of solution =
0.5 L
Step 3: Calculate the molarity of the solution. Molarity (M) =
Number of moles / Volume of solution in liters Molarity (M) = 0.256
moles / 0.5 L Molarity (M) = 0.512 M
Therefore, the molarity of the solution containing 15 grams of
sodium chloride dissolved in 500 mL of water is 0.512 M.
Question 5
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample is
0.100 M.Question 5: A student wants to determine the concentration
of acetic acid in a vinegar sample. The student performed a titration
with 0.100 M sodium hydroxide solution. It took 25.0 mL of the
sodium hydroxide solution to reach the equivalence point. Calculate
the concentration of acetic acid in the vinegar sample. The balanced
chemical equation for the reaction is:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O (l)
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid and sodium hydroxide as provided:
CH3COOH (aq) + NaOH (aq) -¿ CH3COONa (aq) + H2O(l)
2. Determine the mole ratio between acetic acid and sodium hy-
droxide from the balanced equation. The ratio is 1:1.
3. Calculate the moles of sodium hydroxide used in the titration:
Moles of NaOH = (Volume of NaOH solution used (L)) x (Molarity
of NaOH) Moles of NaOH = (25.0 mL / 1000) L x 0.100 mol/L Moles
of NaOH = 0.00250 mol
4. Since the mole ratio between acetic acid and sodium hydroxide
is 1:1, the moles of acetic acid present in the vinegar sample are also
0.00250 mol.
5. Calculate the concentration of acetic acid in the vinegar sample:
Concentration of acetic acid = (Moles of acetic acid) / (Volume of
vinegar sample in liters) Assuming the volume of the vinegar sample
used in the titration is 25.0 mL (0.0250 L), Concentration of acetic
acid = 0.00250 mol / 0.0250 L Concentration of acetic acid = 0.100
M
Therefore, the concentration of acetic acid in the vinegar sample
is 0.100 M.
Question 6
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
5
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.Question
6: A sample of unknown compound is found to have a mass of 3.50
grams. When the compound is burned, it produces 6.30 grams of
carbon dioxide (CO2) and 1.80 grams of water (H2O). Determine the
empirical formula of the compound.
Step-by-step solution: 1. Calculate the moles of carbon (C) in the
compound: - Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol
(O) = 44.01 g/mol - Moles of CO2 = 6.30 g / 44.01 g/mol 0.143
moles of C
2. Calculate the moles of hydrogen (H) in the compound: - Molar
mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H2O = 1.80 g / 18.02 g/mol 0.100 moles of H
3. Determine the ratio of moles of C to moles of H in the com-
pound: - Divide moles of C by moles of H to find the simplest whole-
number ratio - Ratio of C:H 0.143 / 0.100 1.43:1 - Since we want
whole numbers, we round to the nearest whole number - Ratio of
C:H = 1:1
4. Write the empirical formula of the compound using the whole-
number ratio of C to H: - Empirical formula = CH
Therefore, the empirical formula of the compound is CH.
Question 7
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
6
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.Question
7: A 0.250 g sample of a compound containing only nitrogen and oxy-
gen was burned in excess oxygen, yielding 0.415 g of nitrogen dioxide
(NO2). What is the empirical formula of the compound?
Step-by-step solution: 1. Calculate the moles of nitrogen dioxide
(NO2) produced: - The molar mass of NO2 = 46.01 g/mol (from the
periodic table) - Moles of NO2 = mass of NO2 / molar mass of NO2
= 0.415 g / 46.01 g/mol 0.009 mol
2. Determine the moles of nitrogen in the compound: - From the
balanced chemical equation, 1 mole of nitrogen is present in 1 mole
of NO2. - Therefore, moles of nitrogen = 0.009 mol
3. Find the moles of oxygen in the compound: - Since the com-
pound contains only nitrogen and oxygen, the remaining moles of the
compound after the nitrogen is accounted for are due to oxygen. -
Moles of oxygen = total moles of compound - moles of nitrogen =
total moles of compound - 0.009 mol
4. Calculate the molar ratio of nitrogen to oxygen: - The molar
ratio of nitrogen to oxygen can be determined using the moles of
nitrogen and moles of oxygen.
5. Determine the empirical formula of the compound: - Write the
empirical formula using the molar ratios found in step 4.
Therefore, the empirical formula of the compound can be deter-
mined using the calculated molar ratios of nitrogen and oxygen.
Question 8
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
7
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.Question 8: A student performs an experiment to determine the
concentration of acetic acid in vinegar. The student titrates a 25.00
mL sample of vinegar with 0.100 M sodium hydroxide (NaOH) solu-
tion. It took 18.75 mL of the NaOH solution to reach the equivalence
point. Calculate the concentration of acetic acid in the vinegar.
Step-by-step solution: 1. Write the balanced chemical equation for
the reaction between acetic acid (CH3COOH) and sodium hydroxide
(NaOH): CH3COOH + NaOH
CH3COONa + H2O
2. Determine the moles of NaOH used in the titration: Moles
of NaOH = Molarity of NaOH
Ö
Volume of NaOH (in L) Moles of
NaOH = 0.100 mol/L
Ö
0.01875 L Moles of NaOH = 0.001875 mol
3. From the balanced chemical equation, it is a 1:1 molar ratio
between acetic acid and NaOH. Therefore, the moles of acetic acid
equal the moles of NaOH used in the titration.
4. Calculate the concentration of acetic acid in the vinegar: Con-
centration of acetic acid = Moles of acetic acid / Volume of vinegar
(in L) Volume of vinegar = 25.00 mL = 0.02500 L Concentration of
acetic acid = 0.001875 mol / 0.02500 L Concentration of acetic acid
= 0.075 M
Therefore, the concentration of acetic acid in the vinegar is 0.075
M.
Question 9
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The em-
pirical formula of the compound is H2O, which is water.Question 9:
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A sample of a compound contains 2.00 g of hydrogen and 16.00 g of
oxygen. Determine the empirical formula of the compound.
Step-by-step solution: 1. Determine the number of moles of each
element: - Moles of hydrogen:
moles of hydrogen =2.00 g
1.008 g/mol = 1.98 mol
- Moles of oxygen:
moles of oxygen =16.00 g
16.00 g/mol = 1.00 mol
2. Determine the mole ratio between hydrogen and oxygen: Divide
each number of moles by the smallest number of moles (1.00 mol): -
Hydrogen: 1.98 mol / 1.00 mol = 1.98 - Oxygen: 1.00 mol / 1.00 mol
= 1.00
3. Round the mole ratios to the nearest whole number: - Hydro-
gen: 1.98 2 - Oxygen: 1.00 1
4. Write the empirical formula using the mole ratios: The empir-
ical formula of the compound is H2O, which is water.
Question 10
Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.Question
10: A sample of aluminum metal is reacted with excess hydrochloric
acid to produce aluminum chloride and hydrogen gas. If 5.00 grams
of aluminum react with 20.0 grams of hydrochloric acid, what mass
of aluminum chloride is produced? Convert the masses to moles and
determine the limiting reactant.
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Step-by-step Solution: 1. Write the balanced chemical equation
for the reaction: 2Al + 6HCl -¿ 2AlCl3 + 3H2
2. Calculate the moles of aluminum: Molar mass of aluminum
(Al) = 26.98 g/mol 5.00 g Al x (1 mol Al / 26.98 g) = 0.185 mol Al
3. Calculate the moles of hydrochloric acid: Molar mass of hy-
drochloric acid (HCl) = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46
g/mol 20.0 g HCl x (1 mol HCl / 36.46 g) = 0.548 mol HCl
4. Determine the limiting reactant by comparing the moles of
aluminum and hydrochloric acid. The reactant that produces less
product is the limiting reactant. Aluminum: 0.185 mol Al x (2 mol
AlCl3 / 2 mol Al) = 0.185 mol AlCl3 Hydrochloric acid: 0.548 mol
HCl x (2 mol AlCl3 / 6 mol HCl) = 0.182 mol AlCl3
5. Since hydrochloric acid produces less aluminum chloride, it is
the limiting reactant.
6. Calculate the mass of aluminum chloride produced: Molar mass
of aluminum chloride (AlCl3) = 26.98 g (Al) + 3(35.45 g) (Cl) =
133.33 g/mol 0.182 mol AlCl3 x 133.33 g/mol = 24.28 g AlCl3
Therefore, 24.28 grams of aluminum chloride is produced.
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