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Chapters 3.1, 3.2, 3.4, 3.5, and 4.5: Composition of Substances and

Solutions

CHEM 101 Fall 2020

Dr. Lauren Genova

1 2

Chapter 3

By the end of this chapter, you will be able to: • Explain the relation between mass, moles, and numbers

of atoms or molecules, and perform calculations that require you to convert between these quantities

• Calculate solution concentrations using molarity

• Perform dilution calculations using the dilution equation

• Compute the percent composition of a compound

• Determine the empirical and molecular formulas of a compound

• Solve combustion analysis problems

Chapter 3: Learning Objectives

Scaling Up to the Real World…

12.0 amu = 0.0000000000000000000000199 g

± 0.0001 g

Can we measure this in real life?

1 carbon atom

12.0 amu

12.0 amu/atom

Chapter 3

The Mole Concept

Mole = standard unit for expressing macroscopic quantities of elements or compounds. Usually abbreviated as mol.

1 mole carbon atoms 6.022 × 1023 carbon atoms

12.01 grams

12.01 grams/mol

1 carbon atom

12.0 amu

12.0 amu/atom

Chapter 3

• A mole is a defined number of “chemical things” just like: – One pair of copper atoms = 2 copper atoms – One dozen water molecules = 12 water molecules – One mole of MgCl2 fomula units = 6.02 × 1023 MgCl2 formula units

• Avogadro’s number: the number of elementary entities (particles, atoms, molecules, formula units, etc.) in a mole of a substance.

= 6.02 × 1023 particles/mol = 6.02 × 1023 atoms/mol = 6.02 × 1023 molecules/mol = 6.02 × 1023 formula units/mol (the term "formula unit" is used instead of "molecule" when we refer to ionic compounds)

Avogadro’s Number Defines a Mole

Chapter 3

6.02 X 1023

Image: https://quizizz.com/admin/quiz/5c530b3d960462001b489154/moles

Molar Mass (Molecular Weight = MW) • Molar mass (= molecular weight, MW) of an element or

compound is the mass in grams of 1 mole of that substance. – Units of molar mass are g/mol (1 amu = 1 g/mol)

• Note: The masses of 1 mole of different elements are different since the masses of the individual atoms are directly different. (Each sample contains 6.02×1023 atoms = 1 mol of atoms.)

Chapter 3

https://quizizz.com/admin/quiz/5c530b3d960462001b489154/moles

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Molar Mass (Molecular Weight = MW) • To find the molar mass of an element, look up

the molar mass on the periodic table – Example: Molar mass of nitrogen

• To find the molar mass of a molecule/ compound, look up the molar masses of all the atoms in the compound and add them together – In other words, the molar mass of the compound is

the sum of the mass of the atoms – Example: Molar mass of NH3

= mass of 1 nitrogen atom + mass of 3 hydrogen atoms = 14.007 g/mol + 3 (1.008 g/mol) = 17.031 g/mol 7

Chapter 3

= 14.007 g/mol

Practice: Calculating Molecular Weight 1.) What is the molecular weight of H2O?

Chapter 3

Practice: Calculating Molecular Weight 2.) Ibuprofen, C13H18O2, is the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular weight of ibuprofen?

Chapter 3

Molar Mass as a Conversion Factor • Once you calculate the molar mass, you can use it as a

conversion factor to convert between grams and moles

• Use Avogadro’s number to convert between moles and particles (atoms, molecules, formula units, etc.)

• You can use the molecular formula itself to convert between types of a given atom in a molecule and total number of molecules.

Let’s practice each of these! 10

Chapter 3

Converting between Grams and Moles • Once you calculate the molar mass, you can use it as a

conversion factor to convert between grams and moles

Chapter 3

Number of grams Number of grams

molar mass

Converting between Grams and Moles: Example

Chapter 3

• According to nutritional guidelines from the USDA, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated requirement of potassium in moles?

4.7 g K × 1 mol K 39.10 g K

= 0.12 mol K

Atomic mass of K = 39.10 amu ∴ Molar mass (MW) of K = 39.10 g/mol

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Converting between Molecules (or Formula Units) and Moles

Chapter 3

• Use Avogadro’s number to convert between moles and particles (atoms, molecules, formula units, etc.)

Avogadro’s number

Converting between Molecules (or Formula Units) and Moles: Example

• Consider the formula for iron (II) sulfate, a common iron supplement. How many moles of iron ions are present in 1.2×1024 formula units of this supplement?

Chapter 3

1.2×1024 formula units Fe(SO4) × 1 mol Fe(SO4)

6.022×1023 formula units Fe(SO4) × 1 mol Fe

1 mol Fe(SO4) = 2.0 mol Fe

Number of atoms

1 molecule Number of molecules

1 molecule

Number of atoms

• You can use the molecular formula itself to convert between types of a given atom in a molecule and total number of molecules.

Chapter 3

Converting between Atoms and Molecules

Number of atoms Number of molecules

(from molecular formula)

Number of atoms

Converting between Atoms and Molecules: Example

• You have 5 molecules of aspirin, C9H8O4. How many atoms of C, H and O are present?

– Answer: 45 C atoms, 40 H atoms, 20 O atoms 16

Chapter 3

5 molecules C9H8O4 × 9 C atoms

1 molecule C9H8O4 = 45 C atoms

5 molecules C9H8O4 × 8 H atoms

1 molecule C9H8O4 = 40 H atoms

5 molecules C9H8O4 × 4 O atoms

1 molecule C9H8O4 = 20 O atoms

Practice: Mole Conversions 1

Chapter 3Chapter 2Chapter 2Poll Question #1

3.) How many molecules are there in 17.1 g of sucrose, C12H22O11? Molar mass of sucrose = 342.0 g/mol

A) 5.85 × 1021 B) 3.01 × 1022 C) 1.20 × 1025 D) 3.52 × 1027 E) None of the above

4.) Ascorbic acid, or vitamin C (C6H8O6) is an essential vitamin. If a typical tablet contains 500.0 mg of vitamin C, how many carbon atoms are present in the tablet? The molar mass of vitamin C is 176.1 g/mol.

A) 1.026 × 1022 carbon atoms B) 3.522 × 1021 carbon atoms C) 1.711 × 1021 carbon atoms D) 7.841 × 1020 carbon atoms E) None of the above

Chapter 3

Practice: Mole Conversions 2

Chapter 3Chapter 2Chapter 2Poll Question #2

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Solution – a homogeneous mixture of two or more substances. • Solute – the substance dissolved in a solution (e.g., sugar, salt,

CuSO4). Typically is the species present in lesser amount. • Solvent – the dissolving medium in a solution (e.g., water, ethanol).

Typically is the species present in greater amount

Concentration – the relative amount of solute in a solution 19

Chapter 3

Solution Definitions

• Molarity (M) = one of the most common units used to measure the concentration of a solution. – Definition: number of moles of solute in one liter of

solution.

• The abbreviation for “molar” is an upper case “M” (read as “molar,” or “moles per liter”), or sometimes square brackets [ ] placed around the chemical formula of the solute.

= moles of solute Molarity liters of solution

Chapter 3

Molarity

part whole

Molarity = mol solute L solution

= 3.0 mol NaCl 12 L

= 0.25 M = [NaCl]

Chapter 3

Molarity: Example • Calculate the molar concentration of a solution containing

3.0 moles of sodium chloride dissolved in 12 L of water.

– Solution:

21 22

Chapter 3

Molarity: Practice 5.) How many moles of sugar are contained in a 10-mL sip of a 0.375 M soft drink? (report your answer in the correct number of sig figs)

• As with other units we have seen, molarity (mol/L) can be used in unit conversion-type problems.

• Strategy: start with what is given to you, and then set up appropriate conversion factors so that you have all parts of the Molarity = mol/L equation.

– Let’s practice this!

Chapter 3

Molarity as a Conversion Factor

6.) If 40.00 g of NaOH are dissolved into enough water to make a 1.00 L of solution, what is the molarity of the resulting solution?

A. 40.0 M B. 4.00 M C. 0.0400 M D. 1.00 M

Chapter 3

Molarity as a Conversion Factor

Chapter 3Chapter 3Chapter 3Chapter 2Chapter 2Poll Question #3

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• Stock solution – a concentrated solution of a substance used to prepare solutions of lower concentration.

• Dilution – the process of lowering the concentration of a solution by adding more solvent.

moles of solute before dilution = moles of solute after dilution

Chapter 3

Preparing Solutions by Dilution

Concentrated (large amt. of solute)

Dilute (small amt. of solute)

Serial dilutions (series of sequential dilutions)

Scientist Spotlight Dr. Sine Haegele

Scientist, Jounce Therapeutics (Cambridge, MA)

Topic: Serial dilutions

Favorite quote: “Everything will be OK in the end. If it’s not OK, it’s not the end.”

Email: [email protected]

• The commonly used equation for this calculation is:

M1V1 = M2V2 or MconcVconc = MdiluteVdilute where M = molarity and V = volume. [I like to assign the “1”’s to the original solution (before dilution) and the “2’s” to the final solution (after dilution).]

• Example: 2 mL of 100 mM KCl is diluted to a final volume of 10 mL. What is the final molar concentration of a solution?

(2 mL) (100 mM) = (M2)(10 mL) 27

Chapter 3

Dilution Calculations moles of solute before dilution = moles of solute after dilution

10 mL 10 mL M2 = 20 mM

M1V1 = M2V2

Chapter 3Chapter 3Chapter 3Chapter 3Chapter 3Chapter 2Chapter 2Poll Question #4

Dilutions 7.) What volume of 12.0 M commercial stock HCl should you use to prepare 240.0 L of 0.10 M HCl?

A) 0.12 L B) 1.2 L

C) 2.0 L D) 2.4 L

Percent Composition • By definition, all samples of a pure compound have the

same composition – For a given compound, the ratio of particles (or of moles) of

the various atoms is always the same – The percent of each element by weight in various samples

of any compound is constant • Law of Constant Composition (Law of Definite Proportions)

• If we know the molecular formula for a compound and the molar masses of each atom in that compound, we can determine its mass percent composition (also called weight percent)

Chapter 3

• We will carry out all our calculations based on one mole of compound.

– Step 1: Use molecular formula and atomic masses to convert the number of atoms of each element to grams of that element in 1 mole of compound (g/mol format)

– Step 2: Calculate molar mass of the compound, which is equal to the total mass of 1 mole of that compound

– Step 3: Mass % of element = mass of element in 1 mole of compound

total mass of 1 mole of compound or: _part_

whole 30

Chapter 3

Calculating Percent Composition from Molecular Formula

× 100

mailto:[email protected]

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8a.) Calculate the weight % of each element in Ca3(PO4)2.

Step 1: Count the number of atoms represented by Ca3(PO4)2: ___ Ca ___ P ___ O

Step 2: Calculate the molar mass of Ca3(PO4)2: 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 310.2 g/mol

Step 3: Calculate the % composition for each element by dividing each part by the whole.

3(40.08 g/mol) Ca * 100 = 38.76% Ca 2(30.97 g/mol) P * 100 = 19.97% P 310.2 g/mol Ca3(PO4)2 310.2 g/mol Ca3(PO4)2

8(16.00 g/mol) O * 100 = 41.26% O 310.2 g/mol Ca3(PO4)2

Chapter 3

Let’s pause here to check our work: Add the % composition for each element in the compound. They should add to 100% (±1%). 38.76% + 19.97% + 41.26% = 99.99% ✓

3 Ca 2 P 8 O

Calculating Percent Composition from Molecular Formula: Example

8b.) What mass of P is present in 454 g of Ca3(PO4)2?

Step 1: Count the number of atoms represented by Ca3(PO4)2: ___ Ca ___ P ___ O

Step 2: Calculate the molar mass of Ca3(PO4)2: 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 310.2 g/mol

Step 3: Calculate the % composition for each element by dividing each part by the whole.

3(40.08 g/mol) Ca * 100 = 38.76% Ca 2(30.97 g/mol) P * 100 = 19.97% P 310.2 g/mol Ca3(PO4)2 310.2 g/mol Ca3(PO4)2

8(16.00 g/mol) O * 100 = 41.26% O 310.2 g/mol Ca3(PO4)2

Chapter 3

Step 4: Mass of P in 454 g Ca3(PO4)2 = 19.97% * 454 g = 0.1997 * 454 g = 90.7 g P

Calculating Percent Composition from Molecular Formula: Example

3 Ca 2 P 8 O

9.) What is the approximate percent by mass of hydrogen in C2H6?

A) 6% B) 20% C) 25% D) 75% E) 80%

Chapter 3

Calculating Percent Composition from Molecular Formula

Chapter 3Chapter 3Chapter 2Chapter 2Poll Question #5

• Another application of calculating % composition is when you are give the masses of each “part” and the mass of the “whole”. – Example 1: A 5.53 g sample of a laboratory solution contains

1.27 g acid. What is the mass % of acid in the sample?

_part_ __1.27 g acid__ whole 5.53 g solution

– Example 2: A solution of CaCl2 in water forms a mixture that is 40.5% CaCl2 by mass. If the total mass of the mixture is 703.1 g, what masses of CaCl2 and water were used?

Chapter 3

Calculating Percent Composition from Mass

× 100 = × 100 = 23.0%

40.5% = part whole×100% 40.5% =

part 703.1 g×100%

40.5% 100% =

part 703.1 g

0.405 × 703.1 g = 285 g CaCl2 703.1 g total (CaCl2 + water) – 285 g CaCl2 = 418 g H2O

• The experimentally-measured weight % composition and the molar mass of a compound can be used to determine empirical and molecular formulas

• When might this be useful? – Newly discovered compounds must be characterized by

molar mass and % composition to confirm identity

Chapter 3

Determining Empirical and Molecular Formulas from Percent Composition

• Empirical formula – formula showing the composition of a compound given as the simplest whole-number ratio of atoms

• Molecular formula – formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in the molecule of the compound

• Example: If a compound’s empirical formula is B2H5, its molecular formula could be: B2H5, B4H10, B6H15, etc. – How would compounds of these molecular formulas differ

experimentally? • Different chemical and physical properties • Different molar masses! 36

Chapter 3

Review: Empirical vs. Molecular Formulas

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Chapter 3

Determining the Empirical Formula from Percent Composition: Example

10a.) Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

Strategy: 1.) Change each of the percentages to grams (assume you have 100 g of the compound)

2.) Calculate the number of moles of each element in the compound.

3.) Divide each number of moles by the least number of moles (this is the mole ratio)

4.) Multiply up to the nearest whole number only if needed (i.e., if you end up with a fraction of an atom, which isn’t possible.

• If very close to a whole number, just round (1.99 = 2; 4.01 = 4)

10a.) Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

Chapter 3

Determining the Empirical Formula from Percent Composition: Example

Strategy: 1.) Change each of the percentages to grams (assume you have 100 g of the compound)

• Let’s assume we have 100 g of ascorbic acid. • That means there are 40.92 g C, 4.58 g H, and 54.50 g O

(40.92 g + 4.58 g + 54.40 g = 100 g total)

2.) Calculate the number of moles of each element in the compound.

40.92 g C × 1 mol C 12.01 g C

= 3.41 mol C 4.58 g H× 1 mol H 1.008 g H

= 4.54 mol H 54.5 g O × 1 mol O 16 g O

= 3.41 mol O

10a.) Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

Chapter 3

Determining the Empirical Formula from Percent Composition: Example

Strategy: 3.) Divide each number of moles by the least number of moles (this is the mole ratio)

4.) Multiply up to the nearest whole number only if needed (i.e., if you end up with a fraction of an atom, which isn’t possible)

• We can multiply each subscript by 3: C1*3H1.33*3O1*3 = C3H4O3

40.92 g C × 1 mol C 12.01 g C

= 3.41 mol C 4.58 g H× 1 mol H 1.008 g H

= 4.54 mol H 54.5 g O × 1 mol O 16 g O

= 3.41 mol O

3.41 mol C 3.41 mol C

=1 4.54 mol H 3.41 mol H

=1.33 3.41 mol O 3.41 mol O

=1 What we have so far:

C1H1.33O1

10b.) The molar mass of ascorbic acid (C3H4O3) is 176.12 g/mol. What is the molecular formula of ascorbic acid?

Strategy: Empirical formula mass * x = molar mass, where x is a multiplying factor applied to the subscripts of each element in the empirical formula.

1.) Calculate the mass of the empirical formula

2.) Divide the molar mass by the empirical formula to achieve x (the whole number multiplier)

x = molar mass ÷ empirical mass

3.) Multiply the subscripts of each element in the empirical formula to arrive at the molecular formula.

Empirical formula * x = molecular formula 40

Chapter 3

Determining the Molecular Formula from the Empirical Formula: Example

10b.) The molar mass of ascorbic acid (C3H4O3) is 176.12 g/mol. What is the molecular formula of ascorbic acid?

Strategy: Empirical formula mass * x = molar mass, where x is a multiplying factor applied to the subscripts of each element in the empirical formula.

1.) We already know that the empirical formula of ascorbic acid is C3H4O3. And so, let’s calculate the mass of the empirical formula!

– There are 3 C, 4 H, and 3 O. So, the mass of C3H4O3 = 3(12.01 g/mol) + 4(1.008 g/mol) + 3(16.00 g/mol) = 88.062 g/mol.

2.) Equation: Empirical formula mass * x = molar mass • Plug in what we know: 88.082 g/mol * x = 176.12 g/mol • Solve for x (multiplier) 88.082 g/mol 88.082 g/mol 3.) Molecular formula = C3*2H4*2O3*2 = C6H8O6 41

Chapter 3

x = 2

Determining the Molecular Formula from the Empirical Formula: Example

41 42

Chapter 3

Application: Combustion Analysis • It is possible to determine the empirical formula of an organic

compound (containing carbon, hydrogen, and sometimes oxygen) through combustion analysis.

(CuO is used to ensure complete combustion of all carbon to CO2 and all hydrogen to H2O)

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Chapter 3

Application: Combustion Analysis • It is possible to determine the empirical formula of an organic

compound (containing carbon, hydrogen, and sometimes oxygen) through combustion analysis. – A sample of known mass is placed

in the furnace and heated in the presence of pure oxygen (O2).

– The heat from the furnace converts the carbon in the compound to CO2 and the hydrogen into H2O, which are collected in “traps” that are weighed before and after combustion.

• All of the carbon in the original sample ends up in the CO2 • All of the hydrogen in the original sample ends up in the water

(H2O). • Any oxygen in the sample can be deduced via conservation of

mass (by subtraction).

43 44

Chapter 3

How to Solve a Combustion Analysis Problem Strategy: Determining the Empirical and Molecular Formulas of hydrocarbons (CXHY) • Step 1: Find the number of moles of each element:

– Convert mass of CO2 and mass of H2O to moles of each compound

– Convert moles of CO2 to moles of carbon; convert moles of H2O to moles of hydrogen

• Step 2: Divide by the smallest number of moles to find the empirical formula (multiply up to the nearest whole number if needed.)

• Step 3: Calculate molecular formula (if asked to solve for this) – Empirical mass * x = molar mass; solve for x; multiply subscripts

of each element in the empirical formula.

11.) Upon combustion, a hydrocarbon (compound containing only hydrogen and carbon) produces 1.83 g of CO2 and 0.901 g of H2O. What is the empirical formula for the compound?

Chapter 3

Combustion Analysis: CXHY Example

0.901 g H2O × 1 mol H2O

18.016 g H2O × 2 mol H 1 mol H2O

= 0.100 mol H

C: 0.0416 mol H: 0.100 mol ÷ 0.0416 mol

= 1.00 C = 2.40 H

Multiplying up to the nearest whole number…. 2.4*2 = 4.8 X 2.4*3 = 7.2 X 2.4*4 = 9.6 X 2.4*5 = 12 ✓

Step 1: Find moles of each element

Step 2: Divide by the least number of moles to find empirical formula

C1H2.4 are we finished? No! Can’t have a fraction of an atom!

C1*5H2.4*5 = C5H12

(Each mole of CO2 corresponds to 1 mole of C atoms in the hydrocarbon sample)

(Each mole of H2O corresponds to 2 moles of H atoms in the hydrocarbon sample)

1.83 g CO2 × 1 mol CO2

44.011 g CO2 × 1 mol C 1 mol CO2

= 0.0416 g Cmol C

45 46

Chapter 3

How to Solve a Combustion Analysis Problem Strategy: Determining the Empirical and Molecular Formulas of organic compounds (CXHYOZ) Just start with one additional step: • Step 1: Calculate the mass of oxygen by computing the

masses of carbon and hydrogen:

Total mass of compound – mass C – mass H = mass O

All subsequent steps are the same as in the hydrocarbon case! • Step 2: Find the number of moles of each element • Step 3: Divide by the smallest number of moles to find the

empirical formula (multiply up to the nearest whole number if needed.)

• Step 4: Calculate molecular formula (if asked to solve for this)

12.) 1.516 g of a compound containing carbon, hydrogen and oxygen (CXHYOZ) is subjected to combustion analysis. The results show that 2.082 g of CO2 and 1.705 g of H2O were produced.

a.) What is the empirical formula for the compound? b.) If the molecular weight of the compound is 160.2 g/mol, what is the molecular formula of the compound?

Chapter 3

Combustion Analysis: CXHYOZ Example

2.082 g CO2 × 1 mol CO2

44.011 g CO2 × 1 mol C 1 mol CO2

× 12.011 g C 1 mol C

= 0.568 g C

1.705 g H2O × 1 mol H2O

18.016 g H2O × 2 mol H 1 mol H2O

× 1.008 g H 1 mol H

= 0.191 g H

1.516 g CXHYOZ - 0.568 g C - 0.191 g H= 0.757 g O

0.568 g C× 1 mol C 12.011 g C

= 0.0473 mol C

0.191 g H× 1 mol H 1.008 g H

= 0.189 mol H

0.757 g O× 1 mol O 16.00 g O

= 0.0473 mol O

C: 0.0473 mol H: 0.189 mol O: 0.0473 mol

÷ 0.0473 mol = 1 C = 4 H = 1 O

Empirical mass * x = molar mass (12.011 g + 4(1.008 g) + 16.00 g)x = 160.2 g/mol x = 5 Molecular formula = C1*5H4*5O1*5 = C5H20O5

Step 1: Find mass of oxygen Step 2: Find moles of each element

Step 3: Divide by the least number of moles to find empirical formula

CH4O

Step 4: Calculate molecular formula