Module 4...

A chemical equation is a recipe or equation for a chemical reaction.  (remember a chemical reaction has reactants –starting material- that are rearranged to form one or more products).  The number of each element on the reactant and the product side must be equal.  This is the Law of Conservation of Matter .  Since the number of each element must be the same on the reactant and the product side, the equations must be balanced.  A balanced equation will generally have coefficients in front of the chemical formulas in order to make the equation balance.  SO:

C(s) + O2(g) → CO2(g)

Reads as: solid carbon reacts with gaseous oxygen to form gaseous carbon dioxide (more on naming chemicals next module)

There are no coefficients (other than the implied "1") for each of these chemicals.  There is 1 carbon on the reactant and product side.  There are 2 O's on each side as well.  This is a balanced equation.

For:  2NO(g) + O2(g) → 2NO2(g)

Reads as: gaseous nitrogen monoxide reacts with gaseous oxygen to form gaseous nitrogen dioxide.

There is a coefficient of "2" in front of the Nitrogen monoxide (on the reactant side) as well as in front of the Nitrogen dioxide on the product side. (**can you count the number of each element on each side?)

Types of chemical reactions

There are 4 generic types of reactions:

1. Combination or Synthesis reaction: A + B → C

1. There are 2 reactants that combine to form one product.

2. Decomposition reaction C → A + B

1. There is 1 reactant that breaks into smaller parts. This type of reaction can have several products.

3. Single Displacement reaction: AB + C → CB + A

1. One of the reactants exchanges with one component of the other reactant. For example: Cu + AgNO3 → Ag + Cu(NO3)2

4. Double Displacement reaction: AB + CD → CB + AD

1. The cations switch places. For example: AgNO3 + NaCl →AgCl + NaNO3

Balancing Chemical Reactions (image attribution: redbubble.com)

Chemical equations are the chemist "recipe".  The balanced chemical equation shows the relative proportions of each reactant (on the left) and how much of each product (on the right) will be produced.  A balanced chemical equation obeys the law of conservation of matter: Matter is neither lost nor gained… only converted. 

When there is an unbalanced equation coefficients are added to the front of each chemical formula until the same numbers of each atom are on the reactant and product side.  **You CANNOT change the subscripts of the chemical formulas **

Sometimes it is easiest to make a chart

CH4(g) + O2(g) → CO2(g) + H2O(g)

It is easier to start with an element that is only contained in one compound on each side (ie. Carbon is only in one compound on the reactant side, CH4; and in one compound on the product side, CO2).  Since carbon is balanced, it has the same number of atoms on the reactant side and the product side, I will move to hydrogen.  Hydrogen has 4 on the reactant side and 2 on the product side; 2 * 2 = 4 so a coefficient of "2" is placed in front of H2O, water, which changes the hydrogen and oxygen for the product side.

CH4(g) + O2(g) → CO2(g) + 2 H2O(g)

Now the oxygen needs to be balanced.  Since the oxygen is alone and existing as a diatomic element the coefficient can be placed in front of oxygen gas on the reactant side to become:

CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g)

 The equation is now balanced.

Test your understanding of the topics by completing numbers 1-13 in Chapter 8 of Burdge's Atoms First.

 Stoichiometry

You can see how the balanced equation obeys the law of conservation of matter by analyzing it in terms of molar mass.

  CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g)

16.0g + 64.0g = 80.0g = 44.0g + 36.0g

Because each side, reactants and products, have equal masses this equation obeys the law of conservation of matter.

Stoichiometry

The coefficients (numbers in front of the chemicals) are the molar ratios of the chemicals. That is called stoichiometry. Just as the subscripts in a chemical formula tell you the molar ratio of the elements in a compound (H2O = 2 moles H and 1 mole O) the coefficients tell you the moles of the compounds in the reaction.

  CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g)

With words this reaction reads: 1 mole methane gas reacts with 2 moles oxygen gas react to form 1 mole carbon dioxide gas and 2 moles of gaseous water.

Quantities in Chemical Reactions

The balanced chemical equation yields the stoichiometric ratio of products and reactants. Knowing those ratios, the quantities of reactants can be used to predict the quantity of products that can be produced. Also, the quantities of reactants can be calculated if there is an expected yield of products. For example,

How many moles of H2(g) can be formed from the reaction of 3.0 moles of Na with excess H2O(l)?

2 Na(s) + 2H2O(l) → 2 NaOH(aq) + _H2(g)

Start with the quantity given, 3.0 moles Na. Keep the units with the numbers. Use the stoichiometric ration found in the balanced chemical equation and convert moles of Na to moles of H2(g).

Remember that the absence of a number in the balanced equation implies a 1. I put an underscore in front of H2 to show that is where the 1 came from. The measured quantity in the question, 3.0, has 2 significant figures therefore the answer will have two significant figures. The stoichiometric ratio is considered an exact number therefore has infinite significant figures.

Another example:

How many grams of Na(s) is needed to produce 0.494g H2(g)?

Start with what you are given in the problem. 99% of the time in this class, when you see grams, convert to moles. The molar mass of H2(g) is 2.02 g/mol. The number stays with grams. That is read "2.02 grams per 1 mole H2) Just to recap: H2 = H + H = 1.01g/mol + 1.01g/mol = 2.02 g/mol

The measured quantity of H2 has three significant figures. The molar mass of H2 and Na could vary with the significant figures therefore to determine the significant figures in the answer, only the sig figs of the measured quantity will be considered. 0.494 has 3 sig figs therefore the answer has 3 sig figs.

Limiting Reactant

The concept of limiting reactant isn't difficult. When making lunch for a crowd you might have 12 apples, 16 sandwiches, and 24 cookies. If 1 lunch can be said to be equivalent to:

1 sandwich + 1 apple + 4 cookies = 1 lunch

You likely don't need to do calculations to see that you can make a total of 6 complete lunches before you run out of cookies. (I like cookies.) To determine this mathematically, calculate the total number of lunches that could be made from each component if you had an excess of everything else.

The calculations for this problem look like this:

This is an example of a limiting reactant (sometimes Limiting Reagent) problem. The cookies will run out first. That is the limiting reactant. 6 lunches, which is the maximum that can be made from the cookies, is the theoretical yield. In a limiting reactant problem, you are also likely to be given an actual yield. Sometimes this is due to experimental error, or simply because there are few reactions that yield 100% of the product. In the case of making lunches, that error is because I eat cookies as I make the lunches. If a total of 4 complete lunches can be made, then we can calculate the percent yield.

% yield = (Actual yield / Theoretical yield) * 100

In lunches, the actual yield = 4 lunches. Theoretical yield = 6 lunches.

% yield = (4 lunches / 6 lunches) * 100 = 66.7% <= in reality, we'll need to watch those sig figs.

With a chemical example:

What is the % yield of H2 when 0.405g of H2 is collected from a reaction of 11.2g Na and 9.00g H2O?

2 Na(s) + 2H2O(l) → 2 NaOH(aq) + _H2(g)

1. Solve the "how much product can I produce" problem for each of the reactants:

2. Compare the results to see which reactant produces less product. The sodium will produce less product therefore the Sodium is the limiting reactant. **remember the limiting reagent is ALWAYS a reactant

3. Identify the theoretical yield. The theoretical yield = 0.494 g H2

4. Calculate the percent yield. The % yield = (Actual yield / Theoretical yield) * 100 = (0.405g/0.494g) * 100 = 82.0%

Test your understanding by completing problems 24-59 in Chapter 8 of Burdge's Atoms First.

 Combustion Analysis

Combustion analysis can be used to determine the relative amounts of carbon, hydrogen, and oxygen contained in hydrocarbons. Combustion is the breaking down of a compound in the presence of oxygen gas to form carbon dioxide and water. *It will always have O2, H2O, CO2. Typically, combustion is shown in chemistry with hydrocarbons or organic molecules. For example, the combustion of glucose:

C6H12O6 + 6 O2 → 6 H2O + 6 CO2

Combustion reactions can be used to determine a molecule's empirical formula. When something like glucose is burned in a combustion analysis chamber, the carbon and hydrogen can be measured. The oxygen is then derived with a calculation. For example, if 18.8g of glucose is combusted in an experiment and 27.6g of CO2 and 11.3 g H2O are produced, we can calculate the empirical formula by analyzing the carbon and hydrogen collected.

 Steps to solve a combustion analysis problem

1. Convert the grams to moles of the element in the parent compound.

This calculation shows that the parent compound contained 7.53g C and 1.26g H. We can calculate O from the original mass of the parent compound and the two calculated quantities above.

Compound = C + H + O = 18.8g = 7.53g + 1.26g + mass O

Simple addition and subtraction yields the mass of oxygen is 10.0 g.

2. Convert the calculated masses of the elements into moles.

This calculation gives us the mole ratios of the elements in the parent compound. The moles of C and O are close enough that they can be seen as 1: 1.

3. Make a pseudo chemical formula with the calculated moles from 2.

C0.627H1.25O0.626

4. Divide all of the subscripts by the subscript of the smallest value.

C0.627/0.626    H1.25/0.626    O0.626/0.626

C1H2O1

This is the empirical formula of the parent compound.

5. If given the molar mass of the parent compound, the molecular formula can be determined.

· Find the empirical molar mass of CH6O = 12.01 + 2(1.01) + 16.00 =30.03 g/mol

· If the molar mass was given as "about 180 g/mol" then find the multiplicative factor from empirical to molecular by dividing the molecular mass by the empirical mass.

180 / 30.03 = ~ 6

Finally, Use the multiplicative factor (6) and multiply the subscripts of the empirical formula to find the molecular formula.

C1H2O1 * 6 = C6H12O6

Try this:  What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10g of the compound yields 6.59g CO2 and 2.70g H2O and its molar mass is about 84 g/mol?

Test your understanding by completing problems 14-23 in Chapter 8 of Burdge's Atoms First.