AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Rhyme to remember order of steps to convert % composition

empirical formula:
Percent to mass
Mass to mole
Divide by small
Times till whole
• Finding the empirical mass percent of elements from the empirical formula:
• If we have the empirical formula, we know how many moles of each element is present in one mole of same.
• Then we use molar masses (or atomic weights) to convert to grams of each element.
• We divide the number of grams of element by grams of 1 mole of sample to get the fraction of each element in 1 mole of sample.
• Multiply each fraction by 100 to convert to a percent.
Sample Exercise 3.13 (p. 99)
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass.
What is the empirical formula of vitamin C?
(C
3
H
4
O
3
)
Practice Exercise 1
A 2.144-g sample of phosgene, a compound used as a chemical warfare agent during World War I, contains
0.260 g of carbon, 0.347 g of oxgen, and 1.537 g of chlorine. What is the empirical formula of this substance?
(a) CO
2
Cl
6
, (b) COCl
2
, (c) C
0.022
O
0.022
Cl
0.044
, (d) C
2
OCl
2
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Practice Exercise 2
A 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain
3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen.
What is the empirical formula of this substance?
(C
4
H
4
O)
• To get the molecular formula from the empirical formula, we need to know the molecular weight, MW.
• The ratio of molecular weight (MW) to formula weight (FW) of the empirical formula must be a whole number.
Sample Exercise 3.14 (p. 100)
Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C
3
H
4
. The experimentally determined molecular weight of this substance is 121 amu.
What is the molecular formula of mesitylene?
(C
9
H
12
)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Practice Exercise 1
Cyclohexane, a commonly used organic solvent, is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula?
(a) C
6
H, (b) CH
2
, (c) C
5
H
24
, (d) C
6
H
12
, (e) C
4
H
8
Practice Exercise 2
Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. a) What is the empirical formula of ethylene glycol? (CH
3
O) b) What is its molecular formula? (C
2
H
6
O
2
)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
• A sample containing C, H and O is combusted in excess oxygen to produce CO
2
and H
2
O.
• The amount of CO
• The amount of H
2
2
gives the amount of C originally present in the sample.
O gives the amount of H originally present in the sample.
• Watch stoichiometry: 1 mol H
2
O contains 2 mol H.
• The amount of O originally present in the sample is given by the difference between the amount of sample and the amount of C and H accounted for.
• More complicated methods can be used to quantify the amounts of other elements present, but they rely on analogous methods.
Sample Exercise 3.15 (p. 101)
Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO
2
and 0.306 g H
2
O. Determine the empirical formula of isopropyl alcohol.
(C
3
H
8
O)
Practice Exercise 1
The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a 2.2.03-g sample of this compound produces 4.401 g CO
2
and 1.802 g H
2
O. A separate experiment shows that it has a molar mass of 88.1 g g/mol. Which of the following is the correct molecular formula for dioxane?
(a) C
2
H
4
O, (b) C
4
H
2
O
2
, (c) CH
2
, (d) C
4
H
8
O
2
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Practice Exercise 2 a) Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms.
Combustion of a 0.225 g sample of this compound produces 0.512 g CO
2
and 0.209 g H
2
O. What is the empirical formula of caproic acid? (C
3
H
6
O) b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula? (C
6
H
12
O
2
)
Hydrates
Anhydrous = dry, does not contain water
Hydrate = a compound formed by the combination of water with some other substance in which the water retains its molecular state as H
2
O
Determining Hydrate Formulas
1. Find the formula of a hydrate from experimental mass data: a) Calculate the number of moles of the anhydrous salt (mass  mole conversion) b) Calculate the mass of H
2
O by subtracting mass of anhydrous salt from mass of hydrated salt c) Calculate the moles of H
2
O (mass  mole conversion) d) Calculate ratio: # mol H
2
O:# mol anhydrous salt
e)  correct formula = salt
 x H
2
O ( x = mole ratio)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
2. Find the formula from mass % data: a) Use the method of using 100 g (from 100 %) to give mass in grams of the salt and water b) Use mass  mole conversion to calculate the # moles of the anhydrous salt and the # moles of water c) Calculate the mole ratio of H
2
O:anhydrous salt
3. Find the % H
2
O in a hydrate from its formula: Use mass percent.
Practice Problems – Hydrates
1.
A mass of 2.50 g of blue hydrated copper sulfate (CuSO
4
• x H
2
O) is placed in a crucible and heated.
After heating, 1.59 g white anhydrous copper sulfate (CuSO
4
) remains. What is the formula for the hydrate? Name the hydrate. (CuSO
4
• 5 H
2
O)
2.
A hydrate is found to have the following percent composition: 48.8% MgSO
4
and 51.2% H
2
O.
What is the formula and the name for this hydrate? (MgSO
4
• 7 H
2
O)
3. The formula for a hydrate is SnCl
2
• 2 H
2
O. What is the mass percent of water in this hydrate?
(15.97% H
2
O)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations

The limiting reactant is the reactant present in the smallest stoichiometric amount. In other words, it’s the reactant you’ll run out of first
• The one or more reactants which are completely consumed are called the limiting reactants or limiting reagents .
• Reactants present in excess are called excess reactants or excess reagents .
Chemical Equation:
1 25.0 mL
2 25.0 mL
3 25.0 mL
4 25.0 mL
5 25.0 mL
6 25.0 mL
2
0.50 g
1.00 g
1.50 g
2.00 g
2.50 g
3.00 g
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Sample Exercise 3.18 (p. 108)
The most important commercial process for converting N
2
from the air into nitrogen-containing compounds is based on the reaction of N
2
and H
2
to form ammonia (NH
3
):
N
2(g)
+ 3 H
2(g)
 2 NH
3(g)
How many moles of NH
3
can be formed from 3.0 mol of N
2
, and 6.0 mol of H
2
?
(4.0 mol NH
3
)
Practice Exercise 1
When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction
2 CH
3
OH
(l)
+ 3 O
2(g)

2 CO
2(g)
+ 4 H
2
O
(g)
, what is the excess reactant and how many moles of it remains at the end of the reaction?
(a) 9 mol CH
3
OH
(l)
, (b) 10 mol CO
2(g)
, (c) 10 mol CH
3
OH
(l)
, (d) 14 mol CH
3
OH
(l)
, (e) 1 mol O
2(g)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Practice Exercise 2
Consider the reaction 2 Al
(s)
+ 3 Cl
2(g)

2 AlCl
3(s)
.
A mixture of 1.50 mol of Al and 3.00 mol of Cl
2
are allowed to react. a) What is the limiting reactant? (Al) b) How many moles of AlCl
3
are formed? (1.50 mol) c) How many moles of the excess reactant remain at the end of the reaction? (0.75 mol Cl
2
)
Sample Exercise 3.19 (p. 108)
Consider the following reaction that occurs in a fuel cell:
2 H
2(g)
+ O
2(g)

2 H
2
O
(l)
This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed?
(1300 g H
2
O)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
Practice Exercise 1
Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar cells: Ga
(l)
+ As
(s)

GaAs
(s)
If 4.00 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?
(a) 4.94 g As, (b) 0.56 g As, (c) 8.94 g Ga, or (d) 1.20 g As.
Practice Exercise 2
A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:
Zn
(s)
+ 2 AgNO
3(aq)

2 Ag
(s)
+ Zn(NO
3
)
2(aq) a) Which reactant is limiting? (AgNO
3
) b) How many grams of Ag will form? (1.59 g) c) How many grams of Zn(NO
3
)
2
will form? (1.39 g) d) How many grams of the excess reactant will be left at the end of the reaction? (1.52 g Zn)
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AP Chemistry Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations
• The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield .
• The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:
Percent yield = actual yield___ x 100%
theoretical yield
Sample Exercise 3.20 (p. 109)
Adipic acid, H
2
C
6
H
8
O
4
, is used to produce nylon. It is made commercially by a controlled reaction between cyclohexane (C
6
H
12
) and O
2
:
2 C
6
H
12(l)
+ 5 O
2(g)

2 H
2
C
6
H
8
O
4(l)
+ 2 H
2
O
(g) a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane, and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (43.5 g H
2
C
6
H
8
O
4
) b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid?
(77.0%)
Practice Exercise 1
If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, to form 7.7 g of titanium(IV) chloride in a combination reaction, what is the percent yield of the product?
(a) 65%, (b) 96%, (c) 48%, or (d) 86%
Practice Exercise 2
Imagine that you are working on ways to improve the process by which iron ore containing Fe
2
O
3
is converted into iron. In your tests you carry out the following reaction on a small scale:
Fe
2
O
3(s)
+ 3 CO
(g)

2 Fe
(s)
+ 3 CO
2(g) a) If you start with 150 g of Fe
2
O
3
as the limiting reagent (reactant), what is the theoretical yield of Fe?
(105 g Fe) b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? (83.7%)
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