1. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations
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2. Law of Conservation of Mass
“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”
--Antoine Lavoisier, 1789
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3. 3.1 Chemical Equations
Chemical equations are concise representations of chemical reactions.
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4. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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5. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.
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6. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.
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7. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.
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8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.
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9. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule.
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10. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule
Coefficients tell the number of molecules.
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11. 3.1 Chemical Equations Indicating the States of Reactants and Products
• The physical state of each reactant and product may be added to the equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
• Reaction conditions occasionally appear above or below the reaction arrow (e.g., "" is often used to indicate the addition of heat).
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12. 3.1 Practice Problems
Balance the following equations by providing the missing coefficients:
_Fe(s) + _O2(g) → _Fe2O3(s)
4, 3, 2
_C2H4(g) + _O2(g) → _CO2(g) + _H2O(g)
1, 3, 2, 2
_Al(s) + _HCl(aq) → _AlCl3(aq) + _H2(g)
2, 6, 2, 3
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13. 3.2 Reaction Types – Simple Patterns of Chemical Reactivity
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14. 3.1 Combination and Decomposition Reactions
In combination reactions two or more substances react to form one product.
Combination reactions have more reactants than products.
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15. Consider the reaction: 2Mg(s) + O2(g)  2MgO(s)
Since there are fewer products than reactants, the Mg has combined with O2 to form MgO.
Note that the structure of the reactants has changed.
Mg consists of closely packed atoms and O2 consists of dispersed molecules.
MgO consists of a lattice of Mg2+ and O2– ions.
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16. Combination Reactions
In this type of reaction two or more substances react to form one product.
Examples:
2 Mg (s) + O2 (g)  2 MgO (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
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17. Decomposition Reactions
In decomposition reactions one substance undergoes a reaction to produce two or more other substances.
Decomposition reactions have more products than reactants Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
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18. Another Decomposition Reaction
Consider the reaction that occurs in an automobile air bag:
2 NaN3(s)  2Na(s) + 3N2(g)
Since there are more products than reactants, the sodium azide has decomposed into sodium metal and nitrogen gas.
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19. Combustion Reactions
Combustion reactions are rapid reactions that produce a flame .
Most often involve hydrocarbons reacting with oxygen in the air.
Most combustion reactions involve the reaction of O2(g) from air.
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
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20. Practice exercise for 3.2
Write balanced chemical equations for the following reactions:
(a) Solid mercury(II) sulfide decomposes into its component elements when heated.
HgS(s) → Hg(l) + S(s)
(b) The surface of aluminum metal undergoes a combination reaction with oxygen in the air.
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
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21. Practice exercise for 3.2
Write the balanced equation for the reaction that occurs when ethanol, C2H5OH(l), is burned in air.
Answer: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
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22. 3.3 Formula Weights
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23. Formula Weight (FW)
A formula weight is the sum of the atomic weights for the atoms in a chemical formula.
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
Formula weights are generally reported for ionic compounds.
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24. 3.3 Examples: Formula Weight
Example: FW (H2SO4)
= 2 (A.W.of H) + 1(A.W. of S) + 4(A.W. of O)
= 2(1.0 amu) amu + 4(16.0 amu)
= 98.1 amu.
Example: MW (C6H12O6)
= 6(12.0 amu) + 12 (1.0 amu) + 6 (16.0 amu)
= amu.
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25. 3.3 Practice Formula Weight
Calculate the formula weight of
(a) Al(OH)3
Answer: (a) 78.0 amu
(b) CH3OH.
Answer:(b) 32.0 amu
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26. Molecular Weight (MW)
A molecular weight is the sum of the atomic weights of the atoms in a molecule.
For the molecule ethane, C2H6, the molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
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27. 3.3 Examples: Molecular Mass
Example: MW (C6H12O6)
= 6(12.0 amu) + 12 (1.0 amu) + 6 (16.0 amu)
= amu.
Note: Formula weight of the repeating unit (formula unit) is used for ionic substances.
Example: FW (NaCl)
= 23.0 amu amu
= 58.5 amu.
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28. 3.3 Percent Composition
Percentage composition is obtained by dividing the mass contributed by each element (number of atoms times AW) by the formula weight of the compound and multiplying by 100 :
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100
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29. Percent Composition So the percentage of carbon in ethane is…C2H6
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%
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30. % Mass Problem
Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
%N = 28/164.1 x 100 %
Answer: 17.1%
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31. 3.4 Avogadro’s Number Moles
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32. Avogadro’s Number 6.02 x 1023 1 mole of 12C has a mass of 12 g.
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33. The Mole
The mole (abbreviated "mol") is a convenient measure of chemical quantities.
1 mole of something = x 1023 of that thing.
This number is called Avogadro’s number.
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34. Molar Mass
By definition, the molar mass is the mass of 1 mol of a substance (i.e., g/mol).
The molar mass of an element is the mass number for the element that we find on the periodic table.
The mass of 1 mole of 12C = 12 g. Exactly.
The molar mass of a molecule is the sum of the molar masses of the atoms:
Example: The molar mass of N2 = 2 x 14 (molar mass of N). = 28 g
The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).
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35. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale.
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36. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.
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37. 3.4 Practice Problem
Without using a calculator, arrange the following samples in order of increasing number of O atoms: 1 mol H2O, 1 mol CO2, 3  1023 molecules O3.
Answer:
1 mol H2O (6  1023 O atoms)
3  1023 molecules O3 (9  1023 O atoms)
1 mol CO2 (12  1023 O atoms)
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38. 3.4 Practice Problem Converting Moles to Atoms
Calculate the number of H atoms in mol of C6H12O6.
0.350 mol of C6H12O6 X 6.2 x 1023 molecule C6H12O6 X 12 H atoms
1 mol of C6H12O molecule C6H12O6.
= 2.53 x 1024 H atoms
How many oxygen atoms are in
(a) 0.25 mol Ca(NO3)2
Answer: (a) 9.0  1023
(b) 1.50 mol of sodium carbonate?
Answer (b) 2.71  1024
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39. What is the mass in grams of 1.000 mol of glucose, C6H12O6?
Calculating Molar Mass
What is the mass in grams of mol of glucose, C6H12O6?
Solution
Analyze We are given a chemical formula and asked to determine its molar mass.
Plan The molar mass of a substance is found by adding the atomic weights of its component atoms.
Because glucose has a formula weight of amu, one mole of this substance has a mass of g. In other words, C6H12O6 has a molar mass of g/mol.
Practice Exercise
Calculate the molar mass of Ca(NO3)2.
Answer: g/mol

40. 3.4 Practice Problem Sample Exercise 3.10 Converting Grams to Moles
Calculate the number of moles of glucose (C6H12O6) in g of C6H12O6.
The molar mass of a substance provides the factor for converting grams to moles. The molar mass of C6H12O6 is g/mol
5.380 x 1mol/180.0 g =
Answer: mol of C6H12O6
Practice Exercise
How many moles of sodium bicarbonate (NaHCO3) are in 508 g of NaHCO3?
Answer: 6.05 mol NaHCO3
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41. 3.4 Practice Problem Converting Moles to Grams Practice Exercise
Calculate the mass, in grams, of mol of calcium nitrate
.433 mol Ca(NO3)2 x g Ca(NO3)2
1mol Ca(NO3)2
= 71.1 grams Ca(NO3)2
Practice Exercise
What is the mass, in grams, of
(a) 6.33 mol of NaHCO3
(b) 3.0  10–5 mol of sulfuric acid?
Answer: (a) 532 g, (b) 2.9  10–3 g
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42. 3.4 Calculating the Number of Molecules and Number of Atoms from Mass
(a) How many glucose molecules are in 5.23 g of C6H12O6?
5.23 g C6H12O6 X 1 mol C6H12O6 X x 1023 molecules
180.0 g C6H12O mol C6H12O6
= x 1022 molecules of C6H12O6
(b) How many oxygen atoms are in this sample?
1.75 x 1022 molecules of C6H12O6 x Oxygen atom
1molecule of C6H12O6
= 1.05 x 1023 Oxygen atoms
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43. 3.4 Practice Probem
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44. 3.4 Practice Probem
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45. 3.4 Practice Problems
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46. 3.5 Empirical Formulas from Analyses
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47. Calculating Empirical Formulas
Recall that the empirical formula gives the relative number of atoms of each element in the molecule.
Finding empirical formula from mass percent data:
We start with the mass percent of elements (i.e., empirical data) and calculate a formula.
Assume we start with 100 g of sample.
The mass percent then translates as the number of grams of each element in 100 g of sample.
From these masses, the number of moles can be calculated (using the atomic weights from the periodic table).
The lowest whole-number ratio of moles is the empirical formula
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48. Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition.
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49. Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.
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50. Calculating Empirical Formulas
Assuming g of para-aminobenzoic acid,
C: g x = mol C
H: g x = 5.09 mol H
N: g x = mol N
O: g x = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g
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51. Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: =  7
H: =  7
N: = 1.000
O: =  2
5.105 mol
mol
5.09 mol
1.458 mol
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52. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
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53. Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.
C is determined from the mass of CO2 produced.
H is determined from the mass of H2O produced.
O is determined by difference after the C and H have been determined.
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54. Elemental Analyses
Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.
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55. 3.5 Calculating Empirical Formula
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?
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56. 3.5 Calculating Empirical Formula
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?
Solve We first assume, for simplicity, that we have exactly 100 g of material (although any mass can be used). In 100 g of ascorbic acid, therefore, we have:
g C, 4.58 g H, 54,50 g O2 .
Second, we calculate the number of moles of each element:
Third, we determine the simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles, 3.406:

57. 3.5 Calculating Empirical Formula
Solution (continued)
The ratio for H is too far from 1 to attribute the difference to experimental error; in fact, it is quite close to 1 1/3. This suggests that if we multiply the ratio by 3, we will obtain whole numbers:
The whole-number mole ratio gives us the subscripts for the empirical formula
PA g sample of methyl benzoate, a compound used in the manufacture of perfumes, contains g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
Answer: C4H4O

58. 3.5.1 Sample Problem
An organic compound was found to contain only C, H, & Cl. When a 1.50 sample of the compound was completely combusted in air, 3.52 g of CO2 was formed
In a separate experiment the chlorine in a 1.00 gram sample was converted to 1.27 g of AgCl.
Determine the empirical formula for this compound.
Plan: Because different sample sizes were used to analyze the different elements, calculate mass % of each element in the sample.
Calculate mass % C from CO2.
Calculate Mass % Cl from AgCl.
Get Mass % H by subtraction.
Calculate mole ratios & the empirical formula
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59. 3.5.1 Sample Problem(cont) Calculate mass % C from CO2.
3.52 g CO2 x 1 mol CO2 x 1 mol C x g C
44.01 g CO2 1 mol CO mol C =.961 g C
.961 g C x 100 % = % C
1.50 g sample
Calculate Mass % Cl from AgCl.
1.27 g AgCl x 1 mol AgCl x 1 mol Cl x g Cl
143.3 g AgCl 1 mol AgCl 1 mol Cl = .314 g Cl
.314 g Cl x 100 % = % C
1.00 g sample
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60. 3.5.1 Sample Problem(cont) Get Mass % H by subtraction.
% H = 100 – ( ) = 4.54% H
Calculate mole ratios & the empirical formula
Assume 100% g Sample
64.04 g C x 1 mol C/12.01 g C = 5.33 mol C /0.886 = 6.02
31.24 g Cl x 1 mol Cl/35.45 g Cl = mol Cl /0.886 = 1.00
4.54 g H x 1 mol H /1.008 g H = 4.50 mol H 4.50/ = 5.08
The empirical formula is C6H5Cl
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61. 3.6 Quantitative Information from Balanced Equations
The coefficients in a balanced chemical equation give the relative numbers of molecules (or formula units) involved in the reaction.
The stoichiometric coefficients in the balanced equation may be interpreted as:
the relative numbers of molecules or formula units involved in the reaction or
the relative numbers of moles involved in the reaction.
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62. 3.6 Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products.
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63. 3.6 Quantitative Information from Balanced Equations
The molar quantities indicated by the coefficients in a balanced equation are called stoichiometrically equivalent quantities.
Stoichiometric factors (or molar ratios) may be used to convert between quantities of reactants and products in a reaction.
It is important to realize that the stoichiometric ratios are the ideal proportions in which reactants are needed to form products.
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64. 3.6 Quantitative Information from Balanced Equations
A balanced reaction equation often provides more stoichiometric factors (or molar ratios) than needed to solve any particular stoichiometric problem. Often only one or two of them are relevant in a given problem.
The number of grams of reactant cannot be directly related to the number of grams of product.
To get grams of product from grams of reactant:
convert grams of reactant to moles of reactant (use molar mass),
convert moles of one reactant to moles of other reactants and products (use the stoichiometric ratio from the balanced chemical equation),
convert moles back into grams for desired product (use molar mass).
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65. 3.6 Stoichiometric Calculations
Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).
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66. Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.
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67. Sample Exercise 3.16 Calculating Amounts of Reactants and Products
How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6?
C6H12O6(s) + 6 O2(g)→6 CO2(g) + 6 H2O(l)
Solution
Analyze We are given the mass of a reactant and are asked to determine the mass of a product in the given equation.
Plan The general strategy, as outlined in Figure 3.13, requires three steps. First, the amount of C6H12O6 must be converted from grams to moles. Second, we can use the balanced equation, which relates the moles of C6H12O6 to the moles of H2O: 1 mol C6H12O mol H2O. Third, we must convert the moles of H2O to grams.
Solve First, use the molar mass of C6H12O6 to convert
from grams C6H12O6 to moles C6H12O6:
Second, use the balanced equation to convert moles of C6H12O6 to moles of H2O:
Third, use the molar mass of H2O to convert from
moles of H2O to grams of H2O:
The steps can be summarized in a diagram like that in Figure 3.13:
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68. 3.7 Limiting Reactants
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69. How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients.
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).
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70. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.
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71. Limiting Reagent
The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. 2

72. Limiting Reactants
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 (in this case, the H2).
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73. Limiting Reactants
In the example below, the O2 would be the excess reagent.
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74. Limiting Reagent
Zinc metal reacts with hydrochloric acid by the following reaction.
If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? 2

75. Limiting Reagent
Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reagent.
Since HCl is the limiting reagent, the amount of H2 produced must be 0.26 mol. 2

76. Theoretical and Percent Yield
The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants.
The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). 2

77. Theoretical Yield
The theoretical yield is the maximum amount of product that can be made.
In other words it’s the amount of product possible as calculated through the stoichiometry problem.
This is different from the actual yield, which is the amount one actually produces and measures.
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78. Sample Exercise 3.18 Calculating the Amount of Product Formed from
a Limiting Reactant
The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):
N2(g) + 3 H2(g)→2 NH3(g)
How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?
Solution
Analyze We are asked to calculate the number of moles of product, NH3, given the quantities of each reactant, N2 and H2, available in a reaction. Thus, this is a limiting reactant problem.
Plan If we assume that one reactant is completely consumed, we can calculate how much of the second reactant is needed in the reaction. By comparing the calculated quantity with the available amount, we can determine which reactant is limiting. We then proceed with the calculation, using the quantity of the limiting reactant.
Solve The number of moles of H2 needed for complete consumption of 3.0 mol of N2 is:
Because only 6.0 mol H2 is available, we will run out of
H2 before the N2 is gone, and H2 will be the limiting reactant. We use the quantity of the limiting reactant, H2, to calculate the quantity of NH3 produced:
Comment The table on the right summarizes
this example:

79. Sample Exercise 3.18 Calculating the Amount of Product Formed from
a Limiting Reactant
Consider the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react.
(a) Which is the limiting reactant?
(b) How many moles of AlCl3 are formed?
(c) How many moles of the excess reactant remain at the end of the reaction?
Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl2

80. Sample Exercise 3.19 Calculating the Amount of Product Formed from
a Limiting Reactant
Consider the following reaction that occurs in a fuel cell:
2 H2(g) + O2 (g) → 2 H2O (g)
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?
Solution
Analyze We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem.
Plan We must first identify the limiting reagent. To do so, we can calculate the number of moles of each reactant and compare their ratio with that required by the balanced equation. We then use the quantity of the limiting reagent to calculate the mass of water that forms.
Solve From the balanced equation, we have the following stoichiometric relations:
Using the molar mass of each substance, we can calculate the number of moles of each reactant:

81. Sample Exercise 3.19 Calculating the Amount of Product Formed from
a Limiting Reactant
Solution (continued)
Thus, there are more moles of H2 than O2. The coefficients in the balanced equation indicate, however, that the reaction requires 2 moles of H2 for every 1 mole of O2. Therefore, to completely react all the O2, we would need 2  47 = 94 moles of H2. Since there are only 75 moles of H2, H2 is the limiting reagent. We therefore use the quantity of H2 to calculate the quantity of product formed. We can begin this calculation
with the grams of H2, but we can save a step by starting with the moles of H2 that were calculated previously in the exercise:
Check The magnitude of the answer seems reasonable. The units are correct, and the number of significant figures (two) corresponds to those in the numbers of grams of the starting materials.
Comment The quantity of the limiting reagent, H2, can also be used to determine the quantity of O2 used (37.5 mol = 1200 g). The number of grams of the excess oxygen remaining at the end of the reaction equals the starting amount minus the amount consumed in the reaction, 1500 g – 1200 g = 300 g.

82. Sample Exercise 3.19 Practice Problem
A strip of zinc metal with a mass of 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 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)
Practice Exercise
Which reactant is limiting?
How many grams of Ag will form?
How many grams of Zn(NO3)2 will form?
How many grams of the excess reactant will be left at the end of the reaction?
Answers: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn

83. 2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(g)
Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent
Yield for a Reaction
Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2:
2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(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?
(b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid?
Solution
Analyze We are given a chemical equation and the quantity of the limiting reactant (25.0 g of C6H12). We are asked first to calculate the theoretical yield of a product (H2C6H8O4) and then to calculate its percent yield if only 33.5 g of the substance is actually obtained.
Plan (a) The theoretical yield, which is the calculated quantity of adipic acid formed in the reaction, can be calculated using the following sequence of conversions:
g C6H12 → mol C6H12 → mol H2C6H8O4 → g H2C6H8O4

84. Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent
Yield for a Reaction
Solution (continued)
(b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theoretical yield using Equation 3.14.
Solve
Check Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than 100% as necessary.

85. Exercise 3.20 Practice Problem
Imagine that you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron. In your tests you carry out the following reaction on a small scale:
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
If you start with 150 g of Fe2O3 as the limiting reagent, what is the theoretical yield of Fe?
(b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield?
Answers: (a) 105 g Fe, (b) 83.7%
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