1. Chemistry, The Central Science, 10th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

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

3. Chemical Equations
Concise representations of chemical reactions

4. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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.

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.

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.

8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

9. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule

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

11. Reaction Types

12. Combination Reactions
Two or more substances react to form one product
Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s)

13. 2 Mg (s) + O2 (g)  2 MgO (s)

14. Decomposition Reactions
One substance breaks down into two or more substances
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)

15. Combustion Reactions Rapid reactions that produce a flame
To combust is to react, or burn, in the presence of oxygen
Most often involve hydrocarbons reacting with oxygen in the air
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
There are two other main types, single and double replacement reactions, that we will discuss later in Chapter 4.

16. Formula Weights

17. Formula Weight (FW)
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
These are generally reported for ionic compounds

18. Molecular Weight (MW)
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

19. PRACTICE EXERCISE
Calculate the formula weight of (a) Al(OH)3 and (b) CH3OH.
Answers:
(a) amu, (b) amu

20. Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100

21. Percent Composition So the percentage of carbon in ethane is…
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%

22. Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C12H22O11.

23. Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
PRACTICE EXERCISE
Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
Answer:
17.1%

24. Moles

25. Avogadro’s Number
6.02 x 1023
1 mole of 12C has a mass of 12 g

26. Molar Mass
By definition, these are 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 formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

27. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale

28. 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

29. Calculate the number of H atoms in 0.350 mol of C6H12O6.
SAMPLE EXERCISE 3.8 Converting Moles to Number of Atoms
Calculate the number of H atoms in mol of C6H12O6.
Solve:
H atoms =

30. PRACTICE EXERCISE
How many oxygen atoms are in (a) 0.25 mol Ca(NO3)2 and (b) 1.50 mol of sodium carbonate?
Answer:
(a) 9.0  1023, (b) 2.71  1024

31. Finding Empirical Formulas

32. Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition

33. 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.

34. 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

35. 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

36. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2

37. SAMPLE EXERCISE 3.13 Calculating an 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?

38. SAMPLE EXERCISE 3.13 continued
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: 1 3
The ratio for H is too far from 1 to attribute the difference to experimental error; in fact, it is quite close to 1–. 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:

39. PRACTICE EXERCISE
A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
Answer:
C4H4O

40. 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

41. SAMPLE EXERCISE 3.15 Determing Empirical Formula by Combustion Analysis
Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.
Solution
Analyze: We are told that isopropyl alcohol contains C, H, and O atoms and given the quantities of CO2 and H2O produced when a given quantity of the alcohol is combusted. We must use this information to determine the empirical formula for isopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample.
Plan: We can use the mole concept to calculate the number of grams of C present in the CO2 and the number of grams of H present in the H2O. These are the quantities of C and H present in the isopropyl alcohol before combustion. The number of grams of O in the compound equals the mass of the isopropyl alcohol minus the sum of the C and H masses. Once we have the number of grams of C, H, and O in the sample, we can then proceed as in Sample Exercise 3.13: Calculate the number of moles of each element, and determine the mole ratio, which gives the subscripts in the empirical formula.
Solve: To calculate the number of grams of C, we first use the molar mass of CO2, 1 mol CO2 = 44.0 g CO2, to convert grams of CO2 to moles of CO2. Because there is only 1 C atom in each CO2 molecule, there is 1 mol of C atoms per mole of CO2 molecules. This fact allows us to convert the moles of CO2 to moles of C. Finally, we use the molar mass of C, 1 mol C = 12.0 g C, to convert moles of C to grams of C. Combining the three conversion factors, we have:

42. SAMPLE EXERCISE 3.14 Determining a Molecular Formula
Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4 . The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene?
Solve: First, we calculate the formula weight of the empirical formula, C3H4
Next, we divide the molecular weight by the empirical formula weight to obtain the multiple used to multiply the subscripts in C3H4 :
Only whole-number ratios make physical sense because we must be dealing with whole atoms. The 3.02 in this case results from a small experimental error in the molecular weight. We therefore multiply each subscript in the empirical formula by 3 to give the molecular formula: C9H12.
Check: We can have confidence in the result because dividing the molecular weight by the formula weight yields nearly a whole number.

43. SAMPLE EXERCISE 3.14 continued
PRACTICE EXERCISE
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? (b) What is its molecular formula?
Answers:
(a) CH3O, (b) C2H6O2

44. SAMPLE EXERCISE 3.15 continued
The calculation of the number of grams of H from the grams of H2O is similar, although we must remember that there are 2 mol of H atoms per 1 mol of H2O molecules:
The total mass of the sample, g, is the sum of the masses of the C, H, and O. Thus, we can calculate the mass of O as follows:
We then calculate the number of moles of C, H, and O in the sample:
To find the empirical formula, we must compare the relative number of moles of each element in the sample. The relative number of moles of each element is found by dividing each number by the smallest number, The mole ratio of C : H : O so obtained is 2.98 : 7.91 : The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C3H8O.
Check: The subscripts work out to be moderately sized whole numbers, as expected.

45. SAMPLE EXERCISE 3.15 continued
PRACTICE EXERCISE
(a) Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO2 and g H2O. What is the empirical formula of caproic acid? (b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?
Answers:
(a) C3H6O, (b) C6H12O2

46. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products

47. Stoichiometric Calculations
From 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)

48. SAMPLE EXERCISE 3.17 Calculating Amounts of Reactants and Products
Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide?
We are given the grams of LiOH and asked to calculate grams of CO2 We can accomplish this task by using the following sequence of conversions:
Solve:
Check: Notice that , 24  2 = 48, and 44/48 is slightly less than 1. Thus, the magnitude of the answer is reasonable based on the amount of starting LiOH; the significant figures and units are appropriate, too. ≈

49. SAMPLE EXERCISE 3.17 continued
PRACTICE EXERCISE
Propane, C3H8, is a common fuel used for cooking and home heating. What mass of oxygen is consumed in the combustion of 1.00 g of propane?
Answers:
3.64 g

50. Limiting Reactants

51. 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)

52. 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

53. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount

54. 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)

55. Limiting Reactants
In the example below, the O2 would be the excess reagent

56. SAMPLE EXERCISE 3.18 continued
PRACTICE EXERCISE
Consider the reaction
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

57. SAMPLE EXERCISE 3.19 continued
PRACTICE EXERCISE
A strip of zinc metal having 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:
(a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) 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

58. Theoretical Yield
The theoretical yield is the 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, the amount one actually produces and measures

59. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
A comparison of the amount actually obtained to the amount it was possible to make
Actual Yield
Theoretical Yield
Percent Yield = x 100

60. SAMPLE EXERCISE 3.20 Calculating the Theoretical Yield and 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:
(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?

61. SAMPLE EXERCISE 3.20 continued
Solve:
Check: Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than 100% as necessary.