1. Chapter 3 Chemical Reactions and Reaction Stoichiometry
Lecture Presentation
Chapter 3 Chemical Reactions and Reaction Stoichiometry
James F. Kirby
Quinnipiac University
Hamden, CT

2. Stoichiometry The study of the mass relationships in chemistry
Based on the Law of Conservation of Mass (Antoine Lavoisier, 1789)
“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

3. Chemical Equations
Chemical equations are concise representations of chemical reactions.

4. What Is in a Chemical Equation?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Reactants appear on the left side of the equation.

5. What Is in a Chemical Equation?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Products appear on the right side of the equation.

6. What Is in a Chemical Equation?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
The states of the reactants and products are written in parentheses to the right of each compound.
(g) = gas; (l) = liquid; (s) = solid;
(aq) = in aqueous solution

7. What Is in a Chemical Equation?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Coefficients are inserted to balance the equation to follow the law of conservation of mass.

8. Why Do We Add Coefficients Instead of Changing Subscripts to Balance?
Hydrogen and oxygen can make water OR hydrogen peroxide:
2 H2(g) + O2(g) → 2 H2O(l)
H2(g) + O2(g) → H2O2(l)

9. Three Types of Reactions
Combination reactions
Decomposition reactions
Combustion reactions

10. Combination Reactions
In combination reactions two or more substances react to form one product.
Examples:
2 Mg(s) + O2(g) MgO(s)
N2(g) + 3 H2(g) NH3(g)
C3H6(g) + Br2(l) C3H6Br2(l)

11. Decomposition Reactions
In a decomposition reaction one substance breaks down into two or more substances.
Examples:
CaCO3(s) CaO(s) + CO2(g)
2 KClO3(s) KCl(s) + O2(g)
2 NaN3(s) Na(s) + 3 N2(g)

12. Combustion Reactions Examples:
Combustion reactions are generally rapid reactions that produce a flame.
Combustion reactions most often involve oxygen in the air as a reactant.
Examples:
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(g)

13. 3.3 Formula Weight (FW)
A formula weight is the sum of the atomic weights for the atoms in a chemical formula.
This is the quantitative significance of a formula.
The formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.08 amu)
+ Cl: 2( amu)
amu

14. 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( amu)
+ H: 6( amu)
amu

15. Ionic Compounds and Formulas
Remember, ionic compounds exist with a three-dimensional order of ions. There is no simple group of atoms to call a molecule.
As such, ionic compounds use empirical formulas and formula weights (not molecular weights).

16. 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)
× 100

17. Percent Composition So the percentage of carbon in ethane, C2H6, is
(2)( amu)
( amu)
amu
amu =
× 100
= %

18. Practice
Calculate the percent of nitrogen, by mass, in calcium nitrate?
2. Calculate the percentage of potassium, by mass in K2PtCl6?
Answers:
17.1%
16.1%

19. Avogadro’s Number
In a lab, we cannot work with individual molecules. They are too small.
6.02 × 1023 atoms or molecules is an amount that brings us to lab size. It is ONE MOLE.
One mole of 12C atoms has a mass of g.

20. Molar Mass
A molar mass is the mass of 1 mol of a substance (i.e., g/mol).
The molar mass of an element is the atomic weight for the element from the periodic table. If it is diatomic, it is twice that atomic weight.
The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

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

22. 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.

23. Practice How many sulfur atoms are in 0.45 mol BaSO4?
How many oxygen atoms are in 0.25 mol Ca(NO3)2

24. Determining Empirical Formulas
One can determine the empirical formula from the percent composition by following these three steps.

25. Determining Empirical Formulas— an Example
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.

26. Determining Empirical Formulas— an Example
Assuming g of para-aminobenzoic acid,
C: g × = mol C
H: g × = 5.09 mol H
N: g × = mol N
O: g × = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g

27. Determining Empirical Formulas— an Example
Calculate the mole ratio by dividing by the smallest number of moles:
5.105 mol
mol
5.09 mol
1.458 mol
C: = ≈ 7
H: = ≈ 7
N: = 1.000
O: = ≈ 2

28. Determining Empirical Formulas— an Example
These are the subscripts for the empirical formula:
C7H7NO2

29. Determining a Molecular Formula
Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula.
If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula.

30. Determining a Molecular Formula— an Example
The empirical formula of a compound was found to be CH. It has a molar mass of 78 g/mol. What is its molecular formula?
Solution:
Whole-number multiple = 78/13 = 6
The molecular formula is C6H6.

31. Practice
1. Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula?
2. Cyclohexane is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula?

32. Answers
1. C3H4O3
© 2015 Pearson Education, Inc.

33. Combustion Analysis
Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14.
C is determined from the mass of CO2 produced.
H is determined from the mass of H2O produced.
O is determined by the difference after C and H have been determined.

34. Sample Exercise 3.15 Determining an Empirical Formula by Combustion Analysis
Isopropyl alcohol, 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.
Plan Because all of the carbon and hydrogen in the sample is converted to CO2 and H2O, we can use dimensional analysis to calculate the mass C and H in the sample of isopropyl alcohol.
The mass of O in the compound equals the mass of the original sample
minus the sum of the C and H masses.
Once we have the C, H, and O masses, we can calculate moles of C, H, and O in isopropyl alcohol and determine the empirical formula.

35. Solve Because all of the carbon in the sample is converted to CO2, we can use dimensional analysis and the following steps to calculate the mass C in the sample. Using the values given in this example, the mass of C is
Because all of the hydrogen in the sample is converted to H2O, we can use dimensional analysis and the following steps to calculate the mass H in the sample. We use three significant figures for the atomic mass of H to match the significant figures in the mass of H2O produced.
The mass of the sample, g, is the sum of the masses of C, H, and O. Thus, the O mass is
Mass of O = mass of sample – (mass of C + mass of H)
= g – (0.153 g g) = g O
The number of moles of C, H, and O in the sample is therefore

36. To find the empirical formula, we must compare the relative number of moles of each element in the sample
The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C3H8O.

37. Quantitative Relationships
The coefficients in the balanced equation show
relative numbers of molecules of reactants and products.
relative numbers of moles of reactants and products, which can be converted to mass.

38. Stoichiometric Calculations
Stoichiometry compares two DIFFERENT materials, using the MOLE RATIO from the balanced equation!

39. An Example of a Stoichiometric Calculation
How many grams of water can be produced from 1.00 g of glucose?
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
There is 1.00 g of glucose to start.
The first step is to convert it to moles.

40. An Example of a Stoichiometric Calculation
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Then convert moles of one substance in the equation to moles of another substance.
The MOLE RATIO comes from the balanced equation.

41. An Example of a Stoichiometric Calculation
The last step is to convert moles of water (substance B) into grams of water, using molar mass.

42. Stoichiometry Practice
Sodium hydroxide reacts with carbon dioxide to form sodium carbonate and water: 2 NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l) How many grams of Na2CO3 can be prepared from 2.40 g of NaOH?

43. 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).

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

45. Limiting Reactants
The limiting reactant is used in all stoichiometry calculations to determine amounts of products and amounts of any other reactant(s) used in a reaction.

46. 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
The LIMITING REACTANT determines the theoretical yield.
This is different from the actual yield, which is the amount one actually produces and measures.

47. Percent Yield actual yield theoretical yield Percent yield = × 100
One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield):
Percent yield = × 100
actual yield
theoretical yield

48. Sample Exercise 1 The reaction: 2 H2(g) + O2(g) → 2 H2O(g)
Is used to produced electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2 (g) and 1500 g of O2 (g).
How many grams of water can form?
What is the limiting reactant?
What mass of excess reactant is left when the reaction is completed?

49. 2 C6H12 (l) + 5 O2(g) → 2 H2C6H3O4(l) + 2 H2O(g)
Sample Exercise 2
Adipic acid, H2C6H8O4, used to produce nylon, is made commercially by a reaction between cyclohexane (C6H12) and O2:
2 C6H12 (l) + 5 O2(g) → 2 H2C6H3O4(l) + 2 H2O(g)
Assume that you carry out this reaction with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid?
If you obtain 33.5 g of adipic acid, what is the percent yield for the reaction?