1. Mass Relationships in Chemical Reactions
Chapter 3
Mass Relationships in Chemical Reactions

2. Atomic mass is the mass of an atom in atomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H = amu
16O = amu

4. Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things.
The atomic masses of its two stable isotopes, (69.09 percent) and (30.91 percent), are amu and amu, respectively.
Calculate the average atomic mass of copper. The relative abundances are given in parentheses.
Each isotope contributes to the average atomic mass based on its relative abundance.
Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope.
(0.6909) (62.93 amu) + (0.3091) ( amu) = amu

6. Avogadro’s number (NA)
The Mole (mol): A unit to count numbers of particles
Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly grams of 12C
1 mol = NA = x 1023
Avogadro’s number (NA)

7. Avogadro’s number (NA)
6.022 x 1023 = NA = 1mol

8. Molar mass is the mass of 1 mole of in grams marbles atoms
eggs
shoes
Molar mass is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)

9. One Mole of: S C Hg Cu Fe

10. 1 12C atom
12.00 amu
12.00 g
6.022 x C atoms =
1.66 x g
1 amu x M
= molar mass in g/mol
NA = Avogadro’s number

11. 3.2
Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea diving tanks, and balloons.
How many moles of He atoms are in 6.46 g of He?
Thus, there are 1.61 moles of He atoms in 6.46 g of He.
Check
Because the given mass (6.46 g) is larger than the molar mass of He, we expect to have more than 1 mole of He.

12. Moles to g How many grams of gold (Au) are in 15.3 moles of Au?
1mole of Au = g
15.3 moles of Au = (197.0 g/1mol)x 15.3 mol
= g
=3.01x103 g

13. Mass and Moles of a Substance
Mole calculations
Suppose we have grams of iron (Fe). The atomic mass of iron is 55.8 amu. How many moles of iron does this represent? 2

14. Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 5.75 moles of magnesium (molar mass = 24.3 g/mol). What is its mass? 2

15. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K = x 1023 atoms K
1 mol K
39.10 g K x x
6.022 x 1023 atoms K
1 mol K =
0.551 g K
8.49 x 1021 atoms K

16. Let's Practice Using these Tools
1. What is the mass in grams of 2.5 mol Ca
2. Calculate the number of atoms in 1.7 mol B.
3. Calculate the number of atoms in 5.0 g Al.
4. Calculate the mass of 5,000,000 atoms of Au.

17. formula mass (amu) = molar mass (grams)
Molecular Weight and Formula Weight
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na
22.99 amu
1Cl
amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = amu
1 mole NaCl = g NaCl

18. Do You Understand Formula Mass?
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2 P
2 x 30.97
8 O
+ 8 x 16.00
amu

19. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2

20. Mass and Moles of a Substance
Mole calculations
This same method applies to compounds. Suppose we have grams of H2O (molar mass = 18.0 g/mol). How many moles does this represent? 2

21. Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = g/mol). (the correct term is molar mass, but this is a common usage) What is its mass? 2

22. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
1 mol H atoms x =
72.5 g C3H8O
5.82 x 1024 atoms H

23. Heavy
Light

25. Determining Chemical Formulas
The percent composition of a compound is the mass percentage of each element in the compound.
We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is, 2

26. Mass Percentages from Formulas
Let’s calculate the percent composition of butane, C4H10.
First, we need the molecular mass of C4H10.
Now, we can calculate the percents. 2

27. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%

28. Determining Chemical Formulas
Determining the formula of a compound from the percent composition.
The percent composition of a compound leads directly to its empirical formula.
An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. 2

29. Percent Composition and Empirical Formulas
Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O percent.
nK = g K x
= mol K
1 mol K
39.10 g K
nMn = g Mn x
= mol Mn
1 mol Mn
54.94 g Mn
nO = g O x
= mol O
1 mol O
16.00 g O

30. Percent Composition and Empirical Formulas
nK = , nMn = , nO = 2.532
K : ~
1.0
0.6330
0.6329
Mn :
0.6329
= 1.0
O : ~
4.0
2.532
0.6329
KMnO4

31. g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
g C
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
g H
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
See p
Divide by smallest subscript (0.25)
Empirical formula C2H6O

32. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula?
In other words, give the smallest whole-number ratio of the subscripts in the formula
Cx HyOz 2

33. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Our grams of benzoic acid would contain:
This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge. 2

34. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Our grams of benzoic acid would contain:
now it’s not too difficult to see that the smallest whole number ratio is 7:6:2.
The empirical formula is C7H6O2 . 2

35. Determining Chemical Formulas
Determining the “true” molecular formula from the empirical formula.
For example, suppose the empirical formula of a compound is CH2O and its “true” molecular weight is 60.0 g.
The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g.
This would imply that the “true” molecular formula is actually the empirical formula doubled (60g/30g = 2), or
C2H4O2 2

36. 3.11
A sample of a compound contains percent nitrogen and percent oxygen by mass, as determined by a mass spectrometer.
In a separate experiment, the molar mass of the compound is found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass of the compound.

37. 3.11
Strategy
To determine the molecular formula, we first need to determine the empirical formula. Comparing the empirical molar mass to the experimentally determined molar mass will reveal the relationship between the empirical formula and molecular formula.
Solution
We start by assuming that there are 100 g of the compound. Then each percentage can be converted directly to grams; that is, g of N and g of O.

38. 3.11 Let n represent the number of moles of each element so that
Thus, we arrive at the formula N2.174O4.346, which gives the identity and the ratios of atoms present. However, chemical formulas are written with whole numbers.
Try to convert to whole numbers by dividing the subscripts by the smaller subscript (2.174). After rounding off, we obtain NO2 as the empirical formula.

39. empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
3.11
The molecular formula might be the same as the empirical formula or some integral multiple of it (for example, two, three, four, or more times the empirical formula).
Comparing the ratio of the molar mass to the molar mass of the empirical formula will show the integral relationship between the empirical and molecular formulas.
The molar mass of the empirical formula NO2 is
empirical molar mass = g + 2(16.00 g) = g

40. 3.11
Next, we determine the ratio between the molar mass and the empirical molar mass
The molar mass is twice the empirical molar mass. This means that there are two NO2 units in each molecule of the compound, and the molecular formula is (NO2)2 or N2O4. The actual molar mass of the compound is two times the empirical molar mass,
that is, 2(46.01 g) or g, which is between 90 g and 95 g.

41. Chemical Reactions and Chemical Equations
A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products

42. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO

43. H2 is released when Na is placed in water.
The Experimental Basis of a Chemical Equation
We know that a chemical equation represents a chemical change. The following is evidence for a reaction:
Release of a gas.
CO2 is released when acid is placed in a solution containing CO32- ions.
H2 is released when Na is placed in water.
Formation of a solid (precipitate.)
A solution containing Ag+ ions is mixed with a solution containing Cl- ions.

44. Chemical Reactions: Equations
Writing chemical equations
A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas.
For example, the burning of sodium and chlorine to produce sodium chloride is written
The reactants are starting substances in a chemical reaction. The arrow means “yields.” The formulas on the right side of the arrow represent the products. 2

45. Chemical Reactions: Equations
In many cases, it is useful to indicate the states of the substances in the equation.
Writing chemical equations
When you use these labels, the previous equation becomes 2

46. Chemical Reactions: Equations
The law of conservation of mass dictates that the total number of atoms of each element on both sides of a chemical equation must match. The equation is then said to be balanced.
Writing chemical equations
Consider the combustion of methane to produce carbon dioxide and water. 2

47. Chemical Reactions: Equations
Writing chemical equations
For this equation to balance, two molecules of oxygen must be consumed for each molecule of methane, producing one molecule of CO2 and two molecules of water.
Now the equation is “balanced.” 2 2

48. Balancing Chemical Equations
Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C2H6
NOT
C4H12

49. Balancing Chemical Equations
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O

50. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O

51. Balancing Chemical Equations
Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4 C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
Products
4 C
12 H
14 O

52. Chemical Reactions: Equations
Balance the following equations. 2

53. Chemical Reactions: Equations
Balance the following equations. 2 6 6 2 9 3 4 2

54. Stoichiometry: Quantitative Relations in Chemical Reactions
Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.
It is based on the balanced chemical equation and on the relationship between mass and moles.
Such calculations are fundamental to most quantitative work in chemistry. 2

55. Amounts of Reactants and Products
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units

56. Molar Interpretation of a Chemical Equation
The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships.
For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen. 2

57. Molar Interpretation of a Chemical Equation
This balanced chemical equation shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
1 molecule N molecules H molecules NH3
Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. 2

58. Calculations Using the Chemical Equation7
We will learn in this section to calculate quantities of reactants and products in a chemical reaction.
Need a balanced chemical equation for the reaction of interest.
Keep in mind that the coefficients represent the number of moles of each substance in the equation.

59. Molar Interpretation of a Chemical Equation
Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2.
Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple. 2

60. Limiting Reagent: Reactant used up first in the reaction.
2NO + O NO2
NO is the limiting reagent
O2 is the excess reagent

61. 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

62. 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

63. Reaction Yield Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100

64. 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

65. Theoretical and Percent Yield
To illustrate the calculation of percentage yield, recall that the theoretical yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2.
If the actual yield of the reaction had been 0.22 g H2, then 2

66. Summary Atomic mass unit (amu) Avagadro’s number
Percent composition by mass of a compound,
Empirical formula, molecular formula
Chemical Equations
Stoichiometry and limiting reagents
Theoretical yield, actual yied, and percent yield

67. Homework Practice all the examples worked out in the book
3.2,3.5,3.9,3.13,3.14,3.16,3.22,3.24,3.25,3.26,3.38,3.40,3.48,3.50,3.52,3.59,3.60,3.66,3.67,
3.68,3.89