1. Mass Relationships in Chemical Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

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
1g ≈ x 1023 amu.
3.1

3. Average atomic mass of lithium:
Average atomic mass = (sum of isotope mass x percentage abundance)/100
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x x 7.016
100
= amu
3.1

4. Calcium has 6 isotopes. Their masses & percentage abundances are given in the table below. Calculate the average atomic mass or atomic weight of Ca.
Isotope
Isotope mass (amu)
Percentage Abundance
Ca-40
96.941
Ca-42
0.647
Ca-43
0.135
Ca-44
2.086
Ca-46
0.004
Ca-48
0.187
{( x ) + ( x 0.647) + (0.135 x ) + (2.086 x ) + (.004 x ) + ( x 0.187)}amu/100
=4007.8amu/100
= amu

5. Average atomic mass (6.941)

6. Avogadro’s number (NA)
Dozen = 12
Pair = 2
A mole (mol) of any substance is a collection of an avogadro’s number (NA) of particles of that substance.
Analogy: A dozen (doz) of any substance is a collection of 12 particles of that substance.
1 mol = NA = x 1023
Avogadro’s number (NA)
3.2

7. 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 12C atom = amu
1 mole 12C atoms = x 1023 atoms = g
1 mole 12C atoms = g 12C
Average mass of 1 Li atom = amu
Molar mass of Li = mass of 1 mole lithium atoms = g
For any element,
numerical value of atomic mass (amu) = numerical value of molar mass (grams)
3.2

8. One Mole of: S C Hg Cu Fe
3.2

9. 1 g = 6.022 x 1023 amu = molar mass in g/mol M NA = Avogadro’s number
3.2

10. g Mercury has
How many moles of mercury? mol
How many atoms of Mercury? 4.50 x 1022 atoms
x 1022 atoms of Fe is
How many moles of Fe? mol Fe
What is the corresponding mass? 4.79 g Fe
g Zinc has
How many moles of Zinc?
How many atoms of Zinc?
x 1022 atoms of Cu is
How many moles of Cu?
What is the corresponding mass?

11. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol of K = g of K
1 mol of K = x 1023 atoms of 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 of K
3.2

12. 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
The numerical value of molecular mass in amu = the numerical value of molar mass in grams
1 molecule of SO2 weighs amu
1 mole of SO2 weighs g
1 mole of SO2 =6.022 x 1023 molecules of SO2
3.3

13. You cannot go directly from grams of a compound to moles of A CONSTITUENT ATOM
You have to first calculate the moles of the compound and then from that calculate the moles of a constituent atom:
Grams of compound  Moles of Compound  Moles of a constituent atom
The molecular formula not only tells us how many atoms of each type are present in one molecule of a compound but also how many MOLES of atoms of each type are present in 1MOLE OF THE COMPOUND.
For example, the molecular formula of sugar, C12H22O11, tells us that 1 mole of sugar has 12 moles of C atoms, 22 moles of H atoms & 11 moles of O atoms besides the fact that 12 atoms of C, 22 H atoms & 11 O atoms are present in 1 sugar molecule.

14. Consider a sample of 15.0 g Borazine (B3N3H6)
How many moles of borazine are present? mol
How many molecules of borazine are present? 1.12 x 1023 molecules
How many moles of N atoms are present? mol N
How many N atoms are present? 3.36 x 1023 N atoms
2. A sample of ethanol, C2H5OH has 6.00 x 1023 H atoms.
How many ethanol molecules are present? 1.00 x 1023 molecules
How many moles of ethanol are present?0.166 mol
What is the mass of the sample? 7.65 g

15. Complete the following conversions;
Consider g of H2O
Calculate the following:
The moles of H2O: 2.77 × 10-4
The molecules of H2O: 1.67 × 1020
The moles of H atoms: 5.55 × 10-4
The number of H atoms present: 3.34 × 1020
2. Na atoms in 3.50 mole NaOH
3. grams mass of 2.50 mole O2
4. carbon atoms in 10.5 g C6H12O6
5. A sample of benzene, C6H6 has 3.00 x 1022 H atoms.
How many benzene molecules are present? 5.00 × 1021
How many moles of benzene are present? 8.30 × 10-3
What is the mass of the sample? g

16. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
moles of C3H8O = 72.5 g / g/mol = 1.21 mol
1 mol C3H8O molecules contains 8 mol H atoms
1 mol of H atoms is x 1023 H atoms
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 H atoms
Steps: Convert grams of C3H8O to moles of C3H8O.
2. Convert moles of C3H8O to moles of H atoms.
3. Convert moles of H atoms to number of H atoms.
3.3

17. Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
NaCl
1Na
22.99 amu
1Cl
amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit of NaCl = amu
1 mole of NaCl = g of NaCl
3.3

18. Do You Understand Formula Mass?
What is the formula mass of Ca3(PO4)2 ? amu
1 formula unit of Ca3(PO4)2
3 Ca
3 x g/mol
2 P
2 x g/mol
8 O
+ 8 x g/mol
g/mol = molar mass
Units of grams per mole are the most practical
for chemical calculations!
3.3

19. You have a 10.0 g of Ca3(PO4)2. (Molar mass = 310.18 g/mol)
How many moles are present? mol
How many moles of Ca+2 ions are present? mol
How many phosphate ions are present? = mol = 3.88 x 1022 ions.
2. You have a 10.0 g of (NH4)3PO4 (molar mass = g/mol)
How many moles are present? mol
How many moles of phosphate ions are present? mol
How many ammonium ions are present? 1.21 × 1023

20. 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 (assume you have 1 mole!).
%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%
3.5

21. Calculate percentage composition of
Borazine (B3N3H6; molar mass = 80.52g/mol) 40.28%B, 52.20% N, 7.52% H
Ca3(PO4)2 (Molar mass = g/mol)
38.76 % Ca, % P, % O

22. 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 40.51% 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
To begin, assume for simplicity that you have 100 g of compound!
3.5

23. 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
3.5

24. Complete lab. 4 report 4 (page 39)
Problems 4a, 4b, 5 on prelab 4 (page 41)
Problems 4 & 5 on postlab. 4 (page 43)
Determine the empirical formula of the compound whose percentage composition is
62.1% C, 5.21% H, 12.1% N, and 20.7% O
Step 1: assume 100 g: 62.1 g C, 5.21 g H, 12.1 g N, 20.7 g O
Step 2: (62.1/12.01)mol C = 5.17 mol C; (5.21/1.01) mol H = 5.16 mol H; (12.1/14.01)mol N = mol N; (20.7/16.00) mol O = 1.29 mol O
Step 3: C:H:N:O = 5.17 : 5.16 : : 1.29
= (5.17/0.864) : (5.16/0.864) : 1 : (1.29/0.864) =
= 5.98 : 5.97 : 1: 1.50
= 6 : 6 : 1 : 1.50 (DO NOT ROUND OFF NUMBERS WITH .5; MULTIPLY BY 2 BEFORE ROUNDING OFF)
= 12 : 12 : 2 : 3; C12H12N2O3

25. A sample of an organic compound was found to contain 9. 01g C, 1
A sample of an organic compound was found to contain 9.01g C, 1.01g H, 12.0 g O. Calculate the empirical formula of this compound.
9.01g C = (9.01/12.01) mol C = mol C
1.01 g H = (1.01/1.01)mol H = 1.00 mol H
12.0 g O = (12.0/16.00) mol O = mol O
C : H : O = : 1.00 : 0.750
= (0.750/0.750) : (1.00/0.750) : (0.750/0.750)
= 1.00 : 1.33 : 1.00 = 1 : 1.33 : 1
DO NOT ROUND OFF ANY NO. WITH 0.3 OR 0.33 OR OR 0.5 OR ANY THAT CAN BE ROUNDED OFF TO 0.5.
MULTIPLY THOSE WITH .3, .33 etc. WITH 3 (REMEMBER TO MULTIPLY ALL THE NUMBERS IN THE RATIO)
MULTIPLY THOSE WITH .5 WITH 2 (REMEMBER TO MULTIPLY ALL THE NUMBERS IN THE RATIO)
C : H : O = 1 : 1.33 : 1 = 3 x (1 : 1.33 : 1) = 3 : 3.99 : 3 = 3: 4 : 3
Empirical formula = C3H4O3

26. Determination of Molecular Formulas from Empirical Formulas
Determine the ratio of the molecular weight to empirical formula weight (the molecular weight HAS to be provided): molecular weight/empirical formula weight
Multiply all subscripts in the empirical formula by the above ratio to obtain the molecular formula.
A hydrocarbon has 12.01g C, 3.03g H. Its molecular weight = amu. Calculate the molecular formula of the compound.
12.01 g C, 3.03 g H = (12.01/12.01) mol C, (3.03/1.01) mol H = mol C, 3.00 mol H = 1 mol C, 3 mol H
Empirical formula =CH3
Empirical formula weight = (12.01) +(3 x 1.01) amu = amu
Molecular weight/empirical formula weight = 30.08/15.04 = 2
Therefore molecular formula : C(1 x 2)H(3 x 2) = C2H6

27. Consider the reaction of H2 with O2 to form H2O
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 formulas to show what happens during a chemical reaction.
Consider the reaction of H2 with O2 to form H2O
reactants
products
3.7

28. Balance equations ONLY by putting numbers IN FRONT OF formulas.
Such numbers are called COEFFICIENTS
NEVER CHANGE SUBSCRIPTS – doing so will change the identity of reactants, products and your equation will no longer represent the reaction that you are considering

29. 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
3.7

30. Coefficients: Numbers in front of formulas in a chemical equations.
As you saw in the previous slide, these numbers relate the number of moles of reactants used to the moles of products formed.
To interpret the equation in terms of masses of reactants used and masses of products formed, you have to convert the moles of each species to the corresponding mass by multiplying by the molar mass of that species.
A chemical equation is a ‘recipe’ for the reaction and specifies the relative amounts of reactants required and the relative amounts of products that form.
2 Mg + O2  2 MgO
Ratio of moles of Mg used : moles of O2 used: moles of MgO formed = 2: 1 : 2
Coefficients: 2, 1, 2 define the STOICHIOMETRY of the reaction

31. Stoichiometry Calculations: Amounts of Reactants and Products
Use the fabulous four steps!
Write the 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 the desired units.
3.8

32. Methanol burns in air according to the equation
2 CH3OH + 3 O CO2 + 4 H2O
a. If 209 g of methanol are used up in the combustion,
what mass of water can be produced?
b. Calculate the percentage yield if the combustion yields 230 g of H2O
c. What is the mass of O2 used if all methanol is used up?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g of H2O
3.8

33. Limiting Reactant: Is that reactant that is first used up
Limiting Reactant: Is that reactant that is first used up. When it is used up, the reaction stops because all reactants have to be present for the reaction to occur.
Excess Reactant: A reactant that is remaining even after the reaction has stopped
Theoretical Yield: The maximum mass of a product that is CALCULATED to form when all of the LIMITING REACTANT is consumed. (The limiting reactant predicts the smallest yield)
Actual Yield: An EXPERIMENTAL value: the mass of a product ACTUALLY formed when the experiment is performed
Actual yield ≤ Theoretical Yield
Percentage Yield = (Actual/Theoretical) x 100; ≤ 100%

34. Do You Understand Limiting Reactants?
In a reaction, 124 g of Al are reacted with 601 g of Fe2O3.
2 Al + Fe2O Al2O3 + 2 Fe
Calculate the thoeretical mass of Al2O3 that can form in grams.
What mass of excess reactant remains?
If the actual yield of Al2O3 is 220g, what is the percent yield
Solution to a:
Fabulous Four Steps
Balanced reaction: Done.
Moles of “given” reactants.
Moles of Al = 124 g / g/mol = mol
Moles of Fe2O3 = 601 g / g/mol = mol
3.9

35. 3. Moles of “desired” product, Al2O3.
2 Al + Fe2O Al2O3 + 2 Fe
Moles of Al2O3 = mol Al X 1 mol Al2O3 = mol Al2O3
based on Al mol Al
Moles of Al2O3 = mol Fe2O3 X 1 mol Al2O3 = mole Al2O3 based on Fe2O mol Fe2O3
Keep the smaller answer! Al is the limiting reactant.
4. Grams of Al2O3.
Grams of Al2O3 = mol X g/mol = 235 g
3.9

36. In a certain experiment 2. 50 g of NH3 and 2. 85 g of O2 are taken
In a certain experiment 2.50 g of NH3 and 2.85 g of O2 are taken. The following reaction occurs:
4NH3 + 5 O2  4NO + 6 H2O
1. Determine the theoretical yield of NO? 2.14 g not 4.40g
Limiting reactant = O2, excess reactant: NH3
2. How much excess reactant remains
after the liming reactant is completely consumed? 1.29 g
3. If the actual yield is 2.00g of NO, what is the percentage yield? 93.5%
If only 1 reactant amount is given, it is implied that that reactant is the limiting reactant. There is no need for any further identification.

37. If only 1 reactant amount is given, it is implied that that reactant is the limiting reactant. There is no need for any further identification.
Consider the reaction between benzene, C6H6 and bromine, Br2, forming bromobenzene, C6H5Br:
C6H6 + Br2  , C6H5Br + HBr
If 30.0g of C6H6 is used. What is theoretical yield of C6H5Br? 60.30g
2. If the actual yield of C6H5Br is 56.7 g, what is the % yield? 94.0 %

38. 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
3.10

39. Balancing Chemical Equations
1. 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.
2. 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.
3. Start by balancing those elements that appear in only one reactant and one product.
4. Balance those elements that appear in two or more reactants or products.
5. Coefficients should be whole numbers
6. Check to make sure that you have the same number of each type of atom on both sides of the equation.

40. 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.
2 C2H6
NOT
C4H12
3.7

41. 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
2 CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2 CO2 + 3 H2O
3.7

42. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2 CO2 + 3 H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2 CO2 + 3 H2O 7 2
remove fraction
multiply both sides by 2
2 C2H6 + 7 O2
4 CO2 + 6 H2O
3.7

43. Balancing Chemical Equations
Check to make sure that you have the same number of each type of atom on both sides of the equation.
2 C2H6 + 7 O2
4 CO2 + 6 H2O
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
3.7