Mass Relationships in

Chemical Reactions
Chapter 3

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Chapter 3 Mass Relationships in Chemical
Reactions
• 3. 1 Atomic Mass
• 3. 3 Molecular Mass
• 3. 2 Mole Concept: Avogadro’s Number and Molar
Mass of an Element
• 3. 4 The Mass Spectrometer
• 3. 5 Percent Composition of Compounds
• 3. 6 Experimental Determination of Empirical Formulas
• 3. 7 Chemical Reactions and Chemical Equations
• 3. 8 Amounts of Reactants and Products
• 3. 9 Limiting Reagent Calculations 2

• 3. 10 Reaction Yield
 Chapter -3 Lecture 1
• 3. 1 Atomic Mass
• 3. 3 Molecular Mass

3
 Chapter 2 - Recap
• Atomic structure: Atoms contain protons and neutrons in
the nucleus and electrons outside
• Atomic number and atomic mass: how to find the
number of protons, neutrons and electrons for elements;
isotopes
• The periodic table: how elements are organized
• Ions
• Compounds: ionic and molecular compounds
• Molecular formulas and empirical formulas
• Naming compounds

4
 Chapter 2 - Recap
• Elements are represented by: Atomic symbols
• Examples

• Compounds are represented by chemical formulas

• Examples

Micro World Macro World

atoms & molecules grams

5
 Atomic Mass
Internationally agreed convention:
Atomic mass is the mass of an atom in
atomic mass units (amu)

By definition:
1 atom 12C “weighs” 12 amu
On this scale
Notice that the
reported atomic 1
H = 1.008 amu
masses are not
whole numbers 16
O = 16.00 amu 6
The average atomic mass is the weighted
average of all of the naturally occurring
isotopes of the element.

7
 Average Atomic Mass

Naturally occurring lithium is:

7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)

Average atomic mass of lithium:

7.42 x 6.015 + 92.58 x 7.016

= 6.941 amu
100

8
 Average Atomic Mass: Solved Example
37 35
1. Atom X has two isotopes: 25.0 wt% is X and 75.0% is X. What is the

average molar mass of X?

A. 36.0 amu

B. 72.0 amu

C. 36.5 amu

D. 35.5 amu

E. 34.5 amu

9
Average Atomic Mass and Periodic Table

Average atomic mass (6.941)

10
Formula Mass or Molecular Mass
• The mass of an individual molecule or
formula unit
• Also known as molecular mass or
molecular weight
• Sum of the masses of the atoms in a single
molecule or formula unit
– whole = sum of the parts!
Mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
11
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.

1S 32.07 amu
2O + 2 x 16.00 amu
SO2 SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

12
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.

1Na 22.99 amu

NaCl 1Cl + 35.45 amu
NaCl 58.44 amu

For any ionic compound

formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu

13
What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08
2P 2 x 30.97
8O + 8 x 16.00
310.18 amu

14
 Molecular Mass: Solved Example
2. The molar mass (in amu) of ZnSO4.6H2O =
A.161.5
B. 179.5
C. 269.5
D. 53.5
E. 215.5

15
 Practice for yourself

• Calculate the molecular mass of the following:

– Fe2O3 (Rust)

– C6H8O7 (Citric acid)

– C16H18N2O4 (Penicillin G)
 Chapter 3 – Lecture 2

3. 2 Mole Concept: Avogadro’s Number and

Molar Mass of an Element
3. 4 The Mass Spectrometer

17
 The Mole Concept
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 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA) 18
 One Mole of:

12.01 g/mol 32.07 g/mol

C Hg S

200.6 g/mol
Cu Fe

63.55 g/mol 55.85 g/mol

19
The Utility of Mole Concept

20
 How BIG is a mole?
• 1 mole of sheets of paper?
– 1 mole of sheets of paper could form a million stacks from the surface
of the earth, all that would pass the sun.
• 1 mole of pennies?
– 1 mole of pennies could be distributed to all the currently-living
people of the world so that they could spend a million dollars per hour
every hour (day and night) for the rest of their lives.
• 1 mole of popcorn?
– 1 mole of popcorn kernels could be spread uniformly over the USA if
the thickness of the layer was about 14 km (9 mi).
• 1 mole of donut holes?
– 1 mole of donut holes would cover the earth and be 8 km (5 mi) deep.
How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K

1 mol K 6.022 x 1023 atoms K

0.551 g K x x =
39.10 g K 1 mol K

8.49 x 1021 atoms K

 Mole Concept: Solved Example
What is the mass in grams of 1.00x1012 atoms of lead, Pb?

A.3.44x10–10g
B. 2.07x10-14 g
C. 1.25x1038 g
D. 3.44x10–34 g
E. 1.25x1014g

23
 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 = 6.022 x 1023 atoms H

1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms

72.5 g C3H8O x x x =
60 g C3H8O 1 mol C3H8O 1 mol H atoms

5.82 x 1024 atoms H

24
• Example:
1. Convert 100. g of CuSO4.3H2O (Molar mass = 213.5
amu) to moles.

Ans: 100/213.5 = 0.470 moles

2. How many grams of water are present in 1.00 mole of

CuSO4.3H2O?

Ans: 54.0 g
 Mole Concept: Solved Example
Exercise:
• Which of the following has the greatest mass?
• 2 mol of H2O

• 20 g of H2O

• 6.2 x 1021 molecules of H2O

Mass Spectrometer

Heavy
Light
Light

Heavy
Mass Spectrum of Ne

27
 Chapter 3 – Lecture 3

3. 5 Percent Composition of Compounds

3. 6 Experimental Determination of
Empirical Formulas

28
• Molar Mass of a Compound: Sum of Atomic
Masses of Constituent Elements

Molar mass of C2H6O

= 2(12.01) + 6 (1.008) + 16
= 44.07 g/mol

C2H6O

29
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
%C = x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H = x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O = x 100% = 34.73%
46.07 g
C2H6O 52.14% + 13.13% + 34.73% = 100.0%
30
 Percent Composition: Solved Example
• Example: Which compound has highest percentage of
carbon?
CH4, C2H2O3, C10H22O11
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.

1 mol K
nK = 24.75 g K x = 0.6330 mol K
39.10 g K
1 mol Mn
nMn = 34.77 g Mn x = 0.6329 mol Mn
54.94 g Mn
1 mol O
nO = 40.51 g O x = 2.532 mol O
16.00 g O

32
Percent Composition and Empirical Formulas

nK = 0.6330, nMn = 0.6329, nO = 2.532

0.6330 ~
K: ~ 1.0
0.6329
0.6329
Mn : = 1.0
0.6329
2.532 ~
O: ~ 4.0
0.6329

KMnO4

33
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O

g CO2 mol CO2 mol C gC 6.00 g C = 0.500 mol C

g H2 O mol H2O mol H gH 1.51 g H = 1.50 mol H
g of O = g of sample – (g of C + g of H) 3.99 g O = 0.249 mol O

Divide by smallest subscript (0.25)

Empirical formula C2H6O
 Molecular Formula from % ages of Elements
• Determine the molecular formula of a compound that has the
following %composition by mass: K 24.75%, Mn 34.77%, and O
40.51%. The molar mass of this compound is 142 g/mol

35
 Chapter 3 – Lecture 4

3. 7 Chemical Reactions and Chemical

Equations

36
 Chemical Reactions

A process in which one or more substances is

changed into one or more new substances is a
chemical reaction

A balanced chemical equation represents the

conversion of the reactants to products such that
the number of atoms of each element is conserved.

reactants products
 Chemical Reactions

3 ways of representing the reaction of H 2 with O2 to form H2O

 Example of a Chemical Reaction

►Nickel, a solid metal, is mixed with a colorless solution of

hydrochloric acid in a test tube.
►The nickel is slowly eaten away, the colorless solution turns
green, and a colorless gas bubbles out of the test tube.
 How to “Read” Chemical Equations

2 Mg + O2 2 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 + 32.0 grams O2 makes 80.6 g MgO

NOT
2 grams Mg + 1 gram O2 makes 2 g MgO

40
 Determine # of atoms

• # of hydrogen atoms in : 4 C10H22O11

• # of oxygen atoms in : 3 CuNO3(3H2O)

• # of chlorine atoms in : 2 CH3CH2Cl

• # of oxygen atoms in : 2 CaSO4

 Chemical Equations and Physical States
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)

• CH4 and O2 are the reactants, and CO2 and H2O

are the products
• The (g) after the formulas tells us the state of the
chemical
– (g) = gas; (l) = liquid; (s) = solid
– (aq) = aqueous = dissolved in water
• The number in front of each substance tells us the
numbers of those molecules in the reaction
– called the coefficients
 Balancing Chemical Equations

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)

• This equation is balanced, meaning that there are equal
numbers of atoms of each element on the reactant and
product sides
– to obtain the number of atoms of an element, multiply
the subscript by the coefficient
1C1
4H4
4O2+2
 Balancing Chemical Equations
• When magnesium metal burns in air it produces a
white, powdery compound magnesium oxide
Mg(s) + O2(g)  MgO(s)
1) count the number of atoms on each side
– count polyatomic groups as one “element” if on
both sides
Mg(s) + O2(g)  MgO(s)
1  Mg 1
2O1
 Balancing Chemical Equations

2) pick an element to balance

– avoid element in multiple compounds
– since Mg already balanced, pick O
3) find least common multiple of both sides & multiply
each side by factor so it equals LCM
Mg(s) + O2(g)  MgO(s)

1  Mg 1
1x2O1x2
 Balancing Chemical Equations
4) use factors as coefficients in front of compound containing
the element
Mg(s) + O2(g)  2 MgO(s)

1  Mg 1
1x2O1x2
5) Recount – Mg not balanced now – That’s OK!!
Mg(s) + O2(g)  2 MgO(s)
1  Mg 2
2O2
 Balancing Chemical Equations

6) Repeat – attacking unbalanced element

2 Mg(s) + O2(g)  2 MgO(s)
2 x 1  Mg 2
2O2
Balancing Chemical Equations – A different Way

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.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O

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.
2C2H6 NOT C4H12
48
 Balancing Chemical Equations – A different Way

3. 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

2 carbon 1 carbon multiply CO2 by 2

on left on right
C2H6 + O2 2CO2 + H2O

6 hydrogen 2 hydrogen
multiply H2O by 3
on left on right
C2H6 + O2 2CO2 + 3H2O 49
 Balancing Chemical Equations – A different Way

4. Balance those elements that appear in two or

more reactants or products.
C2H6 + O2 2CO2 + 3H2O multiply O2 by 7
2

2 oxygen 4 oxygen + 3 oxygen = 7 oxygen

on left (2x2) (3x1) on right

C2H6 + 7 O2 remove fraction

2CO2 + 3H2O
2 multiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O

50
 Balancing Chemical Equations – A different Way

5. 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) 4C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)
Reactants Products
4C 4C
12 H 12 H
14 O 14 O
51
Polyatomic Ions and Balancing Chemical Reactions

• Keep polyatomic ions together if they are

present on both sides of the equation.
 Balancing Equation
A) Na + H2O → NaOH + H2 B) C 3 H8 + O 2 → CO2+ H2O

C) HNO3(aq) + Mg(OH)2(aq) -----> Mg(NO3)2(aq) + H2O(l)

D) (NH4)2CO3(aq) + MgCl2(aq) ----> NH4Cl(aq) + MgCO3(s)

53
 Balancing Equation – contd--
E) Al + HCl → AlCl 3 + H2

F) AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)

G) HCl (aq) + CaCO3 (s)  CaCl2 (aq) + CO2 (g) + H2O (l)

54
 Balancing Equation: Practice
• A) 2 Na + 2 H2O → 2 NaOH + H2
• B) C3H8 + 5 O2 → 3 CO2 + 4 H2O
• C) 2HNO3(aq) +Mg(OH)2(aq) -----> Mg(NO3)2(aq) + 2H2O(l)
• D) (NH4)2CO3(aq) + MgCl2(aq) ----> 2 NH4Cl(aq) + MgCO3(s)
• E) 2 Al + 6 HCl → 2 AlCl3 + 3 H2
• (F) AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
• (G) 2HCl (aq) + CaCO3 (s)  CaCl2 (aq) + CO2 (g) + H2O (l)

55
 Chapter 3 - Lecture 5

3. 8 Amounts of Reactants and Products:

Stoichiometry

56
 Stoichiometry
• Stoichiometry: Relates the moles of
products and reactants to each other and to
measurable quantities.
Write a balanced equation
aA +bB cC+dD
Given Given Find Find
Grams of A Moles of A Moles of B Grams of B

Determine molar Use coefficients of Determine molar

mass of A and the balanced mass of B and
then its moles equation and find then its moles
moles of B
 Making Molecules
Mole-to-Mole Conversions

• The balanced equation is the “recipe” for a chemical

reaction
• The equation 3 H2(g) + N2(g)  2 NH3(g) tells us
that 3 molecules of H2 react with exactly 1 molecule
of N2 and make exactly 2 molecules of NH3 or
3 molecules H2  1 molecule N2  2 molecules NH3
– in this reaction
• Since we count molecules by moles
3 moles H2  1 mole N2  2 moles NH3
 Mole-to-Mole Conversions
Sodium chloride, NaCl, forms by the following reaction
between sodium and chlorine. How many moles of
NaCl result from the complete reaction of 3.4 mol of
Cl2? Assume there is more than enough Na.

2 Na(s) + Cl2(g)  2 NaCl(s)

Example:
How many moles of NaCl result
from the complete reaction of
3.4 mol of Cl2 in the reaction
below?
2 Na(s) + Cl2(g)  2 NaCl(s)

• Write down the given quantity and its units.

Given: 3.4 mol Cl2
Example: Information
How many moles of NaCl Given: 3.4 mol Cl2
result from the complete
reaction of 3.4 mol of Cl2 in the
reaction below?
2 Na(s) + Cl2(g)  2 NaCl(s)

• Write down the quantity to find and/or its units.

Find: ? moles NaCl
Example: Information
How many moles of NaCl Given: 3.4 mol Cl2
result from the complete
reaction of 3.4 mol of Cl2 in the Find: ? moles NaCl
reaction below?
2 Na(s) + Cl2(g)  2 NaCl(s)

• Collect Needed Conversion Factors:

according to the equation:

1 mole Cl2  2 moles NaCl
Example: Information
How many moles of NaCl Given: 3.4 mol Cl2
result from the complete
reaction of 3.4 mol of Cl2 in the Find: ? moles NaCl
reaction below? CF: 1 mol Cl2  2 mol NaCl
2 Na(s) + Cl2(g)  2 NaCl(s)

• Write a Solution Map for converting the units :

molCl
mol Cl2 molNaCl
mol NaCl
2

2 mol N aC l
1 mol C l 2
Example: Information
How many moles of NaCl Given: 3.4 mol Cl2
result from the complete
reaction of 3.4 mol of Cl2 in the Find: ? moles NaCl
reaction below? CF: 1 mol Cl2  2 mol NaCl
2 Na(s) + Cl2(g)  2 NaCl(s) SM: mol Cl2  mol NaCl

• Apply the Solution Map:

2 mol NaCl
3.4 mol Cl 2   moles NaCl
1 mol Cl 2
= 6.8 mol NaCl
• Sig. Figs. & Round:
= 6.8 moles NaCl
Example: Information
How many moles of NaCl Given: 3.4 mol Cl2
result from the complete Find: ? moles NaCl
reaction of 3.4 mol of Cl2 in the
CF: 1 mol Cl2  2 mol NaCl
reaction below?
2 Na(s) + Cl2(g)  2 NaCl(s) SM: mol Cl2  mol NaCl

• Check the Solution:

3.4 mol Cl2  6.8 mol NaCl

The units of the answer, moles NaCl, are correct.
The magnitude of the answer makes sense
since the equation tells us you make twice as many
moles of NaCl as the moles of Cl2 .
Example:
• Sodium chloride, NaCl, forms by the following reaction between sodium and
chlorine. How many moles of NaCl result from the complete reaction of 3.4
mol of Cl2? Assume there is more than enough Na.
2 Na(s) + Cl2(g)  2 NaCl(s)
 Making Molecules
Mass-to-Mass Conversions
• we know there is a relationship between the mass and number
of moles of a chemical
1 mole = Molar Mass in grams
• the molar mass of the chemicals in the reaction and the
balanced chemical equation allow us to convert from the
amount of any chemical in the reaction to the amount of any
other
Mass-to-Mass Conversions
Example:
• In photosynthesis, plants convert carbon dioxide and water into glucose,
(C6H12O6), according to the following reaction. How many grams of glucose
can be synthesized from 58.5 g of CO2? Assume there is more than enough
water (water in excess) to react with all the CO2.
Molar Mass C6H12O6 = 180 g/mol; Molar Mass CO2 = 44.01 g/mol

sunlight
6 CO 2(g)  6 H 2 O(l)  
 6 O 2(g)  C 6 H12 O 6(aq)

grams CO2 moles CO2 Moles glucose grams glucose

Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2 in the reaction?
6 CO2(g) + 6 H2O(l) 
6 O2(g) + C6H12O6(aq)

• Write down the given quantity and its units.

Given: 58.5 g CO2
Example: Information
How many grams of glucose
Given: 55.4 g CO2
can be synthesized from 58.5 g
of CO2 in the reaction?
6 CO2(g) + 6 H2O(l) 
6 O2(g) + C6H12O6(aq)

• Write down the quantity to find and/or its units.

Find: ? g C6H12O6
Example: Information
How many grams of glucose
Given: 55.4 g CO2
can be synthesized from 58.5 g
of CO2 in the reaction? Find: g C6H12O6
6 CO2(g) + 6 H2O(l) 
6 O2(g) + C6H12O6(aq)

• Collect Needed Conversion Factors:

Molar Mass C6H12O6 = 6(mass C) + 12(mass H) + 6(mass O)
= 6(12.01) + 12(1.01) + 6(16.00) = 180.2 g/mol
Molar Mass CO2 = 1(mass C) + 2(mass O)
= 1(12.01) + 2(16.00) = 44.01 g/mol
1 mole CO2 = 44.01 g CO2
1 mole C6H12O6 = 180.2 g C6H12O6
1 mole C6H12O6  6 mol CO2 (from the chem. equation)
Example: Information
How many grams of glucose Given: 58.5 g CO2
can be synthesized from 58.5 g
Find: g C6H12O6
of CO2 in the reaction?
6 CO2(g) + 6 H2O(l)  CF: 1 mol C6H12O6 = 180.2 g
6 O2(g) + C6H12O6(aq) 1 mol CO2 = 44.01 g
1 mol C6H12O6  6 mol CO2
• Write a Solution Map:

g mol mol g
CO2 CO2 C6H12O6 C6H12O6
1 mol CO2 1 mol C 6 H12O 6 180.2 g C 6 H12O 6
44.01 g CO2 6 mol CO2 1 mol C 6 H 12O 6
Example: Information
How many grams of glucose Given: 58.5 g CO2
can be synthesized from 58.5 g Find: g C6H12O6
of CO2 in the reaction? CF: 1 mol C6H12O6 = 180.2 g
6 CO2(g) + 6 H2O(l)  1 mol CO2 = 44.01 g
6 O2(g) + C6H12O6(aq) 1 mol C6H12O6  6 mol CO2
SM: g CO2  mol CO2 
mol C6H12O6  g C6H12O6
• Apply the Solution Map:
1 mo le C O 2 1 mo l C 6 H 1 2 O 6 1 8 0 .2 g C 6 H 1 2 O 6
5 8 .5 g C O 2   
4 4 .0 1 g C O 2 6 mo l C O 2 1 mo le C 6 H 1 2 O 6

= 39.9216 g C6H12O6
• Sig. Figs. & Round:
= 39.9 g C6H12O6
Example: Information
How many grams of glucose Given: 58.5 g CO2
can be synthesized from 58.5 g Find: g C6H12O6
of CO2 in the reaction? CF: 1 mol C6H12O6 = 180.2 g
6 CO2(g) + 6 H2O(l)  1 mol CO2 = 44.01 g
6 O2(g) + C6H12O6(aq) 1 mol C6H12O6  6 mol CO2
SM: g CO2  mol CO2 
mol C6H12O6  g C6H12O6
• Check the Solution:
58.5 g CO2 = 39.9 g C6H12O6
The units of the answer, g C6H12O6 , are correct.
It is hard to judge the magnitude.
 Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what
mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar mass coefficients molar mass
CH3OH chemical equation H2O

1 mol CH3OH 4 mol H2O 18.0 g H2O

209 g CH3OH x x x =
32.0 g CH3OH 2 mol CH3OH 1 mol H2O

235 g H2O

76
 Chapter 3 - Lecture 6

3. 9 Limiting Reagent Calculations

3. 10 Reaction Yield

77
Limiting Reagent:
Reactant used up first in
the reaction.

2NO + O2 2NO2

NO is the limiting reagent

O2 is the excess reagent

78
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe MM = 160. g Fe2O3
MM = 102. g Al2O3
Calculate the mass of Al2O3 formed. Molar atomic mass of
Al = 27
Method 1
a. Determine how many grams of Fe2O3 reacts with 124 g of Al

b. Determine which reactant is smaller quantity and which is in excess

c. Using the smallest of the two determine how much of Al2O3 is formed

Method 2
a. Determine how many grams of Al2O3 is formed from 601 g of Fe2O3 and 124
g of Al individually.

b. Determine which reactant gives the smaller quantity of Al2O3 and that would
be the limiting reagent.

79
 In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe MM = 160. g Fe2O3
MM = 102. g Al2O3
Calculate the mass of Al2O3 formed. Molar atomic mass of
Al = 27

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

OR
g Fe2O3 mol Fe2O3 mol Al needed g Al needed

1 mol Al 1 mol Fe2O3 160. g Fe2O3

124 g Al x x x = 367 g Fe2O3
27.0 g Al 2 mol Al 1 mol Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent 80
 Use limiting reagent (Al) to calculate amount of product that
can be formed.

g Al mol Al mol Al2O3 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

1 mol Al 1 mol Al2O3 102. g Al2O3

124 g Al x x x = 234 g Al2O3
27.0 g Al 2 mol Al 1 mol Al2O3

At this point, all the Al is consumed

and Fe2O3 remains in excess.

81
In one process, 124 g of Al are reacted with 601 g of Fe2O3
MM = 160. g Fe O
2Al + Fe2O3 Al2O3 + 2Fe MM = 102. g Al O
2

2
3

3
Molar atomic mass of
Calculate the mass of Al2O3 formed. Al = 27

82
 In one process, 124 g of Al are reacted with 601 g of Fe2O3
MM = 160. g Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe MM = 102. g Al2O3
Molar atomic mass of
Al = 27
Determine excess reagent and how much of the excess
reagent is remaining
g Al mol Al mol Fe2O3 needed g Fe2O3 needed

1 mol Al 1 mol Fe2O3 160. g Fe2O3

124 g Al x x x = 367 g Fe2O3
27.0 g Al 2 mol Al 1 mol Fe2O3

Start with 124 g Al need 367 g Fe2O3

Excess reagent is Fe2O3
Fe2O3 Unreacted = (601-367) g = 234 g
83
Finding Limiting Reactant,
Theoretical Yield and
Percent Yield
 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.

Actual Yield
% Yield = x 100%
Theoretical Yield

85
 Example: When 11.5 g of C are allowed to react with 114.5 g of Cu 2O in the
reaction below, 87.4 g of Cu are obtained. Find the Limiting Reactant,
Theoretical Yield and Percent Yield. Molar Mass Cu O = 143.02 g/mol
2
Molar Mass Cu = 63.54 g/mol
Cu 2O(s)  C(s)  2 Cu(s)  CO(g) Molar Mass C = 12.01 g/mol

C = 122 g
Cu = 101.7
Example:
When 11.5 g of C reacts with 114.5 g
of Cu2O, 87.4 g of Cu are obtained.
Find the Limiting Reactant,
Theoretical Yield and Percent Yield.
Cu2O(s) + C(s)  2 Cu(s) + CO(g)

• Write down the given quantity and its units.

Given: 11.5 g C
114.5 g Cu2O
87.4 g Cu produced
Example: Information
When 11.5 g of C reacts with 114.5 g Given: 11.5 g C, 114.5 g
of Cu2O, 87.4 g of Cu are obtained.
Cu2O
Find the Limiting Reactant,
Theoretical Yield and Percent Yield. 87.4 g Cu produced
Cu2O(s) + C(s)  2 Cu(s) + CO(g)

• Write down the quantity to find and/or its units.

Find: limiting reactant
theoretical yield
percent yield
Example: Information
When 11.5 g of C reacts with 114.5 g Given: 11.5 g C, 114.5 g Cu2O
of Cu2O, 87.4 g of Cu are obtained. 87.4 g Cu produced
Find the Limiting Reactant, Find: Lim. Rct., Theor. Yld., % Yld.
Theoretical Yield and Percent Yield.
Cu2O(s) + C(s)  2 Cu(s) + CO(g)

• Collect Needed Conversion Factors:

Molar Mass Cu2O = 143.02 g/mol
Molar Mass Cu = 63.54 g/mol
Molar Mass C = 12.01 g/mol
1 mole Cu2O  2 mol Cu (from the chem. equation)
1 mole C  2 mol Cu (from the chem. equation)
Example: Information
When 11.5 g of C reacts with 114.5 g Given: 11.5 g C, 114.5 g Cu2O
of Cu2O, 87.4 g of Cu are obtained. 87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
Find the Limiting Reactant,
CF: 1 mol C = 12.01 g; 1 mol Cu =
Theoretical Yield and Percent Yield. 63.54 g; 1 mol Cu2O = 143.08 g;
Cu2O(s) + C(s)  2 Cu(s) + CO(g) 1 mol Cu2O  2 mol Cu; 1 mol
C  2 mol Cu

• Write a Solution Map:

}
g mol mol g
C C Cu Cu smallest
1 mol C 2 mol Cu 63.54 g Cu amount is
12.01 g C 1 mol C 1 mol Cu theoretical
g mol mol g yield
Cu2O Cu2O 2 mol Cu Cu 63.54 g Cu Cu
1 mol Cu 2O
143.08 g Cu 2O 1 mol Cu 2O 1 mol Cu
Example: Information
When 11.5 g of C reacts with Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
114.5 g of Cu2O, 87.4 g of Cu
Find: Lim. Rct., Theor. Yld., % Yld.
are obtained. Find the Limiting CF: 1 mol C = 12.01 g; 1 mol Cu =
Reactant, Theoretical Yield and 63.54 g; 1 mol Cu2O = 143.08 g;
Percent Yield. 1 mol Cu2O  2 mol Cu; 1 mol
C  2 mol Cu
Cu2O(s) + C(s)  2 Cu(s) +
SM: g rct  mol rct  mol Cu  g
CO(g) Cu
• Apply the Solution Map:
1 mo le C 2 mo l C u 6 3 .5 4 g C u
1 1 .5 g C     122 g C u
1 2 .0 1 g C 1 mo l C 1 mo le C u
1 mo le C u 2 O 2 mo l C u 6 3 .5 4 g C u
1 1 4 .5 g C u 2 O     1 0 1 .7 g C u
1 4 3 .0 8 g C u 2 O 1 mo l C u 2 O 1 mo le C u
Example: Information
When 11.5 g of C reacts with Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
114.5 g of Cu2O, 87.4 g of Cu
Find: Lim. Rct., Theor. Yld., % Yld.
are obtained. Find the Limiting CF: 1 mol C = 12.01 g; 1 mol Cu =
Reactant, Theoretical Yield and 63.54 g; 1 mol Cu2O = 143.08 g;
Percent Yield. 1 mol Cu2O  2 mol Cu; 1 mol
C  2 mol Cu
Cu2O(s) + C(s)  2 Cu(s) +
SM: g rct  mol rct  mol Cu  g
CO(g) Cu
• Apply the Solution Map: More Amount
11.5 g C can make 122 g Cu
114.5 g Cu2O can make 101.7 g Cu Limiting Reactant = Cu2O

Theoretical Yield

Least Amount
Example: Information
When 11.5 g of C reacts with 114.5 g Given: 11.5 g C, 114.5 g Cu2O
of Cu2O, 87.4 g of Cu are obtained. 87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
Find the Limiting Reactant,
CF: 1 mol C = 12.01 g; 1 mol Cu =
Theoretical Yield and Percent Yield. 63.54 g; 1 mol Cu2O = 143.08 g;
Cu2O(s) + C(s)  2 Cu(s) + CO(g) 1 mol Cu2O  2 mol Cu; 1 mol
C  2 mol Cu

• Write a Solution Map:

Actual Yield
100%  Percent Yield
Theoretical Yield
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu
are obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s)  2 Cu(s) +
CO(g)
• Apply the Solution Map:

Actual Yield
 100%  Percent Yield
Theoretical Yield
87.4 g Cu
100%  85.9%
101.7 g Cu
Example: Information
When 11.5 g of C reacts with Given: 11.5 g C, 114.5 g Cu2O
114.5 g of Cu2O, 87.4 g of Cu 87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., %
are obtained. Find the Limiting Yld.
Reactant, Theoretical Yield and CF: 1 mol C = 12.01 g; 1 mol Cu =
Percent Yield. 63.54 g; 1 mol Cu2O = 143.08
Cu2O(s) + C(s)  2 Cu(s) + g; 1 mol Cu2O  2 mol Cu;
1 mol C  2 mol Cu
CO(g)
• Check the Solutions:
Limiting Reactant = Cu2O
Theoretical Yield = 101.7 g
Percent Yield = 85.9%
The Percent Yield makes sense as it is less than 100%.
• Limiting Reagent Calculation: Cisplatin is an anti-
cancer agent prepared as follows:

K2PtCl4 + 2 NH3  Pt(NH3)2Cl2 + 2 KCl

• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to

react: (a) which is the limiting reagent? (b) How many
grams of the excess reagent are consumed? (c) How
many grams of cisplatin are formed? Show your work.

MM(K2PtCl4) = 415.08 g/molMM(NH3) = 18.04 g/mol

• Limiting Reagent Calculation: Cisplatin is an anti-cancer agent prepared as follows:

K2PtCl4 + 2 NH3  Pt(NH3)2Cl2 + 2 KCl

• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react: (a) which is the limiting reagent?
(b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin
are formed? Show your work.
MM(K2PtCl4) = 415.08 g/mol MM(NH3) = 18.04 g/mol

97
What is the limiting reagent when 27.0 g of P and 68.0 g of I2
react according to the following chemical equation?
2P(s) + 3I2(s)  2PI3(s)
Ans: I2

98
Determine the number of moles of water produced by the
reaction of 155 g of ammonia and 356 g of oxygen.
4NH3 + 5O2  4NO + 6H2O
Ans: 13.4 moles

99
What is the theoretical yield of PI3 from the reaction of 27.0 g of P
and 68.0 g of I2?
2P(s) + 3I2(s)  2PI3(s)
Ans: 73.5 g

100