Unit 7 Stoichiom2etry Mass to Mass Conversions
Step 1
grams
of A




molar
mass of A
Step 2
moles
of A
Step 3
moles
of B
mole ratio
grams of
B
molar
mass of B
The mole ratio comes from the coefficients in the balanced chemical equation
We use the mole ratio to convert between substances
Molar mass comes from the periodic table
If you are going from grams to mol or mol to grams we attach the infamous 1 to mol.
How you set things up is the same way as we learned earlier this year
Last class we converted from moles of A to moles of B. Now we are going to add extra steps that you guys
know how to from the previous semester.
1. Sodium metal burns in air according to the balanced equation shown below.
4 Na(s) + O2 (g)  2 Na2O(g)
A) How many moles of oxygen are needed to completely react with 9.5g of sodium?
Define substance A(given):
Define Substance B:
Mole ratio of A:B =
Molar mass of A:
Map: g of Na 
g of Na
mol of Na
1 mol of Na
= .10 mol of O2
22.990
g of Na
1

mol of O2
1 mol of O2
4 mol of Na
:
4 Na(s) + O2 (g)  2 Na2O(g)
How many grams of sodium are needed to produce 12.5 g of sodium oxide assuming there is
an abundance of oxygen?
Na2O Define substance B: Na
Define substance A(given):
Mole Ratio A: B = 2:4
Molar Mass A: 61.979 g
Molar Mass B: 22.990 g
Map: g of Na2O 
12. 5g of Na2O
mols of Na2O 
mols of Na
1 mol Na2O
g of Na2O 61.979
4 mols of Na
2 mols of Na2O

g of Na
22.990 g of Na
1 mol of Na
Answer : 9.27 g of Na
B) For letter B, Which steps from the diagram at the top of this page did we complete?
Can you have a problem that asks you to convert from- (you are just looking at what you are passing through)
1.
Moles of A to grams of A?
Convert using: Molar mass of A
2.
Grams of A to moles of A?
Convert using: Molar mass of A
3.
Moles of A to moles of B?
Convert using: Mole Ratio
4.
Grams of A to moles of B?
Convert using: molar mass of A , mole ratio
5.
Moles of A to grams of B?
Convert using: Mole ratio, Molar mass of B
6.
Grams of A to Grams of B?
Convert using: molar mass of A, mole ratio, molar mass of B
7.
2
8.
Given the following equation:
A) How many grams of lithium nitrate will be needed to make 250.0g of lithium sulfate, assuming you have an
adequate amount of lead IV sulfate to do the reaction?
Define substance A:
Li2SO4
Molar Masses of A and B:
Define Substance B:
LiNO3
Li2SO4=
Mole ratio of A:B = 2:4
LiNO3=
SETUP AND SOLVE:
g of Li2SO4

250 g Li2SO4
mol Li2SO4

mol LiNO3

g of LiNO3
1 mol Li2SO4
4 mol LiNO3
68.945
g of LiNO3
109.944 g Li2SO4
2 mol Li2SO4
1 mol LiNO3
Answer: 313 g of LiNO3
B) If 120.00g of lithium nitrate are reacted with excess lead IV sulfate, what is the theoretical yield of lead IV nitrate in
grams?
Define substance A: LiNO3
Define Substance B:Pb(NO3)4
Mole ratio of A:B 4:1
Molar Masses of A and B:
SETUP AND SOLVE:
120 g of LiNO3
Answer: 198 g
3
1 mol LiNO3
1 mol Pb(NO3)4
455.216 g
Pb(NO3)4
68.945
g LiNO3
4 mol LiNO3
1 mol Pb(NO3)4
4
Given the following:
If this reaction with 35.0g of C6H10 how may grams of carbon dioxide will be formed?
Define substance A:
C6H10
Define Substance B:
CO2
Mole ratio of A:B = 2:12
Molar Masses of A and B:
SETUP AND SOLVE: Answer :
35 g of C6H10
5
113 g of CO2
1 mol of C6H10
12 mols of CO2
44.009 g of CO2
82.146 g of C6H10
2 mols of C6H10
1 mol of CO2
6