Performing a
Mass-Mass Stoichiometry
Calculation
A step-by-step tutorial
Question Statement:
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
iron + oxygen  iron (III) oxide
• Step 1
Stoichiometry problems always begin with a
balanced equation. In this problem, we will
need to first write formulas for the
substances above and then determine the
coefficients that balance the equation.
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
The above equation now shows the correct
formulas for the reactants and products and
is also balanced with the correct
coefficients.
The coefficients represent a mole ratio of
4 : 3 : 2 in this reaction.
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
? g
• Step 2
Next, we will convert the starting mass of
15.0 g of Fe into moles.
Here is what we write:
15.0 g

1m ol F e
Fe 
 55.845 g F e





What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
? g
• Step 3
Now, we are ready to use the mole ratio. The
mole ratio converts the substance of known
mass into the substance of unknown mass.
Here is what we write:
15.0 g

1m ol F e
Fe
 55.845 g F e



 

2 m ol F e2 O 3 
4 m ol F e



What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
? g
15.0 g

1m ol F e
Fe
 55.845 g F e



 

2 m ol F e2 O 3 
4 m ol F e



Notice how the ratio in the balanced equation
is transferred into the calculation.
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
15.0 g
? g
Fe

 1 m ol

 55.845

Fe
g Fe





2 m ol F e2 O 3 
4 m ol F e



The ratio is written with the moles of Fe in the
denominator so that the unit label of
“moles Fe” will cancel out.
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
? g
• Step 4
Finally, we convert the moles of iron(III)oxide
into grams using its known molar mass.
Here is what we write:
15.0 g

1m ol F e
Fe
 55.845 g F e



 

2 m ol F e2 O 3   159.866 g F e2 O 3 
4 m ol F e



1m ol F e2 O 3



What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
15.0 g
Fe
? g

 1 m ol

 55.845

Fe
g Fe



 

2 m ol F e2 O3
4 m ol F e

  159.866 g F e O
2 3

 
1 m ol F e2 O3






Once again, the units are placed into the
calculation so that they will cancel out and
leave us with the units we want in the end.
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?
4 Fe + 3 O2  2 Fe2O3
15.0 g
15.0 g
Fe

 1 m ol

 55.845

? g
Fe
g Fe



 

2 m ol F e2 O3
4 m ol F e

  159.866 g F e O
2 3

 
1 m ol F e2 O3






 21.4 g F e2 O3
Performing the calculation results in the mass
of iron(III)oxide that answers the question.
What mass of iron(III)oxide will be produced when 15.0 grams
of iron are reacted with excess oxygen gas?