Chapter 9
Stoichiometry
9.2
chemical calculations
Things you will learn
• You will be able to construct mole ratios from
balanced chemical equations and be able to
apply these in mole-mole calculations.
• You will be able to calculate stoichiometric
quantities (mass, volume, representative
particles, and moles) from balanced
equations.
Baking soda experiment
• Baking soda decomposes with heat to form
sodium carbonate, carbon dioxide, and water
• The skeleton equation is:
• NaHCO3
Na2CO3 + CO2 + H2O
Baking soda experiment
• Baking soda decomposes with heat to form
sodium carbonate, carbon dioxide, and water
• The balanced equation is:
• 2NaHCO3
Na2CO3 + CO2 + H2O
2 Na 2
2C2
6O6
2H2
What we want to find out is how much baking
soda will we have to heat in order to fill a liter
Coke bottle half-full.
Since we have a balanced equation, we can use
the molar road map to find out just how much
we should use.
We will start by putting the given quantity on
the left and the units of what we want to find
on the right, and filling in the middle with
conversion factors.
Given quantity
Mole-mole ratio from
balanced equation
Wanted quantity
Given quantity
.5 L CO2
1 mol CO2
22.4 L CO2
Ratio from balanced equation
2 mol NaHCO3
1 mol CO2
_ g NaHCO3
3.75 g NaHCO3
84 g NaHCO3
1 mol NaHCO3
Wanted quantity
So if we put 3.75 grams of NaHCO3 in a
test tube and heated it, we should get ½
liter of CO2 given off.
What else is given off that might mess up
our calculations?
2NaHCO3
Na2CO3 + CO2 + H2O
So if we put 3.75 grams of NaHCO3 in a
test tube and heated it, we should
generate ½ liter of CO2.
What else is given off that might mess up
our calculations?
2NaHCO3
Na2CO3 + CO2 + H2O
These are solids
These are BOTH gases!
Mole-mole calculations
• 4Al(s) + 3O2(g)
2Al2O3(s)
• Using this balanced equation, what are the
possible mole ratios we can use as conversion
factors?
• Remember, conversion factors are always
equal to 1 and so may be used upside down,
with the numerator in the denominator, and
vice versa
Mole-mole calculations
• 4Al(s) + 3O2(g)
2Al2O3(s)
• Using this balanced equation, what are the
possible mole ratios we can use as conversion
factors?
4 moles Al
3 moles O2
3 moles O2
2 moles Al2O3
4 moles Al
2 moles Al2O3
3 moles O2
4 moles Al
2 moles Al2O3
3 moles O2
2 moles Al2O3
4 moles Al
We use mole-mole conversions (ratios from
balanced equations) to solve problems like:
How many moles of oxygen are required to react
completely with 14.8 mole of aluminum?
We always work from our known (given)
quantity towards our unknown (wanted)
quantity.
14.8 mol Al
given
3 mol O2
4 mol Al
mole-mole
ratio
_ mol O2
unknown
Mole-mole calculations
• 4Al(s) + 3O2(g)
2Al2O3(s)
• How many moles of Al2O3(s) would we get if
we only used 1 mole of Al(s)?
1 mol Al
given
2 mol Al2O3
4 mol Al
mole-mole
ratio
_ mol Al2O3
unknown
Mole-mole calculations
• 4Al(s) + 3O2(g)
2Al2O3(s)
• How many moles of Al2O3(s) would we get if
we only used 1 mole of O2(g)?
1 mol O2
given
2 mol Al2O3
3 mol O2
mole
ratio
_ mol Al2O3
unknown
Mole-mole calculations
• 4Al(s) + 3O2(g)
2Al2O3(s)
• How many moles of O2(g) would it take to
make 5 moles of Al2O3(s) ?
5 mol Al2O3
given
3 mol O2
2 mol Al2O3
mole
ratio
_ mol O2
unknown
Mole-mole calculations
• 4Al(s) + 3O2(g)
2Al2O3(s)
• How many moles of Al(s) would it take to make
5 moles of Al2O3(s) ?
14.8 mol Al
3 mol O2
4 mol Al
given
mole
ratio
_ mol O2
unknown
Mass-mass calculations
• The number of moles of a reactant or product
can be gotten directly from the coefficients of
a balanced equation.
• The amount in grams must be gotten from the
molar mass of each. This adds a couple more
conversion factors, but they’re easy to deal
with.
Calculate the number of grams of ammonia
produced from the reaction of 5.4 grams of
hydrogen with an excess of nitrogen
• N2(g) + H2(g)
NH3(g) (skeleton equation)
Calculate the number of grams of ammonia
produced from the reaction of 5.4 grams of
hydrogen with an excess of nitrogen
• N2(g) + 3H2(g)
2NH3(g) (balanced equation)
Calculate the number of grams of ammonia
produced from the reaction of 5.4 grams of
hydrogen with an excess of nitrogen
• N2(g) + 3H2(g)
2NH3(g)
5.4 g H2
given
Gram-mole
conversion
Mole-mole
conversion
_ g NH3
Gram-mole
conversion
unknown
Calculate the number of grams of ammonia
produced from the reaction of 5.4 grams of
hydrogen with an excess of nitrogen
• N2(g) + 3H2(g)
5.4 g H2
given
1 mol H2
2 g H2
Gram-mole
conversion
2NH3(g)
2 mol NH3
3 mol H2
Mole-mole
conversion
_ g NH3
17 g NH3
1 mol NH3
Gram-mole
conversion
unknown
Calculate the number of grams of ammonia
produced from the reaction of 5.4 grams of
hydrogen with an excess of nitrogen
• N2(g) + 3H2(g)
5.4 g H2
given
1 mol H2
2 g H2
Gram-mole
conversion
2NH3(g)
2 mol NH3
3 mol H2
Mole-mole
conversion
30.6 g NH3
17 g NH3
1 mol NH3
Gram-mole
conversion
unknown
Mole-mole conversions
Ammonia burns in oxygen forming nitrogen
oxide and water. The skeleton formula is:
NH3(g) + O2(g)
NO(g) + H2O(g)
Balance this equation and determine how
many moles of ammonia are consumed if
1.35 moles of oxygen are used.
4NH3(g) + 5O2(g)
1.35 mol O2
4NO(g) + 6H2O(g)
4N4
12 H 12
10 O 10
4 mol NH3
5 mol O2
_mol NH3
Mass-mole conversions
Phosphorus reacts with chlorine to form phosphorus
pentachloride. The skeleton formula is:
P(s) + Cl2(g)
PCl5(s)
Balance this equation and determine how
many moles of chlorine are consumed if
5.50 grams of PCl5 are produced.
2P(s) + 5Cl2(g)
2PCl5(s)
2P2
10 Cl 10
5.50 g PCl5
1 mol PCl5
208 g PCl5
5 mol Cl2
2 mol PCl5
_mol Cl2