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Stoichiometric Calculations
The coefficients in the balanced equation give the
ratio of moles of reactants and products
Stoichiometric Calculations
From the mass of
Substance A you can
use the ratio of the
coefficients of A and B
to calculate the mass of
Substance B formed (if
it’s a product) or used
(if it’s a reactant)
Stoichiometric Calculations
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Sample #1
• CH4 +2O2 2H2O + CO2
How many moles of CH4 are needed to make 13
moles of water?
Sample #2
• CH4 +2O2 2H2O + CO2
• How many moles of water are made from 10g
of Oxygen?
Sample #3
CH4 +2O2 2H2O + CO2
• How many grams of water are produced from
100 moles of CH4?
Sample #4
CH4 +2O2 2H2O + CO2
• How many grams of Oxygen are needed in
order to produce 200 g of carbon dioxide?
Limiting Reactants
The limiting reactant
is the reactant present
in the smallest
stoichiometric amount
Limiting Reactants
• The limiting reactant is the reactant present in the
smallest stoichiometric amount
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2)
Limiting Reactants
In the example below, the O2 would be the excess
reagent
Theoretical Yield
• The theoretical yield is the amount of product
that can be made
– In other words it’s the amount of product possible
as calculated through the stoichiometry problem
• This is different from the actual yield, the
amount one actually produces and measures
Percent Yield
A comparison of the amount actually
obtained to the amount it was possible to
make
Actual Yield
Theoretical Yield
Percent Yield = x 100

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Stoichiometric calculations

  • 1. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products
  • 2. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)
  • 3. Stoichiometric Calculations Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams C6H12O6 + 6 O2  6 CO2 + 6 H2O
  • 4. Sample #1 • CH4 +2O2 2H2O + CO2 How many moles of CH4 are needed to make 13 moles of water?
  • 5. Sample #2 • CH4 +2O2 2H2O + CO2 • How many moles of water are made from 10g of Oxygen?
  • 6. Sample #3 CH4 +2O2 2H2O + CO2 • How many grams of water are produced from 100 moles of CH4?
  • 7. Sample #4 CH4 +2O2 2H2O + CO2 • How many grams of Oxygen are needed in order to produce 200 g of carbon dioxide?
  • 8. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount
  • 9. Limiting Reactants • The limiting reactant is the reactant present in the smallest stoichiometric amount – In other words, it’s the reactant you’ll run out of first (in this case, the H2)
  • 10. Limiting Reactants In the example below, the O2 would be the excess reagent
  • 11. Theoretical Yield • The theoretical yield is the amount of product that can be made – In other words it’s the amount of product possible as calculated through the stoichiometry problem • This is different from the actual yield, the amount one actually produces and measures
  • 12. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield = x 100