Moles and Stoichiometry Review
Most of the things we see around us are far too large to measure with atomic mass units,
including the many compounds and substances we deal with in our daily lives. So instead of
amu, we must measure mass with something larger.
Mole: unit of measurement used to convert from atomic mass units (amu) to grams (g)
→ Discovered by Amedeo Avogadro
,
His number, NA, is 6.02·1023 which is the number of particles in exactly one mole of a pure
substance.
1 molC=12.01 gC
FOR ALL ELEMENTS: atomic mass (amu) will always equal the molar mass (g)
Molar mass: the number of grams in one mole of any substance

Total atomic mass of a molecule (measured in g/mol)
Molar mass of water:
2H= 2(1.01)=2.02g/mol
O=16.00g/mol
18.02g/mol
Mass to Mole Conversions

Molar Mass:
Mass of substance X (1 mol substance/molar mass)= number of moles
Number of moles substance X (molar mass/1 mol substance)= mass of substance

Avogadro’s Number:
Number of atoms of substance X (1 mol substance/6.02·1023 atoms substance)= number of
moles of substance
Number of moles of substance X (6.02·1023 atoms substance/1 mol)= number of atoms of
substance

Molar Volume (22.4L/mol)
Number of moles of substance X (22.4 L substance/1 mol substance)= Volume of substance
(L)
Volume (L) of substance X (1 mol substance/22.4 L substance)= Number of moles of
substance

STOICHIOMETRY
Steps to calculate mass changes in chemical reactions:
1) Write the equation and balance it
2) Convert the amounts of the known substances into moles
3) Calculate the wanted quantity using the coefficients of the balanced chemical equation
4) Convert the moles of the wanted quantity into the desired units.
Sometimes during experiments we measure out more of a substance than we need, so often we
must remember the limiting and excess reactants.
Limiting reactant: the reactant that gives a limit to the amount of the product formed from the
reaction
Excess reactant: the substance that remains after the chemical reaction

How to determine a limiting/excess reactant:
1) Balance the equation
2) Choose one of the reactants and convert grams of one reactant into grams of the other
reactant.
3) Compare the amounts calculates with the amount in the experiment. If the number is
smaller, the second reactant is in excess, but is limiting of the number is larger.
Even when we measure out the reactants properly, the product won’t always form in the
expected amount. To see how much of the product was formed in respect to how much there
should be, we must calculate the percent yield.
Theoretical yield: the amount of the product that would result if all of the limiting reactant was
used in the reaction
Actual yield: the amount of the product that was formed in the experiment
Percent yield= (actual yield/theoretical yield) x 100%
Practice Problems
1) What is the total number of atoms in 9.45 moles of ammonia?
2) If 4.20 mol Al was mixed with 1.75 mol Fe₂O₃, which would be the limiting reactant? How
much iron can be formed from this reaction?
3) From the reaction: B₂H₆ + 6O₂  2HBO₂ + 2H₂O, what mass of O₂ will react completely with
57.8 g of B₂H₆?
4) How many grams of carbon dioxide are there in a container with a volume of 4.50L?
5) For the balanced equation shown below, if the reaction of 20.7 grams of CaCO₃ produces
6.81 grams of CaO, what is the percent yield?
6) How many moles of chlorine must react with 46 g of sodium to form sodium chloride?
7) A certain amount of glucose, C₆H₁₂O₆ burned in oxygen, and produced 22 grams of carbon
dioxide, CO₂. How many grams of water were produced at the same time?
Practice Problems Answers
1) What is the total number of atoms in 9.45 moles of ammonia?
Formula: NH₃
9.45 mol NH₃ x 6.02·1023 atoms NH₃/1 mol NH₃=5.69·1024 atoms NH₃
2) If 4.20 mol Al was mixed with 1.75 mol Fe₂O₃, which would be the limiting reactant? How
much iron can be formed from this reaction?
Reaction: 2Al + Fe₂O₃  Al₂O₃ + 2Fe
4.20 mol Al x 2 mol Fe/2 mol Al =4.20 mol Fe
1.75 mol Fe₂O₃ x 2 mol Fe/1 mol Fe₂O₃=3.50 mol Fe- Limiting reactant
3.50 mol Fe x 55.85 g Fe/1 mol Fe= 195 g Fe
3) From the reaction: B₂H₆ + 6O₂  2HBO₂ + 2H₂O, what mass of O₂ will react completely with
57.8 g of B₂H₆?
57.8 g B₂H₆ x 1 mol B₂H₆/27.68 g B₂H₆ x 6 mol O₂/1 mol B₂H₆ x 32.00 g O₂/1 mol O₂=401 g
O₂
4) How many grams of carbon dioxide are there in a container with a volume of 4.50L?
4.50 L CO₂ x 1 mol CO₂/22.4 L CO₂ x 44.01 g CO₂/1 mol CO₂= 8.84 g CO₂
5) For the balanced equation shown below, if the reaction of 20.7 grams of CaCO₃ produces
6.81 grams of CaO, what is the percent yield?
CaCO₃  CaO+CO₂
20.7 g CaCO₃ x 1 mol CaCO₃/100.09 g CaCO₃ x 1 mol CaO/1 mol CaCO₃ x 56.08 g CaO/1
mol CaO= 11.6 g CaO
6.81 g CaO/11.6 g CaO x 100%= 58.72%
6) How many moles of chlorine must react with 46 g of sodium to form sodium chloride?
2Na + Cl₂  2NaCl
46 g Na x 1 mol Na/1.01 g Na x 2 mol NaCl/2 mol Na x 36.46 g NaCl/1 mol NaCl= 1660.55 g
NaCl
1660.55 g NaCl x 1 mol NaCl/36.46 g NaCl x 1 mol Cl₂/2 mol NaCl= 22.77 mol Cl₂
7) A certain amount of glucose, C₆H₁₂O₆ burned in oxygen, and produced 22 grams of carbon
dioxide, CO₂. How many grams of water were produced at the same time?
C₆H₁₂O₆ + 6O₂  6CO₂ + 6H₂O
22 g CO₂ x 1 mol CO₂/44.01 g CO₂ x 6 mol H₂O/6 mol CO₂ x 18.02 g H₂O/1 mol H₂O= 9.0 g
H₂O