Stoichiometry
In chemistry, stoichiometry focuses on the quantities of substances that enter into and are produced by
chemical reactions. It essentially provides the quantitative relationship between reactants and products in a
chemical reaction. These reactant and product proportions will be determined from chemical formulas,
equations, atomic weights, and molar masses.
Stoichiometry problems are typically solved using the following strategy:
Use Quantity A → Convert to moles of A → Convert to moles of B → Find Quantity B
Example: 4 NH3 + 3 O2
2 N2 + 6H2O
Given that you started with 51.0 g of NH3, how many grams of water will be produced?
This result of this problem is a calculation of theoretical yield, or the maximum amount of product that can
be produced from a specific amount of reactant.
Step 1 - Calculating Molar Mass:
We have to calculate the molar mass of substance A (in our example, NH3) to be able to convert from grams
to moles. Molar mass is defined as the mass of a given substance divided by its amount of substance.
Typically molar mass is expressed in units of grams per mole (g/mol). To calculate molar mass: add together
the atomic weights for each element in the molecule. Refer to the periodic table for accurate values of atomic
weight. If an element in the molecule has a subscript beside its chemical symbol, multiply that specific
atomic weight by that subscript value.
For NH3: 1 N = 14.007 g/mol x 1 = 14.007 g/mol
3 H = 1.008 g/mol x 3 = 3.024 g/mol
14.007 g/mol + 3.024 g/mol = 17.031 g/mol
So the total molar mass for NH3 is
17.031 g/mol
Step 2 - Converting from grams to moles:
We can now use the calculated molar mass to convert from grams to moles of NH3. Use the method of
dimensional analysis to convert these quantities. First start with the value given in the problem.
51.0 g NH3
1
X
1 mole NH3
17.031 g NH3 = ?
51.0 g NH3
1
X
1 mole NH3
17.031 g NH3 = 2.99 moles NH3
Step 3 - Converting moles of A to moles of B:
The previous calculation resulted in moles of A (NH3), which we can now use to find the moles of B (H2O)
produced. This conversion of moles will use stoichiometric coefficients from the balanced equation in the
problem.
2.99 moles NH3
1
X
6 moles H2O
4 moles NH3 = ?
2.99 moles NH3
1
X
6 moles H2O
4 moles NH3 = 4.485 moles H2O
Step 4 - Converting from moles to grams:
The last step involves converting the moles of product (H2O) formed to grams by using the product’s molar
mass. H2O’s molar mass is 18.015 g/mol.
4.485 moles H2O
18.015 g H2O
1
X 1 mole H2O = ?
4.485 moles H2O
18.015 g H2O
1
X 1 mole H2O = 80.80 g H2O