Stoichiometry of Chemical Reactions

Converting from Mass of Substance A to Moles of Substance B

The mole ratio from a balanced chemical equation must be used when it is necessary to convert the mass of one substance to the moles of another substance within a chemical reaction.

Complete these steps to convert the mass of substance A to moles of substance B.

1. Write the balanced chemical equation for the reaction.
2. Convert the mass of substance A to moles of substance A, using its molar mass.
3. Determine the mole ratio between substance A and substance B.
4. Use dimensional analysis to calculate the moles of substance B to the corresponding moles of substance A.

Step-By-Step Example
Converting the Mass of One Substance to the Moles of Another Substance
Determine the number of moles of oxygen gas (O2) required to burn with 60.0 grams of butane (C4H10).
Step 1
Write the balanced equation.
2C4H10+13O28CO2+10H2O2{\rm C}_4{\rm H}_{10}+13{\rm O}_2\rightarrow8{\rm{CO}}_2+10{\rm H}_2\rm O
Step 2
Calculate the molar mass of butane using the number of each type of atom in the formula and the molar mass of each element. The molar mass has the same value as the element's atomic mass listed on the periodic table, but the units are g/mol.
Molar mass of butane=(4)(12.01g/mol)+(10)(1.01g/mol)=58.14g/mol\begin{aligned}{\text{Molar mass of butane}}&=(4)(12.01\rm{\; g/mol})+(10)(1.01\rm{\; g/mol})\\&=58.14\rm{\; g/mol}\end{aligned}
Step 3
Use the molar mass to convert the mass of butane to moles of butane.
Moles of butane=Mass of butaneMolar mass of butane=60.0g58.14g/mol=1.032mol\begin{aligned}{\text{Moles of butane}}&=\frac{{\text{Mass of butane}}}{{\text{Molar mass of butane}}}\\&=\frac{60.0\rm{\; g}}{58.14\rm{\; g/mol}}\\&=1.032\rm{\; mol}\end{aligned}
Step 4
The coefficients of the balanced equation show that the mole ratio of butane to oxygen gas is 2:13. This means 2 mol of butane react with 13 mol of oxygen gas to produce 8 mol of carbon dioxide and 10 mol of water.
Solution
Use dimensional analysis to convert moles of butane to moles of oxygen gas using the mole ratio.
Moles of oxygen=(1.032molC4H10)(13molO22molC4H10)=6.71molO2\begin{aligned}{\text{Moles of oxygen}}&=(1.032\rm{\; mol\; C}_4\rm{H}_{10})\!\left(\frac{13\rm{\; mol\; O}_2}{2\rm{\; mol\; C}_4\rm{H}_{10}}\right)\\&=6.71\rm{\; mol\; O}_2\end{aligned}
The answer is expressed to the correct number of three significant figures.
Step-By-Step Example
Converting from Mass of Substance A to Moles of Substance B
Determine the number of moles of sulfuric acid (H2SO4) required to react with 30.0 grams of iron to form iron(II) sulfate (FeSO4).
Step 1
Write the balanced equation.
Fe+H2SO4H2+FeSO4\rm{Fe}+{\rm H}_2{\rm{SO}}_4\rightarrow{\rm H}_2+{\rm{FeSO}}_4
Step 2
Convert the mass of iron to moles of iron using its molar mass. The molar mass has the same value as iron's atomic mass listed on the periodic table, 55.85 atomic mass units (amu), but the units are grams per mole (g/mol).
Moles of iron=Mass of ironMolar mass of iron=30.0g55.85g/mol=0.5372molFe\begin{aligned}{\text{Moles of iron}}&=\frac{{\text{Mass of iron}}}{{\text{Molar mass of iron}}}\\&=\frac{30.0\rm{\; g}}{55.85\rm{\; g/mol}}\\&=0.5372\rm{\; mol\; Fe}\end{aligned}
Step 3
The balanced equation shows that the mole ratio of iron to sulfuric acid is 1:1. This means 1 mol of iron reacts with 1 mol of sulfuric acid to produce 1 mol of hydrogen gas and 1 mol of iron(II) sulfate.
Solution
Use dimensional analysis to convert moles of iron to moles of sulfuric acid using the mole ratio.
Moles of H2SO4=(0.5372molFe)(1molH2SO41molFe)=0.5372molH2SO4\begin{aligned}{\text{Moles of }{\rm H_{2}SO_{4}}}&=(0.5372\rm{\; mol\; Fe})\!\left(\frac{1\rm{\; mol\; H}_2\rm{SO}_4}{1\rm{\; mol\; Fe}}\right)\\&=0.5372\rm{\; mol\; H}_2\rm{SO}_4\end{aligned}
With the appropriate significant figures, the answer becomes 0.537 mol of sulfuric acid.