# 12.4: Mass-Mass Stoichiometry

I want to send 250 grams of shelled walnuts to a friend (don't ask why - just go with the question). How many walnuts in shells do I need to buy? To figure this out, I need to know how much the shell of a walnut weighs (about $$40\%$$ of the total weight of the unshelled walnut). I can then calculate the mass of walnuts that will give me 250 grams of shelled walnuts and then determine how many walnuts I need to buy.

### Mass to Mass Problems

Mass-mass calculations are the most practical of all mass-based stoichiometry problems. Moles cannot be measured directly, while the mass of any substance can generally be easily measured in the lab. This type of problem is three steps and is a combination of the two previous types.

$\text{mass of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{mass of unknown}$

The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Finally, the moles of the unknown are converted to mass by use of its molar mass.

Example $$\PageIndex{1}$$

Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.

$\ce{NH_4NO_3} \left( s \right) \rightarrow \ce{N_2O} \left( g \right) + 2 \ce{H_2O} \left( l \right)$

In a certain experiment, $$45.7 \: \text{g}$$ of ammonium nitrate is decomposed. Find the mass of each of the products formed.

Solution

Step 1: List the known quantities and plan the problem.

Known

• Given: $$45.7 \: \text{g} \: \ce{NH_4NO_3}$$
• $$1 \: \text{mol} \: \ce{NH_4NO_3} = 1 \: \text{mol} \: \ce{N_2O} = 2 \: \text{mol} \: \ce{H_2O}$$
• Molar mass of $$\ce{NH_4NO_3} = 80.06 \: \text{g/mol}$$
• Molar mass of $$\ce{N_2O} = 44.02 \: \text{g/mol}$$
• Molar mass of $$\ce{H_2O} = 18.02 \: \text{g/mol}$$

Unknown

• Mass $$\ce{N_2O} = ? \: \text{g}$$
• Mass $$\ce{H_2O} = ? \: \text{g}$$

Perform two separate three-step mass-mass calculations as shown below.

$\text{g} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{N_2O} \rightarrow \text{g} \: \ce{N_2O}$

$\text{g} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{H_2O} \rightarrow \text{g} \: \ce{H_2O}$

Step 2: Solve.

$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \frac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \frac{1 \: \text{mol} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \frac{44.02 \: \text{g} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{N_2O}} = 25.1 \: \text{g} \: \ce{N_2O}$

$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \frac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \frac{2 \: \ce{H_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \frac{18.02 \: \text{g} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} = 20.6 \: \text{g} \: \ce{H_2O}$