# What is the percent composition by mass of nitrogen in (NH_4)_2CO_3 (gram-formula mass 96.0 g/mo)?

Jun 8, 2017

29.2% "N"

#### Explanation:

We're asked to find the percentage by mass of $\text{N}$ in ammonium carbonate, ("NH"_4)_2"CO"_3, given that its molar mass is $96.0 \text{g"/"mol}$.

To find the mass of $\text{N}$ in the compound, we multiply its molar mas, $14.01 \text{g"/"mol}$, by however many atoms of $\text{N}$ are in the compound, which is $2$:

$\text{mass of N" = (2)(14.01"g"/"mol") = 28.02"g"/"mol}$

Thus, in one mole of ("NH"_4)_2"CO"_3 (which has a mass of $96.0$ $\text{g}$), there are $28.02$ $\text{g}$ of nitrogen. The percentage by mass of $\text{N}$ is the mass of $\text{N}$ divided by the total mass, then multiplied by $100$:

% "N" = ("mass of N")/("mass of" ("NH"_4)_2"CO"_3) xx 100%

% "N" = (28.02"g N")/(96.0"g" ("NH"_4)_2"CO"_3) xx 100% = color(red)(29.2% "N"