What is the mass percent of nitrogen in ammonium carbonate, $\ce{(NH4)2CO3}$?

To determine the mass percentage of nitrogen, first we have to calculate the molar mass of ammonium carbonate $\ce{((NH4)2CO3)}$.

The molar mass of $\ce{N}$ is $\ce{14.007 g/mol}$

The molar mass of $\ce{H}$ is $\ce{1.008 g/mol}$

The molar mass of $\ce{C}$ is $\ce{12.011 g/mol}$

The molar mass of $\ce{O}$ is $\ce{16.000 g/mol}$

Therefore, the molar mass of ammonium carbonate is

$$\begin{align*} \ce{M((NH4)2CO3)} &= \ce{ (M(N) + M(H) \times 4) \times 2 + M(C) + M(O) \times 3 }\\ &= \ce{ (14.007 + 1.008 \times 4) \times 2 + 12.011 + 16.000 \times 3 }\\ &= \ce{96.089 g/mol} \end{align*}$$