# What is the percentage of ammonium ions in ammonium dichromate?

The formula unit for ammonium dichromate is $\text{(NH"_4)_2"Cr"_2"O"_7}$, and has a molar mass of $252.07 \text{g/mol}$.
The ammonium ion, $\text{NH"_4^+}$, is the cation, and its molar mass is $18.039 \text{g/mol}$. Since there are two moles of ammonium cations in one mole of ammonium dichromate, we multiply $18.039 \text{g/mol} \times 2$, which is equal to $36.078 \text{g/mol}$.
"%NH"_4^+=(36.078 "g/mol")/(252.07 "g/mol")xx100=14.313%