# What is the percent by mass of nitrogen atom in ammonium sulfate, ("NH"_4)_2"SO"_4?

Jul 28, 2015

Percent of nitrogen: 21.2%

#### Explanation:

To get the percent composition of nitrogen is ammonium sulfate, ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$, you need to know the molar mass of the compound and that of elemental nitrogen.

The important thing to notice is that ammonium sulfate contains 2 nitrogen atoms, each belonging to one ammonium ion, $N {H}_{4}^{+}$.

So, the molar mass of ammonium sulfate is 132.14 g/mol and the molar mass of nitrogen is 14.007 g/mol.

You can write the percent composition of nitrogen in ammonium sulfate by dividing the molar mass of the two nitrogen atoms by the molar mass of the compound, and multiplying the result by 100.

(2 * 14.007color(red)(cancel(color(black)("g/mol"))))/(132.14color(red)(cancel(color(black)("g/mol")))) * 100 = color(green)("21.2%")

This means that every 100g of ammonium sulfate contain 21.2 g of nitrogen.

Jul 30, 2015

The %"mass" (percent by mass) is:

${M}_{A} / \left({M}_{{\left(N {H}_{4}\right)}_{2} S {O}_{4}}\right)$

with $A$ as some arbitrary molecule or atom.

Thus, you would need to look up or recall the molar masses of these compounds.

$N \approx 14.007 \text{ amu}$
$H \approx 1.0079 \text{ amu}$
$S \approx 32.065 \text{ amu}$
$O \approx 15.999 \text{ amu}$

${\left(N {H}_{4}\right)}_{2} S {O}_{4} = 2 \cdot \left(14.007 + 4 \cdot \left(1.0079\right)\right) + 32.065 + 4 \cdot 15.999$
$\approx 132.1382 \text{ amu}$

So then:

M_(N)/(M_((NH_4)_2SO_4)) = 14.007/(132.1382) ~~ 10.6% of each $N$ atom

-> 21.2% since it is ${\left(N {H}_{4}\right)}_{2}$ and thus there are two $N$ atoms in there.