# How many grams of (NH_4)_2SO_4 are present in .100 moles of (NH_4)_2SO_4 ?

Feb 23, 2016

Ammonium sulfate has an equivalent weight of $132.14$ $g \cdot m o {l}^{-} 1$.

#### Explanation:

You have a $0.100$ $m o l$ quantity.

$0.100$ $\cancel{m o l}$ $\times$ $132.14$ $g \cdot \cancel{m o {l}^{-} 1}$ $=$ ?? $g$.

Feb 23, 2016

The mass in grams of 0.100 mol $\text{(NH"_4)_2"SO"_4}$ is 13.2 g.

#### Explanation:

The molar mass of $\text{(NH"_4)_2"SO"_4}$ is needed, which is $\text{132.13952 g/mol}$. http://pubchem.ncbi.nlm.nih.gov/compound/6097028section=Top

To determine the mass of 0.100 mol ammonium sulfate, multiply the given mole by its molar mass.

0.100 cancel(mol (NH_4)_2SO_4)xx(132.13952 g (NH_4)_2SO_4)/(1 cancel(mol (NH_4)_2SO_4))=13.2 g (NH_4)_2SO_4"# (rounded to three significant figures)