What is the mass of 0.0914 moles of ammonium sulfate?

Apr 9, 2017

12.06g

Explanation:

The gram-molecular weight of any compound is the sum of the atomic weights of each element. For ammonium sulfate this is ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ = 28+8+32+64 = 132 g/mol. Thus, 0.0914 moles has a mass of 0.0914 mol * 132 g/mol = 12.06g

Apr 10, 2017

The mass of $\text{0.0914 mol}$ ammonium sulfate is $\textcolor{b l u e}{\text{12.1 g}}$ rounded to three significant figures due to $\text{0.0914 mol}$.

Explanation:

In order to determine the mass of $\text{0.0914 mol}$ ammonium sulfate, you must first know its chemical formula, which is $\text{(NH"_4)_2"SO"_4}$. You then use the formula to determine its molar mass. Then multiply its molar mass by its given mass. This will give the mass of $\text{0.0914 mol}$ ammonium sulfate in grams.

Molar Mass of Ammonium Sulfate
Multiply the subscript of each element by its molar mass, which is its atomic weight on the periodic table in grams/mole.

Molar mass of $\text{(NH"_4)_2"SO"_4} :$(2xx14.007 color(white)(.)"g/mol N")+(8xx1.008 color(white)(.)"g/mol H")+(1xx32.06color(white)(.) "g/mol S")+(4xx15.999color(white)(.) "g/mol O")="132.134 g/mol (NH"_4)_2"SO"_4"

Mass of Ammonium Sulfate in 0.0914 mol
Multiply the given mol ammonium sulfate by its molar mass.

0.0914color(red)cancel(color(black)("mol"))xx(132.134"g")/(1color(red)cancel(color(black)("mol")))="12.1 g"("NH"_4)_2"SO"_4"

The mass of $\text{0.0914 mol}$ ammonium sulfate is $\text{12.1 g}$ rounded to three significant figures due to $\text{0.0914 mol}$.