# How many grams are contained in 3.22 moles of dinitrogen tetroxide?

Apr 23, 2016

Molar mass of ${N}_{2} {O}_{3} = 2 \times 14 + 3 \times 16 = 76 g m o {l}^{-} 1$
Mass of 3.22 moles of ${N}_{2} {O}_{3} = 3.22$moles$\times 76 \frac{g}{m o l} = 244.72 g$

May 1, 2016

The formula for dinitrogen tetroxide is ${N}_{2} {O}_{4}$.

Assume that:
n - number of moles
m - mass of substance
M - molar mass

$n = m \div M$

3.22 moles (n) of ${N}_{2} {O}_{4}$ has been provided for you.

Next step is to find out the molar mass (M). Before that, break up the formula, first, to make sure that you know what atoms are present. You can see that Nitrogen and Oxygen are present. If you refer to your periodic table, the molar mass of Nitrogen is 14.0 g/mol and Oxygen 16.0 g/mol.

Therefore, the molar mass (M) of dinitrogen tetroxide is:
$\left[2 \times 14.0 + 4 \times 16.0\right]$ = 92.0 g/mol.

Your last step is to find out the mass.
Look back at the $n = m \div M$ formula and make m as the subject. Flip it over and it'll become $m = M \times n$.

The mass of ${N}_{2} {O}_{4}$ is:
[m = 92.0 g/mol $\times$ 3.22 moles] = 296.2 grams.
Note that this answer is rounded to 3 significant figures.

Overall, 296.2 grams are contained in 3.22 moles of dinitrogen tetroxide.