Question #a4835

Feb 15, 2018

$\text{0.0395 g}$

Explanation:

As you know, Avogadro's constant tells you the number of molecules of ammonia needed in order to have $1$ mole of ammonia.

$6.022 \cdot {10}^{23} \quad {\text{molecules NH"_3 = "1 mole NH}}_{3}$

Now, the molar mass of ammonia, which tells you the mass of exactly $1$ mole of ammonia, is equal to ${\text{17.031 g mol}}^{- 1}$. This means that $1$ mole of ammonia has a mass of $\text{17.031 g}$.

$\text{1 mole NH"_3 = "17.031 g}$

You can thus say that, for ammonia, you have

$6.022 \cdot {10}^{23} \quad \text{molecules NH"_3 = "1 mole NH"_3 = "17.031 g}$

which, of course, implies that

$6.022 \cdot {10}^{23} \quad \text{molecules NH"_3 = "17.031 g}$

Your sample contains $2.38 \cdot {10}^{22}$ molecules of ammonia, so you can say the mass of the sample will be equal to

$2.38 \cdot {10}^{22} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{molecules NH"_3))) * "17.031 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules NH"_3)))) = color(darkgreen)(ul(color(black)("0.0395 g}}}}$

The answer is rounded to three sig figs, the number of significant figures you have for the number of molecules of ammonia.