# What is the molecular mass of a compound if 560mL have a mass of 1.10g at STP?

Apr 29, 2016

#### Answer:

${M}_{r} = 44.6$.

#### Explanation:

We can use the Ideal Gas Law to solve this problem.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $n = \text{mass"/"molar mass} = \frac{m}{M}$, we can write the Ideal Gas Law as

$P V = \frac{m}{M} R T$

We can rearrange this to

$M = \frac{m R T}{P V}$

At STP, $P = \text{1 bar}$ and $T = \text{0 °C}$.

M = ("1.10 g" × 8.314 × 10^"-2" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 0.560 color(red)(cancel(color(black)("L")))) = "44.6 g/mol"

${M}_{r} = 44.6$