# A gas sample weighing 3.78 grams occupies a volume of 2.28 L at STP. What is the molecular mass of the sample?

Feb 21, 2017

${\text{37.6 g mol}}^{- 1}$

#### Explanation:

The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that $1$ mole of any ideal gas occupies $\text{22.7 L}$ under STP conditions to calculate how many moles of gas you have in your sample.

$\textcolor{red}{\underline{\textcolor{b l a c k}{\text{Under STP conditions: " "1 mole of an ideal gas = 22.7 L}}}}$

In your case, you know that your sample of gas occupies $\text{2.28 L}$ under STP conditions, which are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

This means that your sample will contain

2.28 color(red)(cancel(color(black)("L"))) * overbrace("1 mole gas"/(22.7color(red)(cancel(color(black)("L")))))^(color(blue)("molar volume of a gas at STP")) = "0.10044 moles gas"

Now, the molar mass of the gas is the mass of exactly $1$ mole of the gas. In your case, you know that you get $\text{3.78 g}$ for every $0.10044$ moles, which means that you have

1 color(red)(cancel(color(black)("mole"))) * "3.78 g"/(0.10044color(red)(cancel(color(black)("moles")))) = "37.6 g"

Since this is the mass of $1$ mole of gas, you can say that the molar mass of the gas is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar mass = 37.6 g mol}}^{- 1}}}}$

The answer is rounded to three sig figs.

SIDE NOTE A lot of textbooks and online sources still use the old definition of STP conditions, i.e. a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions, $1$ mole of any ideal gas occupies $\text{22.4 L}$. If this is the value of the molar volume of a gas at STP given to you, simply redo the calculations using $\text{22.4 L}$ instaed of $\text{22.7 L}$.