# If a 0.614 g sample of a gas maintains a pressure of 238 mm Hg when contained in a 1.0L flask at 0.0°C, what is the gases molecular weight?

Nov 20, 2015

The molecular weight (molar mass) in g/mol is $\text{44.0 g}$.

#### Explanation:

We first need to use the ideal gas law to determine the moles of the gas. Then we will divide the given mass by the calculated moles.

The equation for the ideal gas law is $P V = n R T$.

Given
$\text{mass"="0.614 g}$
$P = \text{238 mmHg"="238 torr}$
$V = \text{1.0 L}$
$T = \text{0.0"^"o""C"+273.15="273.15 K}$
$R = \text{62.363577 L torr K"^(-1) "mol"^(-1)}$
https://en.wikipedia.org/wiki/Gas_constant

Unknown
moles of gas

Solution
Rearrange the equation for the ideal gas law to isolate $n$, then solve.

$n = \frac{P V}{R T}$

n=((238cancel"torr")xx(1.0cancel"L"))/((62.363577cancel"L" cancel"torr" cancel("K"^(-1)) "mol"^(-1))xx(273.15cancel"K"))="0.01397 mol"

Determine molecular weight in g/mol by dividing the given mass by the calculated moles.

$\text{molecular weight"=(0.614"g")/(0.01397"mol")="44.0 g/mol}$

This could be carbon dioxide gas, as it has a molar mass of $\text{44.0 g/mol}$.