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?

1 Answer
Nov 20, 2015

The molecular weight (molar mass) in g/mol is "44.0 g"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 PV=nRTPV=nRT.

Given
"mass"="0.614 g"mass=0.614 g
P="238 mmHg"="238 torr"P=238 mmHg=238 torr
V="1.0 L"V=1.0 L
T="0.0"^"o""C"+273.15="273.15 K"T=0.0oC+273.15=273.15 K
R="62.363577 L torr K"^(-1) "mol"^(-1)"R=62.363577 L torr K1mol1
https://en.wikipedia.org/wiki/Gas_constant

Unknown
moles of gas

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

n=(PV)/(RT)n=PVRT

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.

"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 "44.0 g/mol".