At STP a sample of gas with a mass of 27.3 grams occupies 8.20 L. What is the molar mass of this gas?

Answer 1

When @ STP, 1 mole of any ideal gas occupies 22.4 L (a good number to memorize so as not to have to use the ideal gas law).

How many moles of gas do we have?

8.20 L x 1 mole / 22.4 L = 0.366 moles

Molar mass = grams / moles

Molar mass = 27.3 g / 0.366 moles

Molar mass = 74.6 g / mole

Answer 2

STP means "Standard Temperature and Pressure) and means T = 273.15 K (which is 0°C) and P = 1 atmosphere. We are given that V = 8.20 L.

The Ideal Gas Law says PV = nRT where R (in units of liters, atmospheres, moles and K) is 0.0820574 (L•atm)/(mol•K). Solving for n, we get n = (PV)/(RT). Plugging things in gives:

n = (1•8.20)/(0.0820574•273.15) = 0.3658 moles

Since molar mass is in units of grams/mol, the molar mass is 27.3/0.3658 = 74.6 g/mol