A STP Sample ......

Question

At STP a sample of gas with a mass of 23.4 grams occupies 1.50 L. What is the molar mass of this gas?

Answer 1

Since the problem states STP, one does not need to use the Ideal Gas Law, as we know that at STP, the volume of 1 mole on any ideal gas = 22.4 liters. Thus, we can simply use this to convert 1.5 L to moles, and then solve for molar mass.

1.5 L x 1 mol / 22.4 L = 0.06696 moles

Molar mass = 23.4 g / 0.06696 mols = 349 g/mol (3 sig. figs.)

Answer 2

Recall the ideal gas law, which is a formula that holds true for gases under ideal conditions, or under STP. The ideal gas formula is PV=nRT, where P is pressure(in atm), V is volume(in L), n is moles(in mol), T is temperature(in Kelvin), and R is a gas constant, which is 0.082. Under STP, or standard temperature and pressure, temperature is always 273.15 K and pressure is always 1 atm.

Plugging in the STP values and what is known from the question, your equation becomes: (1)(1.50)=n(0.082)(273.15), which simplifies to 1.5=22.4n. Divide both sides by 22.4 to get n= 0.067 mol.

Now, the units of molar mass are g/mol. Use the other number given in the question, which is 23.4g, and divide it by mol of the unknown gas you solved for. This gives you 23.4g/0.067 mol, which divides out to 349.23 g/mol.