Practice Exercise: Calculating the Molar Mass of a Gas

In order to calculate the molar mass of a gas, we need to know the mass of the  gas as well as the number of moles of that gas.  

Remember:         molar mass of a gas = mass / number of moles 

Questions

1. Determine the approximate molar mass of a gas if 560 cm³ has a mass of 1.55 g at STP.

2. At 18^oC and 765 torr, 1.29 L of a gas has a mass of 2.71g. Calculate the molar mass of the gas.

3. Determine the volume occupied by 4.0g of oxygen gas at STP.

4. What volume would 15.0 g of argon occupy at 90^oC and 735 torr.

5. Dry air consists of approximately 21% O₂, 78% N₂, and 1% Ar, by moles. What is the average or apparent molar mass of dry air? Calculate the density of dry air at STP.

6. An organic compound has the following combustion analyses: C = 55.8%; H = 7.03%, O = 37.2%. A 1.5 g sample was vaporized and was found to occupy 530 cm³ at 100^oC and 740 torr. What is the molecular formula of the compound?

7. Which of the following would be the formula for the density of an ideal gas ( P is pressure, V is volume, T is temperature, g is mass, D is density, and  MM is molar mass)?

    a)     D = PMM/RT     b) D = gRT/PMM     c) D = PTMM/gRV     d) D = gV/RT

Answers

1. Determine the approximate molar mass of a gas if 560 cm³ has a mass of 1.55 g at STP.

n = (0.560 L)/(22.4mol/L) = 0.025 mol

MM = 1.55g / 0.025mol = 62 g/mol

2. At 18^oC and 765 torr, 1.29 L of a gas has a mass of 2.71g. Calculate the molar mass of the gas.

T = 18 + 273.15 = 291.15 K

P = (765 torr)/(760 torr/atm) = 1.00658 atm

PV= nRT

n = [(1.00658 atm)(1.29L)] / [(0.0821 L^.atm^.K^-1.mol^-1)(291.15K)] = 0.0544 mol

MM = 2.71 g / 0.0544 mol = 49.8 g/mol

3. Determine the volume occupied by 4.0g of oxygen gas at STP.

n O₂ = 4.0g ¸ 32 g/mol = 0.125 mol O₂

V = 0.125 mol O₂  x  22.4 L/mol  =  2.8 L

4. What volume would 15.0 g of argon occupy at 90^oC and 735 torr ?

T = (735 torr) ¸ (760 torr/atm) = 0.967 atm

n = 15.0g ¸ 39.9 g/mol = 0.376 mol

PV = nRT

V = nRT/P = (0.376mol)(0.0821 L^.atm^.K^-1.mol^-1)(363.15K)] ¸ 0.967 atm = 11.6 L

5. Dry air consists of approximately 21% O₂, 78% N₂, and 1% Ar, by moles. What is the average or apparent molar mass of dry air? Calculate the density of dry air at STP.

1 mol of air would have a mass of:    0.21 mol(32.0 g/mol) + 0.78(28.0 g/mol) + 0.01(39.95 g/mol)= 29 g

1 mol of air at STP has a volume of 22.4 L

D = M/V = 29 g / 22.4 L = 1.29 g/L

6. An organic compound has the following combustion analyses: C = 55.8%; H = 7.03%, O = 37.2%. A 1.5 g sample was vaporized and was found to occupy 530 cm³ at 100^oC and 740 torr. What is the molecular formula of the compound?

Use the combustion analysis to determine the empirical formula and the gas information to determine the molecular mass.

Mol C = (0.558)(1.5 g)/(12.0 g/mol) = 0.06975

Mol H = (0.0703)(1.5 g)/(1.0 g/mol) = 0.1095

Mol O = (0.372)(1.5 g)/(16.0 g/mol) = 0.034875

So far...  C _0.06975 H _0.1095 O _0.034875

Dividing each subscript by 2 gives us the empirical formula of C₂H₃O

How many moles of the substance do we have?

n = PV/RT

n = (0.9737)(0.53)/(0.0821)(373) = 0.0168 mol

MM of  C₂H₃O = 2(12) + 3(1) + 1(16) = 43 g

MM of the substance = 1.5 / 0.0168 = 89.3 g

Therefore, the molecular formula is roughly 2 times the empirical formula...

The molecular formula must be C₄H₆O₂

7.  A is the correct answer.