Question

A sample of an unknown gas with a mass of 4.67 goccupies a volume of 1.88 L at a pressure of 785 mmHg and a temperature of 20.0 ∘C. Molar mass=57.9 g/mol. If the unknown gas is composed of 3.86 g of carbon and the rest is hydrogen, what is its molecular formula?

Answer 1

Moles can be expressed as mass/molar mass (mass/MW)
So, the ideal gas law become:

PV = (mass/MW) RT

Just plug in the values and solve for what you need and solve for MW:
P = 785/760 = 1.03 atm
V (given in the question)= 1.88 L
T(given in the question) = 273+20 = 293 K
R (given in the question)= 0.0821
mass(given in the question) = 4.67 g

PV = (mass/MW) RT
MW = mass RT / PV = 4.67 X 0.0821 X 293 / (1.03 X 1.88) = 58.0 g/mol

To get to the molecular formula, we first have to get to the empirical formula. So, we are given from question that unknown's sample g contains 3.86 g C.(total mass = 4.67)

So, the mass H is (4.67 - 3.86) = 0.81 g H

Convert each of those masses to moles:
=>3.86 g C / 12 g/mol = 0.322 mol C
=>0.81 g H / 1 g/mol = 0.81 mol H

Now,we divide each of those by the smaller:

=>C = 0.322/0.322 = 1
=>H = 0.81/0.322 = 2.5

Since H comes out to be 2.5,we multiply both of these by 2 to get the empirical formula of C2H5. Now, it has a formula weight of 29 g/mol. Since we determined the molar mass of the gas to be 58 (twice the formula weight), the empirical formula must be C4H10.