Question

An unknown gas contains 85.63% carbon and 14.37% hydrogen by mass. if 2.00 L of the gas at 298 K and 0.420 atm and weights 1.926 g

A) What is the molar mass of the unknown gas?

B) What is the empirical formula of the gas?

C) What is the molecular formula of the gas?

Answer 1

A) calculate no. of mole of gas by using ideal gas equation

Ideal gas equation

PV = nRT             where, P = atm pressure= 0.420 atm,

V = volume in Liter = 2 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 298 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.420 2)/(0.08205 298) = 0.0343545636 mole

1.926 gm = 0.0343545636 mole then 1 mole = 1.926/ 0.0343545636 = 56.06 gm/mole

Molar mass of gas is 56.06 gm/mol

B)

Calculate atomic ratio by dividing % composition by atomic mass.

Mass ratio of carbon = 85.63/12 = 7.14

Mass ratio of hydrogen = 14.37/1 = 14.37

Now calculate simplest ratio by dividing atomic ratio by smallest atomic ratio

Simplest ratio of carbon = 7.14/7.14 = 1

Simplest ratio of hydrogen = 14.37/7.14 = 2.01 2

Empirical formula of compound = CH₂

C)

Empirical formula of compound = CH₂

empirical formula weight = 12 + 2(1) = 14 gm

n = molecular formula weight/empirical formula weight where, n = no. of carbon atom

= 56.06/14 = 4

molecular formula = C₄H₈