Mole Fractions of gases?

Question

A gas sample is known to be a mixture of methane and butane. A bulb of 200.0 cm3 capacity is filled

with gas to a pressure of 100.0 kPa at 20.0°C. If the weight of the gas in the bulb is 0.3846 g, what is the

mole fraction of butane in the mixture? You may assume ideal gas behavior.

Answer 1

n = pV / RT = 0.987 atm x 0.200 dm^3 /0.0821 x 293.15 K =0.00820 = moles CH4 + C4H10

(Moles butane x 58.1 g/mol ) + (moles methane x 16.043 g/mol) = 0.3846

let x = moles butane and let y = moles methane :

y + x = 0.00820

58.1 x + 16.043 y = 0.3846

solve the system

mole fraction butane = x / 0.00820