A mixture of methane (CH₄) and butane (C₄H₁₀) at one atmosphere pressure and 25°C has a density of 1.695 g/L. Assuming ideal behavior, what is the mass in grams of carbon that are in 1 liter of mixtur

Question

A mixture of methane (CH₄) and butane (C₄H₁₀) at one atmosphere pressure and 25°C has a density of 1.695 g/L. Assuming ideal behavior, what is the mass in grams of carbon that are in 1 liter of the mixture?

Answer 1

PV = nRT => n = PV/RT

<=> (1atm)(1L)

(0.08205L*atm*K-1*mol-1)(298K)

n = 0.0409 mol of total gas

ρ = m*V^-1 => m = ρ*V

<=> (1L)(1.695m*L^-1)

m = 1.695g of gas total

x + y = 1.695, where x = mass of CH4 and y = mass of C4H10

n_CH4 + n_C4H10 = n_total

=> MW = m*n^-1 => n = m*MW^-1

(m(MW^-1)_CH4 + (m*MW-1)_C4H10 = n_total

=> x + y = 1.695 => y = 1.695 - x

(x/MW)_CH4 + [(1.695-x)/MW]_C4H10 = 1.695

x/16.04 + (1.695-x)/58.12 = 1.695

x = 0.285g CH₄

x + y = 1.695 => y = 1.695 - x

<=> 1.695 - 0.285

y = 1.41g C₄H₁₀

mass% = part/whole x 100%

mass% C in CH₄ = 12/16.04

mass% C in CH₄ = 75%

mass% C in C₄H₁₀ = 48/58.12

mass% C in C₄H₁₀ = 82.76%

mass C in CH₄ = mass % * mass

<=> 0.75*0.285

mass C in CH₄ = 0.2141g

mass C in C₄H₁₀ = mass % * mass

<=> 0.8276*1.41

mass C in C₄H₁₀ = 1.1665

mass_total = mass C in CH₄ + mass C in C₄H₁₀

mass_total = 1.381g