Question

A mixture of 50 wt% methane, 35 wt% ethane, and 15 wt% propane. Determeine the mole fraction of methane.

(Can use a basis of 100kg)

b) What is the average molecular weight of the mixture?

for (b) can use M = \sum Yi Mi where M = average molecular weight, Yi= mole fraction of individual substance, Mi = molecular weight of individual substance OR can also use 1/M = \sum Xi / Mi where 1/M = average molecular weight, Xi = mass fraction of individual substance, Mi = molecualr weight of individual substance.

Answer 1

a) Molar fraction:

Methane: 67,5%

Ethane: 25,1%

Propane: 7,4%

b) Molecular weight of the mixture: 25,16 g/mol

Step-by-step explanation:

with a basis of 100 kg:

Moles of methane:

500 g ×
`(1mol)/(16,04 g)` = 31,2 moles

Mass of ethane

350 g ×
`(1mol)/(30,07 g)` = 11,6 moles

Mass of propane

150 g ×
`(1mol)/(44,1 g)` = 3,4 moles

Total moles: 31,2 moles + 11,6 moles + 3,4 moles = 46,2 moles

Molar fraction of n-methane:

`(31,2moles)/(46,2moles)` =67,5%

Molar fraction of ethane:

`(11,6moles)/(46,2moles)` = 25,1%

Molar fraction of propane:

`(3,4moles)/(46,2moles)` = 7,4%

b) Average molecular weight:

`0,5*16,04g/mol + 0,35*30,07g/mol+0,15*44,1g/mol}` = 25,16g/mol

I hope it helps!