Question

A mixture of gases is analyzed and found to have the following composition in mol %: CO2 12.0 CO 6.0 CH4 27.3 H2 9.9 N2 44.8 a) Determine the composition of the gas mixture in weight %. b) Determine the average molecular weight of the gas mixture.

Answer 1

a) CO₂: 21,9%; CO: 7,0%; CH₄: 18,2%; H₂: 0,8%; N₂: 52,1%

b) 24,09 g/mol

Step-by-step explanation:

a) It is posible to obtain the composition of the gas mixture in weight% using molecular mass of each compound, thus:

12% CO₂×
`(44,01g)/(1mol)` = 528,1 g

6% CO×
`(28,01g)/(1mol)` = 168,1 g

27,3% CH₄×
`(16,05g)/(1mol)` = 438,2 g

9,9% H₂×
`(2,02g)/(1mol)` = 20,0 g

44,8% N₂×
`(28g)/(1mol)` = 1254,4 g

The total mass of the gas mixture is:

528,1g + 168,1g + 438,2g + 20,0g + 1254,4g = 2408,8 g

Thus composition of the gas mixture in weight% is:

CO₂:
`(528,1g)/(2408,8g)`×100 = 21,9%

CO:
`(168,1g)/(2408,8g)`×100 = 7,0%

CH₄:
`(438,2g)/(2408,8g)`×100 = 18,2%

H₂:
`(20,0g)/(2408,8g)`×100 = 0,8%

N₂:
`(1254,4g)/(2408,8g)`×100 = 52,1%

b) The average molecular weight of the gas mixture is determined with mole % composition, thus:

0,12×44,01g/mol + 0,06×28,01g/mol + 0,273×16,05g/mol + 0,099×2,02g/mol + 0,448×28g/mol = 24,09 g/mol

I hope it helps!