Question

The rate of diffusion of saturated hydrocarbon CxH2n+2 is 1.206 times that ot CO2 gas under identical condition. Find the molecular mass and value of x for that gas

Answer 1

1) List known and unknown values

Compound A: CO2

Molar Mass: 44.01 g/mol

Rate of effusion: 1.0

Compound B: CxH2n+2

Rate of effusion: 1.206

Molar mass:

2) Graham's law

`(Rate_A)/(Rate_B)=\sqrt[]{(MM_B)/(MM_A)}`

Plug in values and solve for MMB

`(1.0)/(1.206)=\sqrt[]{\frac{MM_B}{44.01\frac{g}{mol_{}}}}`
`((1.0)/(1.206))^2\cdot44.01=MM_B`
`MM_B=30.26\frac{g}{\text{mol}}`

The molar mass of CxH2n+2 is 30.26 g/mol

3) The molecular mass of the unknown compound

The molar mass of CH4 is 16.04 g/mol

`\text{Ratio}=(30.26)/(16.04)=1.89`

The ratio is 2. So, x is equal to 2

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