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) 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

[tex]\frac{Rate_A}{Rate_B}=\sqrt[]{\frac{MM_B}{MM_A}}[/tex]

Plug in values and solve for MMB

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

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

[tex]\text{Ratio}=\frac{30.26}{16.04}=1.89[/tex]

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

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