Question

A gaseous hydrocarbon contains 82.76% of Carbon. Given that Its vapor density Is 29. Find its molecular formula. [C = 12, H =1].

Answer 1

Determination of the molecular formula from vapour density

• Step 1: Determination of the percentage of each element in hydrocarbon
  Percentage Cabon present in gaseous hydrocarbon = 82.76%
  Percentage of Hydrogen present in the gaseous hydrocarbon = 100 - 82.76%
  Percentage of Hydrogen present in the gaseous hydrocarbon = 17.24%
• Step 2: Determination of Empirical formula

S.No.	Element	Molecular mass	Percentage composition	Number of atoms	Simplest ratio	Nearest whole number rounding off
1	C (Carbon)	12	82.76%	$\frac{82.76}{12}=6.89$	$\frac{6.89}{6.89}=1$	$1\times 2=2$
2	H (Hydrogen)	1	17.24%	$\frac{17.24}{1}=17.24$	$\frac{17.24}{6.89}=2.5$	$2.5\times 2=5$

The empirical formula of gaseous hydrocarbon is = ${\mathrm{C}}_2{\mathrm{H}}_5$

• Step 3: Determination of Empirical formula mass
  The empirical formula mass of ${\mathrm{C}}_2{\mathrm{H}}_5$ = $2\times 12+5\times 1$
  The empirical formula mass of ${\mathrm{C}}_2{\mathrm{H}}_5$ = 29
• Step 4: Determination of molecular mass
  The vapour density = 29
  Molecular mass = $2\times \mathrm{Vapour}\mathrm{density}$
  Molecular mass = $2\times 29$
  Molecular mass = 58
• Step 5 Relation between empirical mass and molecular mass
  Molecular mass = $\mathrm{n}\times \mathrm{Empirical}\mathrm{formula}\mathrm{mass}$
  n = $\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}$
  n = $\frac{58}{29}$
  n = 2
• Step 6: Determination of Molecular formula
  Molecular formula = $\mathrm{n}\times \mathrm{Empirical}\mathrm{formula}$
  Molecular formula = ${\mathrm{C}}_4{\mathrm{H}}_{10}$