Question

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

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Solution

Determination of the molecular formula from vapour densityStep 1: Determination of the percentage of each element in hydrocarbonPercentage 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 formulaS.No. ElementMolecular massPercentage compositionNumber of atomsSimplest ratioNearest whole number rounding off 1C (Carbon)1282.76%$\frac{82.76}{12}=6.89$$\frac{6.89}{6.89}=1$$1×2=2$2H (Hydrogen)117.24%$\frac{17.24}{1}=17.24$$\frac{17.24}{6.89}=2.5$$2.5×2=5$The empirical formula of gaseous hydrocarbon is = ${\mathrm{C}}_{2}{\mathrm{H}}_{5}$Step 3: Determination of Empirical formula massThe empirical formula mass of ${\mathrm{C}}_{2}{\mathrm{H}}_{5}$ = $2×12+5×1$ The empirical formula mass of ${\mathrm{C}}_{2}{\mathrm{H}}_{5}$ = 29 Step 4: Determination of molecular mass The vapour density = 29 Molecular mass = $2×\mathrm{Vapour}\mathrm{density}$Molecular mass = $2×29$Molecular mass = 58Step 5 Relation between empirical mass and molecular mass Molecular mass = $\mathrm{n}×\mathrm{Empirical}\mathrm{formula}\mathrm{mass}$n = $\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}$n = $\frac{58}{29}$n = 2Step 6: Determination of Molecular formulaMolecular formula = $\mathrm{n}×\mathrm{Empirical}\mathrm{formula}$Molecular formula = ${\mathrm{C}}_{4}{\mathrm{H}}_{10}$

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