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A Gaseous Hydrocarbon Contains 82.76% of Carbon. Given that Its Vapour Density is 29, Find Its Molecular Formula. - Chemistry

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प्रश्न

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.
[C = 12, H = 1]

योग
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उत्तर

% of carbon = 82.76%
% of hydrogen = 100 - 82.76 = 17.24%

Element % weight Atomic weight Relative No. of moles Simplest Ratio
C 82.76 12 82.76/12 = 6.89 6.89/6.8 = 1 x 2 = 2
H 17.24 1 17.24/1 = 17.24 17.24/6.89 =2.5 x 2 = 5

Empirical formula = C2H5
Empirical formula weight = 2 × 12 + 1 × 5 = 24 + 5 = 29
Vapour density = 29
Relative molecular mass = 29 × 2 = 58
N = `"Relative molecular mass"/"Empirical weight"`
= `58/29` = 2
Molecular formula = n x empirical formula
= 2 x C2H5
=
C4H10

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अध्याय 5: Mole Concept And Stoichiometry - Exercise 11 [पृष्ठ १२५]

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फ्रैंक Chemistry Part 2 [English] Class 10 ICSE
अध्याय 5 Mole Concept And Stoichiometry
Exercise 11 | Q 7 | पृष्ठ १२५

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