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A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]

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Question

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]

Numerical
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Solution

Element Percentage Molecules Simple ratio Ration Simple whole ratio
Carbon 82.76 `82.76/12` = 6.9 `6.89/6.89` = 1 `6.9/6.9` = 1 2
Hydrogen 17.24 `17.24/1` = 17.24 `17.24/6.89` = 2.5 `17.24/6.9` = 2.5 5

Empirical formula = C2H5

Empirical formula mass = (12 × 2) + (1 × 5)

= 24 + 5 

= 29

Vapour density = 29    ...(Given)

Molecular weight = 2 × Vapour density

Molecular weight = 2 × 29

= 58 gm

Molecular formula mass = n × Empirical formula mass

⇒ n = `"Molecular formula mass"/"Empirical formula mass"`

= `58/29`

= 2

Molecular formula = n × Empirical formula

= 2 × C2H5

= C4H10

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Chapter 5: Mole Concept and Stoichiometry - EXERCISE [Page 109]

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Lakhmir Singh Chemistry [English] Class 10 ICSE
Chapter 5 Mole Concept and Stoichiometry
EXERCISE | Q 21. | Page 109

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