A gaseous hydrocarbon contain 20 % hydrogen by mass the vapour density of the gas is 15. The compound does not decolorize bromine vapour and interact or feebly sensitive to common Chemical reagent From the information giving above deduce @ empirical formular , molecular formular , the structural formular of the compound and the name
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A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. [C=12,H=1].
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Element %composition Mol. mass No. of C atoms Simplest Rounding
ratio off
C 82.76 12 6.89 1× 2 2
H 17.24 1 17.24 2.5× 2 5
Number of carbon atoms present =
12
82.76
=6.89
Number of hydrogen atoms present =
1
17.24
=17.24
simplest ratio =
6.89
6.89
=1
simplest ratio =
6.89
17.24
=2.5
∴ Empirical fromula is C
2
H
5
Empirical formula mass= 2×12+5 = 29
∴ V.D. = 29
Mol. mass = 2 × V.D.
=58
n=
Empiricalformulamass
Molecularmass
=
29
58
=2
∴ Molecular formula = C
4
H
10
.
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