# If the molar mass is 56g mol^-1, determine the molecular formula of the hydrocarbon. Name a possible structure for the hydrocarbon. A gaseous hydrocarbon contains 85.7% carbon by mass. The hydrocarbon burns with a yellow, sooty flame and when bubbled through a solution of bromine, the solution is decolourised as a result of the reaction between bromine and the hydrocarbon. The mass ratio of C:H is 85.7:14.3.

Therefore the molar ratio of C:H is (85.7/12) : (14.3/1) =  7.14:14.3

The molar ratio of C:H = 1: (14.3/7.14) = 1:2 (Approximately)

Therefore the empirical formula is (CH2)n

f this burns with yellows sooty flames, it means it has less hydrogen to...

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The mass ratio of C:H is 85.7:14.3.

Therefore the molar ratio of C:H is (85.7/12) : (14.3/1) =  7.14:14.3

The molar ratio of C:H = 1: (14.3/7.14) = 1:2 (Approximately)

Therefore the empirical formula is (CH2)n

f this burns with yellows sooty flames, it means it has less hydrogen to combine with oxygen and it is not a saturated hydrocarbon, it is an unsaturated hydrocarbon.

If this react with bromine, then this should be an alkene or alkyne, but this (CH2)n therefore it is an alkene.

The empirical formula for an alkene is = Cn H2n

let's find the value for n, if molar mass is 56,

n x 12 + 2 x n x 1 = 56

14n = 56

n = 4

Therefore, the formula is C4H8. It is an alkene,

CH3-CH=CH-CH3 or H-C(H)=C(CH3)-CH3

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