# The ratio between the molar mass of an alkene(A) and an alkyne(B) with the same number of carbon atoms is 1.05. Find the molecular formulas of the two hydrocarbons then write the reaction for how we can obtain the alkene A from the alkyne B.

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We start by remembering the general formulas of alkenes and alkynes:

Alkenes - CnH2n

Alkynes - CnH2n-2

We calculate the molar mass of both of these using the fact that the atomic mass of carbon is 12 and for hydrogen 1 :

μA=12n+2n=14n grams/mole

μB=12n+2n-2=14n-2 grams/mole

The ratio then becomes:

μ/ μB=1.05

14n / 14n - 2 = 1.05

14n =(14n - 2) * 1.05

14n = 14.7n - 2.1

14.7n - 14 = 2.1

0.7n = 2.1

n = 3

Therefore, our hydrocarbons are:

A = C3H6 - propene

B = C3H4 - propyne

The reaction to obtain propene from the propyne is:

CH≡C-CH3 + H2 --Ni---> CH2=CH-CH3

The reaction is a hydrogenation one and the condition needed for it Nickel.

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