Question

The vapour density of a hydrocarbon is 30. If on
complete combustion of hydrocarbon, 88 g Co, and
54 9 H,O are obtained then the molecular formula
of hydrocarbon is
(1) GH
(2) C₃H₂2
(3) CH
(4) CH.​

Answer 1

Answer:

The vapour density of a hydrocarbon is 30. If on

complete combustion of hydrocarbon, 88 g Co, and

54 9 H,O are obtained then the molecular formula

of hydrocarbon is

(1) GH

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Explanation:

Answer 2

Answer:

C4H12.

Explanation:

If we take the molecular weight to be x+y with the x and y being the weights of the substances.

Since, in the question 8.8 g of the CO2 is given so the moles will be 8.8/44 which is 0.2 moles.  

Since, for the 1 mole of the hydrocarbon x moles of CO2 is present hence, for 0.2 moles of the CO2 the moles will be = 0.2/x moles of hydrocarbon

 

Since, in the question it is given that 5.4 g of the H2O which is 5.4/18 moles = 0.3 moles  of the water.

Now again for the y/2 moles of the water in the product side there will be 0.3 moles hence for the hydrocarbon 0.6/y moles will be present.

Hence, equating both the reactant and product side 0.2/x = 0.6/y which on solving we will get y=3x.

Since, we know that the molecular weight = 2*vapour density.  

As, the molecular weight is (x+y).

Hence, 12x+ y = 2*30.  

So, 12x + y = 60.

On substituting the value of y in the above equation we will get that x=4  and  y = 12.  

Therefore, the hydrocarbon will be C4H12.