Please help..?

Question

On complete combustion, 4.0g of a hydrocarbon give 11.0g of carbon dioxide and 9.0g of water.

What is the formula of the hydrocarbon?

(1) CH4
(2) C2H4
(3) C2H6
(4) C3H8

The answer is (1) CH4..why? I don't understand...please explain..thanks.

Answer 1

It is CH4 because, there is 1 Carbon and 4 Hydrogens.

Think that 4.0g of this hydrocarbon give 20 grams of output - therefore-: something * 4 = 20.

Obviously the something is 5. The only option for the hydrocarbon to contain 5 parts is the CH4, as this adds up to 5, the rest would give out more.

Answer 2

4.0g hydrocarbon give rise to 11.0 g CO2

The molar mass of CO2 is 44.0.

So 16g hydrocarbon give rise to one mol carbon dioxide.

This means that16g hydrocarbon contain just one carbon atom. This is true for (1) but not for the others.

Checking: you should work out what weight of H2O is formed from the combustion of 16g of (1). Then see if that fits the experimental result.

Answer 3

11 g of CO2 is formed from 11/44 g-mole of C = 0.25 g-atom of C.
9 g of H2O is formed from 9/18 g-mole of of H2 = 0.5 g-mole of H2 or 1g-atom of H.
=> C : H = 0.25 : 1 = 1 : 4
=> hydrocarbon is ( 1 ) CH4

Answer 4

Methane has molecular formula CH4 with molecular weight of 16gm.(12+4(1))

One mole of methane(1*16= 16g) combines with 2 moles of oxygen (2*32 = 64g) and on complete combustion gives one mole of carbondioxide(12 + 2*16 = 44g) and 2 moles of water[2*(2*1+16) = 36g].

CH4 + 2O2 => CO2 + 2H2O

16g of(CH4)+ 64g of (O2) =>44g of (CO2) + 36g of(H2O)

divide by 4 throughout

4g of(CH4)+ 16g of (O2) =>11g of (CO2) + 9g of(H2O)