# chem a level help

1 mole of a hydrocarbon of formula CnH2n was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192g of oxygen. Work out the formula of the hydrocarbon.

i cant find a recent thread of this question.
'CnH2n + X O2 ==> n CO2 + n H2O
balancing oxygen, we have X= 3n/2
192g of oxygen is 6 moles, so 3n/2 = 6 ==> n=4
and the hydrocarbon is C4H8'

why have they done 3n/2?

'192g of oxygen = 12 mole
bCnH2n + aO2 -> xCO2 + yH2O
For every 3 moles of oxygen there are 2 of hydrogen and one of carbon.
:. bCnH2n has 8 moles of hydrogen and 4 of carbon'

why have they said theres 12mol? isnt there 6moles?(192/32, to get 12 they wouldve done 192/16 but thats not right so then they wouldve got 6 mol but that gives c2h4 and thats not right?
thanyou
Original post by miac0328
1 mole of a hydrocarbon of formula CnH2n was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192g of oxygen. Work out the formula of the hydrocarbon.

i cant find a recent thread of this question.
'CnH2n + X O2 ==> n CO2 + n H2O
balancing oxygen, we have X= 3n/2
192g of oxygen is 6 moles, so 3n/2 = 6 ==> n=4
and the hydrocarbon is C4H8'

why have they done 3n/2?

'192g of oxygen = 12 mole
bCnH2n + aO2 -> xCO2 + yH2O
For every 3 moles of oxygen there are 2 of hydrogen and one of carbon.
:. bCnH2n has 8 moles of hydrogen and 4 of carbon'

why have they said theres 12mol? isnt there 6moles?(192/32, to get 12 they wouldve done 192/16 but thats not right so then they wouldve got 6 mol but that gives c2h4 and thats not right?
thanyou

There must be the same number of oxygens on both sides of the equation:

On the LHS, only O2 has oxygen atoms in it and since the reaction coefficient (the big number in front of the formula) is X and there are two oxygen atoms in each O2 molecule, then there must be 2X oxygens on the LHS.

On the RHS, CO2 and H2O both have oxygens in them and using their reaction coefficients, there must be 2n + n = 3n oxygens.

So we have 2X = 3n ==> X = 3n/2

I agree, 192 g of O2 is 6 mol of oxygen (since O2 has a relative mass of 32 g/mol) and therefore the correct continuation would be to equate 6 with 3n/2 and solve for n, which you can use to deduce the formula of the hydrocarbon.
Original post by TypicalNerd
There must be the same number of oxygens on both sides of the equation:

On the LHS, only O2 has oxygen atoms in it and since the reaction coefficient (the big number in front of the formula) is X and there are two oxygen atoms in each O2 molecule, then there must be 2X oxygens on the LHS.

On the RHS, CO2 and H2O both have oxygens in them and using their reaction coefficients, there must be 2n + n = 3n oxygens.

So we have 2X = 3n ==> X = 3n/2

I agree, 192 g of O2 is 6 mol of oxygen (since O2 has a relative mass of 32 g/mol) and therefore the correct continuation would be to equate 6 with 3n/2 and solve for n, which you can use to deduce the formula of the hydrocarbon.

i just dont really understand why o2 is 2x?
why cant we call it x?
Original post by miac0328
i just dont really understand why o2 is 2x?
why cant we call it x?

We are looking at the number of oxygen atoms in XO2. If there are X molecules made up of two atoms, then there are X x 2 = 2X oxygen atoms