molecular formula of combustion?help

adelarge1

complete combustion of 1.1g of a compound of C,H and O gives 2.2g of CO2 and 0.9g of H2O.
The compound has a relative molar mass close to 130.
Find the molecular formula and true molar mass?

ApeXaviour

Whereabouts are you stuck?

If I were doing the question I'd deduce what weight, and then how many moles of O2 were combusted. Then that leaves you with how much oxygen was in the original compound. Convert all to moles and that will give you your ratio of C, H and O. Multiply in integers until the ratios add to about 130.

Any steps in there eluding you feel free ask.

ianobrien

adelarge1 wrote: »

complete combustion of 1.1g of a compound of C,H and O gives 2.2g of CO2 and 0.9g of H2O.
The compound has a relative molar mass close to 130.
Find the molecular formula and true molar mass?

Have you got them numbers right?

The reason I ask is based on complete combustion, the only products are CO2 & H2O.

Calculating the number of CO2 produced is 0.05M (2.2g divided by 44g/mol)

Calculating the number of H2) produced is 0.05M (0.9g divided by 18g/mol)

As the only source of Carbon in the reaction is the starting material, there must be 0.05M of it there (as there is only 0.05M of Carbon starting off)

This means, the 1.1g of material is equivalent to 0.05M. Thus the molecular mass is 22g/mol (1.1g divided by 0.05M)

Now, working off the 0.05M of C and 0.1M of H present, the empirical formula must be CxH2xOy, with y being an uneven number. Assuming a value of 1 for both x & y gives a molecular weight of 30g/mol?

Or have I got this totally wrong?

ClimberC

To me it would make a lot more sense if the O was being used to oxidize a branched alkane? But i am probably missing out on something obvious :rolleyes: