Firstly we need to find the empirical formula of the fuel
To do that
C H O S
Mass 50 15 28 7
Molar mass 12 1 16 32
Mole=m/mm 50/12 15/1 28/16 7/32
4.17 15 1.75 0.22
Dividing by the smallest mole which is 0.22,
We have our empirical formula to be
C19H68O8S.
So it has to undergo combustion
And the complete combustion of any organic fuel is represented by;
CwHxOySz+1/2[2w+x/2+2z-y]O2----------------------wCO2+x/2H2O+zSO2.
So inputing our values using w=19,x=68,y=8 and z=1
We will have
C19H68O8S+33O2----------19CO2+34H2O+SO2
So now using a mass mole relationship between the fuel and oxygen,we have;
If 33 moles O2 is required for the complete combustion of 456g of the fuel
Then the amount of mole of O2 required for the combustion of 1000g of the fuel is;(the 1kg has been converted to 1000g)
=33×1000/456=72.368moles.
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