Empirical formula of hydrocarbon by combustion analysis?

1) The elements making up the unknown substance almost always include carbon and hydrogen. Oxygen is often involved and nitrogen is involved sometimes. Other elements can be involved, but problems with C and H tend to predominate followed by C, H and O and then by C, H, O and N.

2) We must know the mass of the unknown substance before burning it.

3) All the carbon in the sample winds up as CO2 and all the hydrogen in the sample winds up as H2O.

4) If oxygen is part of the unknown compound, then its oxygen winds up incorporated into the oxides. The mass of oxygen in the sample will almost always be determined by subtraction.

5) Often the N is determined via a second experiment and this introduces a bit of complexity to the problem. Nitrogen dioxide is the usual product when nitrogen is involved. Sometimes the nitrogen product is N2, sometimes NH3.

6) Sometimes the problem asks you for the empirical formula and sometimes for the molecular formula (or both). Two points: (a) you have to know the molar mass to get to the molecular formula and (b) you have to calculate the empirical formla first, even if the question doesn't ask for it. A few lines below is a link that goes to a file that discusses how to go from empirical to molecular formulas.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml