# If 21.8g of the hydrocarbon is burnt in air, calculate the mass of carbon dioxide formed. A gaseous hydrocarbon contains 85.7% carbon by mass. The hydrocarbon burns with a yellow, sooty flame and when bubbled through a solution of bromine, the solution is decolourised as a result of the reaction between bromine and the hydrocarbon. if the carbon is 85.7% by mass

then,

C mass:H mass = 85.7:(100-85.7)

= 85.7:14.3

C moles:H moles= (85.7/12):(14.3/1)

= 7.14:14.3

= 1:2 approximately.

Therfore the empirical formula is (CH2)n

Now this should be unstaturated hydrocarbon since it burns with yellow sooty flame and since it reacts with bromine it should be alkene. Alkenes have the formula of CnH2n same as this one.

For combustion of 1 mol of CnH2n

CnH2n + (n+n/2) O2 ----------------> n CO2 + n H2O

Now one mole of CnH2n gives n moles of CO2.

Now n units of CH2 will give you n amount of CO2 also.

that means one CH2 will give you 1 mole of CO@

CH2 -----------> CO2

1 mol                  1 mol

14 g                    44 g

1 g                      (44/14) g

21.8 g                 21.8 * (44/14) g

21.8 g                  68.514 g

Therefore, the amount of CO2 produced by total combustion of 21.8 g of Hydrocarbon is 68.514 g.

Approved by eNotes Editorial Team