5mL of a gas containing only carbon and hydrogen were mixed with an excess of oxygen (30mL) and the mixture exploded by means of an electric spark. After the explosion, the volume of the mixed gases remaining was 25mL. On adding a concentrated solution of potassium hydroxide, the volume further diminished to 15mL, the residual gas being pure oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas.

5mL of a gas containing only carbon and hydrogen were mixed with an ex

1.	5mL of a gas containing only carbon and hydrogen were mixed with an excess of oxygen (30mL) and the mixture exploded by means of an electric spark. After the explosion, the volume of the mixed gases remaining was 25mL. On adding a concentrated solution of potassium hydroxide, the volume further diminished to 15mL, the residual gas being pure oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas.
Answer» Solution :Let the FORMULA of the hydrocarbonbe `C_(x)H_(y)`. Its COMBUSTION can be shows by the following equation: `{:(C_(x)H_(y)+,(x+(y)/(4))O_(2)rarr,xCO_(2)+(y)/(2)H_(2)O),(1vol,(x+(y)/(4))mL,xvol),(5mL,5(x+(y)/(4))mL,5xvol):}` Volume of carbon dioxide produced `= (25 - 15) = 10mL` `5x = 10` or `x = 2` Volume of oxygen used `= (30 - 15) = 15mL` `5[x+(y)/(4)] = 15` or `x +(y)/(4) = (15)/(5) = 3` or `2 +(y)/(4) = 3` or `(y)/(4) = 3 - 2 = 1` or `y = 4` THUS, the molecular formula of the hydrocarbon `= C_(2)H_(4)`