If the reaction you are referring to is 2C2H6 + 7O2 --> 4CO2 + 6H2O then you would find the percent yield as follows:
1) 32g C2H6 * (1mol C2H6/30.08g C2H6)* (4mol CO2/2mol C2H6) *(44.01g CO2/1mol CO2) = 93.64g CO2.
This is your theoretical yield based on the 32g of C2H6 you started with and the molar ratio between C2H6 and CO2 in your balanced equation.
2) Your experiment produced 44g of CO2, so your actual yield is 44g CO2.
% yield = (actual yield/theoretical yield) x 100
(44g CO2/93.64g CO2) x 100 = 46.99%