An unknown compound contains only C, H, and O. Combustion of 4.00g of this compound produced 9.42 g of CO2 and 2.57 g of H2O. What is the empirical formula of the unknown compound.
1) Determine the grams of carbon in 9.42 g CO2 and the grams of hydrogen in 2.57 g H2O. carbon: 9.42 g x (12.011 g / 44.0098 g) = 2.571 g hydrogen: 2.57 g x (2.0158 g / 18.0152 g) = 0.288 g 2) Convert grams of C and H to their respective amount of moles. carbon: 2.571 g / 12.011 g/mol = 0.214 mol hydrogen: 0.288 g / 1.0079 g/mol = 0.286 mol 3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole numbers. carbon: 0.214 mol / 0.214 mol = 1 hydrogen: 0.286 mol / 0.214 mol = 2.9986 = 1.333 4) Multiply these values till obtain the whole numbers: 1x3 = 3; 1.333x3 = 4 We have now arrived at the answer: the empirical formula of the substance is C3H4 Answer: C3H4