A certain alcohol contains only 3 elements; carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol produced 114.6 grams of CO2 and 70.44 grams of H2O. What is the empirical formula of the alcohol?
By using the law of conservation of masses in a chemical reaction we find the amount of O2 in the reactants.
60g + x g of O2 = 114.6 g + 70.44 g
X= 44.16g of 02
Then we calculate the moles of each compound expect for the unknow .
CO2 = 114.6 g / 44g/mol = 2.6 mo
H20= 3.9 mol
O2= 1.38 mol
We write for each coefficient the mole numbers. (except for the unknown, where its coefficient 1 is)
CaHbOc + 1.38 O2 = 2.6 CO2 + 3.9 H2O
We calculate a, b and c.
a= 2.6 (Ca… + … = 2.6CO2 + …)
b= 3.9 * 2 = 7.8
c+ 1.38 * 2 = 2.6 * 2 + 3.9
We divide a,b and c by 2.6 , which is the smallest coefficient.
ANSWER = C1H3O3