**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?**

**Answer:**

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 .

Moles

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

c= 7.72

C2.6H7.8O7.72

We divide a,b and c by 2.6 , which is the smallest coefficient.

ANSWER = C1H3O3