Question

An unknown compound contains only C, H, and O. Combustion of 8.50 g of this compound produced 20.0 g CO2 and 5.46 g H2O. What is the empirical formula of the unknown compound?

Answer 1

The answer to your question is: C₃H₄O

Step-by-step explanation:

Data

CxHyOz = 8.5 g

CO2 = 20 g

H2O = 5.46 g

Reaction

CxHyOz + O2 ⇒ CO2 + H2O

CO2

MW = 44g

44g CO2 ----------------- 12g of C

20g CO2 ---------------- x

x = (20 x 12) / 44

x = 5.45 g of C

# of moles = n = 5.45 / 12 = 0.454 mol of C

H2O

MW = 18 g

18 g H2O ------------------- 2g of H

5.46 g -------------------- x

x = (5.46 x 2) / 18 = 0.61 g of H

n = 0.61 / 1 = 0.61 moles of H2

Mass of O2

mass CxHyOz = mass CO2 + mass H2 + mass O2

mass O2 = 8.5 - 5.45 - 0.61

mass O2 = 2.44g

n = 2.44 / 16 = 0.153 mol of O2

Now, divide by the lowest number of moles

0.454 mol of C/ 0.153 = 2.97 ≈ 3

0.61 moles of H2/ 0.153 = 3.99 ≈ 4

0.153 mol of O2/ 0.153 = 1

Then, the empirical formula is: C₃H₄O