Answer:
The answer to your question is: C₃H₄O
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
