Answer:
The answer to your question is C₃H₃O
Explanation:
Data
Combustion of a compound C, H, O
mass = 6.10 g
mass CO2 = 14.9 g
mass of water = 6.10 g
Reaction
Cx Hy Oz + O2 ⇒ CO2 + H2O
Process
1.- Calculate the moles of carbon
44 g of CO2 -------------- 12 g of carbon
14.9 g of CO2 ------------- x
x = (14.9 x 12) / 44
x = 4.06 g
12 g of C ------------------ 1 mol
4.06 g ------------------- x
x = (4.06 x 1) / 12
x = 0.34 moles
2.- Calculate the moles of hydrogen
18 g of water ------------- 1 g of hydrogen
6.10 g of water ---------- x
x = (6.10 x 1) / 18
x = 0.33 g
1 g of H ---------------- 1 mol of H
0.33 g ---------------- x
x = (0.33 x 1) / 1
x = 0.33 moles of H
3.- Calculate the mass of oxygen
mass of Oxygen = 6.10 - 4.06 - 0.33
= 1.71 g
16 g of O --------------- 1 mol of O
1.71 g of O ------------- x
x = (1.71 x 1) / 16
x = 0.11 moles
4.- Divide by the lowest number of moles
Carbon = 0.34 / 0.11 = 3
Hydrogen = 0.33 / 0.11 = 3
Oxygen = 0.11 /0.11 = 1
5.- Write the empirical formula
C₃H₃O
