Answer:
The answer to your question is below
Explanation:
Data
mass of the sample = 1.357 g
mass of carbon dioxide = 1.989 g
mass of water = 0..8143 g
Process
-Moles of Carbon
Molar mass of CO₂ = 44 g
44 g of CO₂ -------------- 12 g of C
1.989 g of CO₂ --------- x
x = (1.989 x 12) / 44
x = 0.542 g of C
12 g of C ------------------ 1 mol
0.542 g of C ------------- x
x = (0.542 x 1) / 12
x = 0.0452 moles
-Moles of Hydrogen
Molar mass of water = 18 g
18 g of H₂O --------------- 2 g of H
0.8143 g of H₂O --------- x
x = (0.8143 x 2) / 18
x = 0.0904 g of H
1 g of H ------------------ 1 mol of H
0.0904 g of H --------- x
x = (0.0904 x 1) / 1
x = 0.0904 moles of H
-Moles of Oxygen
Mass of Oxygen = 1.357 - 0.542 - 0.0904
= 0.7246 g
16 g of O ---------------- 1 mol
0.7246 g of O --------- x
x = (0.7246 x 1) / 16
x = 0.04528
2.- Divide by the lowest number of moles
Carbon = 0.0452/0.0452 = 1
Hydrogen = 0.0904/0.0452 = 2
Oxygen = 0.0452 /0.0452 = 1
3.- Write the empirical formula
CH₂O
4.- Find the molecular formula
-Molar mass CH₂O = 12 + 2 + 16
= 30 g
-Divide the molecular mass by the molar mass 60/30 = 2
-Write the molecular formula
2(CH₂O) = C₂H₄O₂
