Answer:
The answer to your question is C₃H₆O
Explanation:
Data
mass of sample = 23.2 g
mass of carbon dioxide = 52.8 g
mass of water = 21.6 g
empirical formula = ?
Process
1.- Calculate the mass and moles of carbon
44 g of CO₂ --------------- 12 g of C
52.8 g --------------- x
x = (52.8 x 12)/44
x = 633.6/44
x = 14.4 g of C
12 g of C ------------------ 1 mol
14.4 g of C --------------- x
x = (14.4 x 1)/(12)
x = 1.2 moles of C
2.- Calculate the grams and moles of Hydrogen
18 g of H₂O --------------- 2 g of H
21.6 g of H₂O ------------- x
x = (21.6 x 2) / 18
x = 2.4 g of H
1 g of H -------------------- 1 mol of H
2.4 g of H ----------------- x
x = (2.4 x 1)/1
x = 2.4 moles of H
3.- Calculate the grams and moles of Oxygen
Mass of Oxygen = 23.2 - 14.4 - 2.4
= 6.4 g
16 g of O ---------------- 1 mol
6.4 g of O -------------- x
x = (6.4 x 1)/16
x = 0.4 moles of Oxygen
4.- Divide by the lowest number of moles
Carbon = 1.2 / 0.4 = 3
Hydrogen = 2.4/ 0.4 = 6
Oxygen = 0.4 / 0.4 = 1
5.- Write the empirical formula
C₃H₆O
