Answer:
The answer to your question is C₃H₄O₃
Explanation:
Data
CxHyOz
mass of sample = 0.150 g
mass of CO₂ = 0.225 g
mass of H₂O = 0.0614 g
Reaction
CxHyOz + O₂ ⇒ CO₂ + H₂O
Process
1.- Calculate the moles of C
44 g of CO₂ ----------------- 12 g of C
0.225 g ---------------- x
x = (0.225 x 12) / 44
x = 0.0614 g of C
12 g of C -------------------- 1 mol
0.0614 g of C --------------- x
x = 0.0051 moles of Carbon
2.- Calculate the moles of hydrogen
18 g of H₂O ------------------ 2 g of H
0.0614 g --------------- x
x = 0.0068 g of H
1 g of H ----------------------- 1 mol of H
0.0068 g --------------------- x
x = 0.0068 moles of H
3.- Calculate the mass of Oxygen
Mass of oxygen = 0.150 - 0.0614 - 0.0068
= 0.0818 g
16 g of O ------------------- 1 mol
0.0818 g -------------------- x
x = (0.0818 x 1) / 16
x = 0.0051 moles of O
4.- Divide by the lowest number of moles
Carbon 0.0051 / 0.0051 = 1
Hydrogen 0.0068 / 0.0051 = 1.33
Oxygen 0.0051 / 0.0051 = 1
Multiply these numbers by 3
Carbon 3
Hydrogen = 4
Oxygen = 3
5.- Write the empirical formula
C₃H₄O₃
