Answer:
The answer to your question is: C₄H₁₀O
Explanation:
Data
CxHyOz
mass sample : 1.376 g
mass CO₂ = 3.268 g
mass H₂O = 1.672 g
Process
Reaction
CxHyOz + O₂ ⇒ CO₂ + H₂O
1.- Calculate the moles and mass of carbon
Molecular mass CO₂ = 44g
44 g of CO₂ -------------- 12 g of C
3.268 g of CO₂ -------- x
x = (3.268 x 12) / 44
x = 0.891 g of Carbon
12 g of carbon ----------- 1 mol
0.891 g of C ---------- x
x = (0.891 x 1) / 12
x = 0.0743 moles of carbon
2.- Calculate the moles and mass of hydrogen
18 g of water --------------- 2 g of H
1.672 g of H₂O ------------ x
x = (1.672 x 2) / 18
x = 0.186 g of hydrogen
1 g of hydrogen ------------ 1 mol of H
0.186 g of H ------------ x
x = (0.186 x 1) / 1
x = 0.186 moles of H
3.- Calculate the mass of Oxygen and its moles
Mass of Oxygen = 1.376 - 0.891 - 0.186
= 0.299 g of O₂
Moles of Oxygen
16 g of Oxygen ---------------- 1 mol
0.299 g of O ----------------- x
x = (0.299 x 1) / 16
x = 0.019 moles of Oxygen
4.- Divide by the lowest number of moles
Carbon 0.0743/ 0.019 = 3.9 ≈ 4.0
Hydrogen 0.186/ 0.019 = 9.7 = 10
Oxygen 0.019/ 0.019 = 1
5.- Write the empirical formula
C₄H₁₀O
for pure water.
at 25 °C.