Answer:
The answer to your question is C₁.₄ H₆ N₁Cl₁
Explanation:
Data
Compound CxHyNzCla
mass of the compound= 1 g
mass of AgCl = 1.95 g
mass of CO₂ = 0.900 g
mass of H₂O = 0.735 g
Empirical formula = ?
Process
1.- Calculate the moles of Chlorine
Molar weight of AgCl = 107.87 + 35.45 = 143.32 g
143.32 g of AgCl ------------- 35.45 g of Cl
1.95 g of AgCl ------------- x
x = (1.95 x 35.45) / 143.32
x = 0.482 g of Cl
35.45 g of Cl --------------- 1 mol
0.482 g of Cl ------------- x
x = (0.482 x 1) / 35.45
x = 0.0136 moles of Cl
2.- Calculate the moles of Carbon
Molar weight of CO₂ = 12.01 + 32 = 44.01
44.01 g --------------------- 12.01 g of C
0.900 g ------------------- x
x = (0.900 x 12.01) / 44.01
x = 0.246 g of C
12 g of C --------------- 1 mol
0.246 g of C --------- x
x = (0.246 x 1)/12
x = 0.020 moles
3.- Calculate the moles of Hydrogen
molar weight of H₂O = 18.01
18.01 g of H₂O -------------- 2 g of H
0.735 g of H₂O ------------ x
x = (0.735 x 2) / 18.01
x = 0.082 g of H
1 g of H ----------------------- 1 mol
0.082 g of H ---------------- x
x = (0.082 x 1) / 1
x = 0.082 moles
4.- Calculate the moles of Nitrogen
Mass of Nitrogen = 1 - 0.482 - 0.246 - 0.082
= 0.19 g
14 g of N ----------------- 1 mol
0.19 g of N -------------- x
x = (0.19 x 1) / 14
x = 0.014
5.- Divide by the lowest number of moles
Carbon 0.020 / 0.014 = 1.4
Hydrogen 0.082 / 0.014 = 5.9
Nitrogen 0.014 / 0.014 = 1
Chlorine 0.014 / 0.014 = 1
6.- Write the empirical formula
C₁.₄ H₆ N₁Cl₁