Answer :
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
Answer:
The empirical formula is C3H12N2Cl2
Explanation:
Step 1: Data given
When 1.00 g of X is dissolved in water and allowed to react with excess AgNO3, all the chlorine (present as chloride) precipitates and 1.95 g of AgCl is collected.
When 1.00 g of X undergoes complete combustion,0.900 g of CO2 and 0.735 g of H2O are formed.
Atomic mass of C = 12.01 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of N = 14.01 g/mol
Atomic mass of Cl = 35.45 g/mol
Atomic mass of Ag = 107.87 g/mol
Step 2: Calculate moles of Cl
Moles AgCl = Mass AgCl / molar AgCl
Moles AgCl = 1.95 grams / 143.32 g/mol
Moles AgCl = 0.0136 moles
For 1 mol AgCl we have 1 mol Cl
Moles Cl = 0.0136 moles Cl
Mass Cl = 0.0136 moles * 35.45 g/mol = 0.482 grams Cl
Step 3: Calculate moles CO2
1 g reacts with O2 to produce 0.9 g CO2
Moles CO2 = 0.9 grams / 44.01 g/mol = 0.02045 moles
For 1 mol CO2 we have 1 mol C
Moles C = 0.02045 moles C
Mass C = 0.02045 moles *12.01 g/mol = 0.2456 grams C
Step 4: Calculate moles H
Moles H2O = 0.735 grams / 18.02 g/mol
Moles H2O = 0.0408 moles H2O
In 1 mol H2O we have 2 moles H
Moles H = 2*0.0408 = 0.0816 moles H
Mass H = 0.0824 grams H
Step 5: Calculate mass N
Mass N = 1.00 grams - 0.482 grams Cl - 0.2456 grams C -0.0824 grams H
Mass N = 0.190 grams N
Step 6: Calculate moles N
Moles N = 0.190 grams N /14.01 g/mol
Moles N = 0.0136 moles N
Step 7: Calculate mol ratio
We divide by the smallest amount of moles
C: 0.02045 moles / 0.0136 moles = 1.5
H: 0.0816 moles / 0.0136 moles = 6
N: 0.0136 moles / 0.0136 moles = 1
Cl: 0.0136 moles/ 0.0136 moles = 1
The empirical formula is C3H12N2Cl2