Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react w

Question

4 years ago

8

Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react w

ith 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. What is the empirical formula of X? [C = 12.01,
H = 1.01, N = 14.01, Cl = 35.45, Ag = 107.87].

Chemistry

2 answers:

natka813 [3] · Answer 1 · 2022-01-29T10:35:56+03:00

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

Alexus [3.1K] · Answer 2 · 2022-01-29T12:41:02+03:00

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