Answer to Question #165371 in Organic Chemistry for Oranade

Question

An organic compound is composed of Carbon, hydrogen, oxygen and Chlronie.

(a) Describe a simple chemical test to show the presence of chlorine in the compound.

1.57g of the compounds on analysis yielded 1.76g of carbon dioxide, 0.54g of water and 2.87g of silver chloride. Find the empirical formula of the compound.

Answer 1

Solution:

Silver nitrate test. To a small portion of an organic compound add to it silver nitrate solution (AgNO₃). White precipitate (AgCl), soluble in ammonium hydroxide, indicates the presence of chlorine in the organic compound.

Schematic equations:

C_xH_yO_zCl_n + .... → xCO₂ + (y/2)H₂O + nAgCl

CO₂ → C

H₂O → 2H

AgCl → Cl

Convert to moles:

Moles of CO₂ = Mass of CO₂ / Molar mass of CO₂

The molar mass of CO₂ is 44.01 g mol^-1.

Hence,

n(CO₂) = 1.76 g / 44.01 g mol^-1 = 0.03999 mol = 0.04 mol

n(C) = n(CO₂) = 0.04 mol (according to the shematic equation above)

Moles of H₂O = Mass of H₂O / Molar mass of H₂O

The molar mass of H₂O is 18.0153 g mol^-1.

Hence,

n(H₂O) = 0.54 g / 18.0153 g mol^-1 = 0.02997 mol = 0.03 mol

n(H) = 2 × n(H₂O) = 2 × 0.03 mol = 0.06 mol (according to the shematic equation above)

Moles of AgCl = Mass of AgCl / Molar mass of AgCl

The molar mass of AgCl is 143.32 g mol^-1.

Hence,

n(AgCl) = 2.87 g / 143.32 g mol^-1 = 0.02000 mol = 0.02 mol

n(Cl) = n(AgCl) = 0.02 mol (according to the shematic equation above)

Convert to grams:

The molar mass of C is 12.0107 g mol^-1.

The molar mass of H is 1.00784 g mol^-1.

The molar mass of Cl is 35.453 g mol^-1.

The molar mass of O is 15.999 g mol^-1.

m(C) = n(C) × M(C) = 0.04 mol × 12.0107 g mol^-1 = 0.4804 g

m(H) = n(H) × M(H) = 0.06 mol × 1.00784 g mol^-1 = 0.0605 g

m(Cl) = n(Cl) × M(Cl) = 0.02 mol × 35.453 g mol^-1 = 0.70906 g

m(compound) = m(C) + m(H) + m(O) + m(Cl)

Hence,

m(O) = m(compound) - m(C) - m(H) - m(Cl)

m(O) = 1.57 g = 0.4804 g - 0.0605 g - 0.70906 g = 0.32004 g

n(O) = m(O) / M(O) = 0.32004 g / 15.999 g mol^-1 = 0.02000 mol = 0.02 mol

Thus: n(C) = 0.04 mol; n(H) = 0.06 mol; n(Cl) = 0.02 mol; n(O) = 0.02 mol.

Divide all moles by the smallest of the results:

C: 0.04 mol / 0.02 mol = 2

H: 0.06 mol / 0.02 mol = 3

O: 0.02 mol / 0.02 mol = 1

Cl: 0.02 mol / 0.02 mol = 1

The empirical formula of the compound is C₂H₃OCl

Answer: The empirical formula of the compound is C₂H₃OCl.