4.72 g of a compound of carbon, hydrogen, and silver was burned in an atmosphere of oxygen, yielding 7.95 g of CO₂, 1.02 g of H₂O, and 3.02 g of a mixture of silver and silver oxide. Because the production of this mixture yielded an indeterminate value for the amount of silver, another sample of the compound weighing 8.12 g was reacted with a solution of NaCl, yielding 5.57 g of AgCl. What is the empirical formula of the compound?

Question

Quizplus · Accepted Answer

Step 1: Calculate the amount of Ag produced
From the second reaction: 8.12 g compound reacted to form 5.57 g AgCl
Molar mass of AgCl = 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl)

5.57 g AgCl x (1 mol AgCl/143.32 g) x (1 mol Ag/1 mol AgCl) x (107.87 g/1 mol) = 4.16 g Ag produced

Step 2: Calculate the amount of O in CO and H2O produced 
7.95 g COS + 1.02 g H2SO4 x (1 mol/60.06 g) x 3 mol O/1 mol H2SO4 = 0.06 g O
(0.06 g O / 16 g/mol) = 0.00375 mol O

Step 3: Calculate the amount of C and H in the compound 
4.72 g compound - 7.95 g COS - 1.02 g H2SO4 = 1.65 g remaining
1.65 g / (12.01 g/mol + 1.01 g/mol) = 0.124 mol C
1.65 g / (1.01 g/mol + 32 g/mol + 1.01 g/mol) = 0.0245 mol H

Step 4: Calculate the empirical formula 
Divide the number of moles of each element by the smallest number of moles to get the simplest whole number ratio of elements.

C: 0.124 mol / 0.0245 mol = 5.06 ≈ 5
H: 0.0245 mol / 0.0245 mol = 1
Ag: 4.16 g / 107.87 g/mol / 0.0245 mol = 7.84 ≈ 8

Empirical formula: C5H8Ag

472 G of a Compound of Carbon, Hydrogen, and Silver