Question

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?

Answer 1

do the silver one first
Ag 108 g/mol
Cl 35.5 g/mol
AgCl = 143.5 g/mol
so 5.57 g AgCl has (108/143.5)5.57 = 4.19 g of Ag

so 8.12 g stuff has 4.19 g Ag
so 4.72 g stuff has 2.44 g Ag

so we have 4.72-2.44 = 2.28 g of C and H

we got 7.95 g of CO2
C = 12 g/mol
O2 = 32 g/mol
CO2 = 44 g/mol
12/44 *7.95 = 2.17 g C

we got 1.02 g of H2O
H2 = 2 g/mol
O = 16 g/mol
H2O = 18 g/mol
H2 = 2/18 * 1.02 = .113= g H2

so our compound had
2.17 g of C
.113 g of H2
2.44 g of Ag
good, that checks with total of 4.72
now chemistry
2.17/12 = .181 mol C
.113/2 =.0565 mol H2=.113 mol H
2.44/108 = .0226 mol Ag

5 H for every Ag .113/.0226
8 C for every Ag .181/.0226

so
C8 H5 Ag