Answer
$78.238\%$
Work Step by Step
Silver is a cation of charge +1 and carbonate is an anion of charge -2, so the salt has the formula-unit:
$Ag_2CO_3$
Calculate the molar mass of this salt:
Atomic weights: Ag - 107.868, C - 12.011, O - 15.999
$M=2\cdot 107.868+12.011\cdot 3\cdot 15.999=275.744\ g/mol$
Calculate the mass of silver atoms per mole of the formula-unit:
$M_{Ag}=2\cdot 107.868=215.736\ g/mol$
The percent by mass of silver in this compound is the ratio of the mass of silver atoms per mole of the formula-unit divided by the molar mass of the formula-unit, in percent form:
$\%m_{Ag}=\frac{M_{Ag}}{M}\cdot100.0\%$
$\%m_{Ag}=\frac{215.736\ g/mol}{275.744\ g/mol}\cdot100.0\%$
$\%m_{Ag}=78.238\%$