## Chemistry and Chemical Reactivity (9th Edition)

$84.28\%$
Number of moles of $CO_2$: $0.558\ g/44.001\ g/mol=0.0127\ mol$ From stoichiometry, that's the number of moles of calcium carbonate. $0.0127\ mol\times 100.087\ g/mol=1.269\ g$ Mass fraction: $1.269\ g/1.506\ g=84.28\%$