If 35.80kg of CaCO3 contain 4.296kg C, 14.33kb Ca, and 17.17kg O, what is the percent composition of CaCO3?

for the CaCO3 we have % C = 4.296/ 35.80 = 12.0%
% Ca = 14.33 / 35.80 = 40,0%
% O = 17.17 / 35.80 = 48%

moles of C in 100 g of compound = 60 g / 12 g/ mol =5 .0 mole
moles of H = 4,480/1 = 4,480 moles
moles of O = 35.53 / 16 = 2.221moles
molar ratio = 5: 4.48: 2.221
divide by 2.221 and we get
2.25: 2.017: 1
* 4 and we have 9:8:4
empirical formula of aspirin is C9H8O4