You are given a crushed sample that is a mixture of limestone (caco3), lime (cao), and sand. the calcium carbonate, limestone, is the only material present in the material that will decompose when heated. you subject a 6.4734 g sample of the mixture to strong heating and after the sample reaches a constant mass (no more mass is lost with additional heating), the sample has a final weight of 4.3385 g. what is the percentage of calcium carbonate present in the original mixture? (f. wt. caco3 = 100.1)

Answer is: percentage of calcium carbonate is 75%.
Chemical reaction: CaCO₃ → CaO + CO₂.
m₁(sample) = 6,4734 g.
m₂(sample) = 4,3385 g.
m(CO₂) = m₁(sample) - m₂(sample).
m(CO₂) = 6,4734 g - 4,3385 g.
m(CO₂) = 2,1349 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 2,1349 g ÷ 44 g/mol.
n(CO₂) = 0,0485 mol.
From chemical reaction: n(CO₂) : n(CaCO₃) = 1 : 1.
n(CaCO₃) = 0,0485 mol.
m(CaCO₃) = 0,0485 mol · 100,1 g/mol.
m(CaCO₃) = 4,854 g.
percentage of calcium carbonate = 4,854 g ÷ 6,4734 g · 100%.
percentage of calcium carbonate = 75%.

It's not the same dearthe first one is in grams per mole while the other is not