Question

A sample of limestone and other soil materials was heated, and the limestone decomposed to give calcium oxide and carbon dioxide. CaCO3(s) → CaO(s) + CO2(g) A 6.498 g sample of limestone-containing material gave 2.44 g of CO2, in addition to CaO, after being heated at a high temperature. What was the mass percent of CaCO3 in the original sample? ___%

Answer 1

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = 2.44 g

mol of CO2 = (mass)/(molar mass)

= 2.44/44.01

= 0.0554 mol

From balanced chemical reaction, we see that

when 1 mol of CO2 forms, 1 mol of CaCO3 is reacts

mol of CaCO3 reacts = moles of CO2

= 0.0554 mol

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = number of mol * molar mass

= 0.0554*100.09

= 5.5492 g

mass % of CaCO3 = mass of CaCO3 * 100 / mass of sample

= 5.5492 * 100 / 6.498

= 85.4 %

Answer: 85.4 %