Reacting Moles and Formula Mass Calculations

Consider the equation: CaCO3 \rightarrow CaO + CO2, where Ca = 40, C = 12, and O = 16. The relative formula mass of calcium carbonate (CaCO3) is calculated as follows: Mr(CaCO3) = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100. A student heated 50g of calcium carbonate. To find the number of moles (n), the formula n = \frac{m}{Mr} is used, where m is the mass in grams and Mr is the relative formula mass. For 50g of CaCO3, n = \frac{50}{100} = 0.5 moles. Therefore, the student heated 0.5 moles of calcium carbonate.

From the balanced equation, 1 mole of CaCO3 decomposes to produce 1 mole of CaO. Thus, if the student started with 0.5 moles of CaCO3, they would expect to have 0.5 moles of CaO at the end of the experiment. The relative formula mass of CaO is calculated as Mr(CaO) = 40 + 16 = 56. Using the formula m = n \times Mr, the mass of 0.5 moles of CaO is m = 0.5 \times 56 = 28g.

The relative formula mass of CO2 is calculated as Mr(CO2) = 12 + (2 \times 16) = 12 + 32 = 44. Since 0.5 moles of CaCO3 were heated, 0.5 moles of CO2 would be produced. The mass of CO2 is m = 0.5 \times 44 = 22g. The percent yield is calculated as: \% \text{ Yield} = (\frac{\text{Actual Yield}}{\text{Theoretical Yield}}) \times 100 where Actual Yield = 21g and Theoretical Yield = 28g. Therefore, \% \text{ Yield} = (\frac{21}{28}) \times 100 = 0.75 \times 100 = 75\%. Hence, the % yield of calcium oxide is 75%.