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 $m = 0.5 \times 56 = 28$g.

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 $m = 0.5 \times 44 = 22$g. 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%.