25. A compound contains 40% calcium, 12% carbon, and 48% orygen by mass. What is the empincal formula of this compound? CaCO_(2) CaC_(2)O_(4) CaCO_(3) CaC_(3)O_(6)

To determine the empirical formula of the compound, we need to convert the given percentages into moles and find the simplest whole number ratio of the elements.<br /><br />1. **Convert percentages to grams**: Assume we have 100 grams of the compound. This means we have:<br /> - 40 grams of calcium (Ca)<br /> - 12 grams of carbon (C)<br /> - 48 grams of oxygen (O)<br /><br />2. **Convert grams to moles**:<br /> - Moles of Ca = \(\frac{40 \text{ g}}{40.08 \text{ g/mol}} \approx 0.998 \text{ mol}\)<br /> - Moles of C = \(\frac{12 \text{ g}}{12.01 \text{ g/mol}} \approx 0.999 \text{ mol}\)<br /> - Moles of O = \(\frac{48 \text{ g}}{16.00 \text{ g/mol}} = 3.00 \text{ mol}\)<br /><br />3. **Find the simplest whole number ratio**:<br /> - Divide each by the smallest number of moles calculated:<br /> - Ca: \( \frac{0.998}{0.998} = 1 \)<br /> - C: \( \frac{0.999}{0.998} \approx 1 \)<br /> - O: \( \frac{3.00}{0.998} \approx 3 \)<br /><br />The simplest whole number ratio is approximately 1:1:3 for Ca:C:O.<br /><br />Therefore, the empirical formula of the compound is \(\mathrm{CaCO}_{3}\).