What is the empirical formula of a molecule that is 40% Oxygen and 60% Magnesium? Mg_(2)O Mg_(3)O MgO_(2) MgO

To determine the empirical formula of a compound given the percentage composition, we need to follow these steps:<br /><br />1. **Convert the percentage composition to grams**:<br /> - Assume we have 100 grams of the compound.<br /> - Oxygen: \( 40\% \) of 100 grams = 40 grams<br /> - Magnesium: \( 60\% \) of 100 grams = 60 grams<br /><br />2. **Convert grams to moles**:<br /> - Use the molar masses of the elements:<br /> - Molar mass of Oxygen (O) = 16.00 g/mol<br /> - Molar mass of Magnesium (Mg) = 24.31 g/mol<br /><br /> Calculate the moles of each element:<br /> - Moles of Oxygen: \( \frac{40 \text{ g}}{16.00 \text{ g/mol}} = 2.5 \text{ moles} \)<br /> - Moles of Magnesium: \( \frac{60 \text{ g}}{24.31 \text{ g/mol}} \approx 2.47 \text{ moles} \)<br /><br />3. **Determine the simplest whole number ratio**:<br /> - Divide each mole value by the smallest number of moles calculated:<br /> - Ratio for Oxygen: \( \frac{2.5}{2.47} \approx 1.01 \)<br /> - Ratio for Magnesium: \( \frac{2.47}{2.47} = 1 \)<br /><br />4. **Write the empirical formula**:<br /> - The ratio of Magnesium to Oxygen is approximately 1:1.<br /><br />Therefore, the empirical formula of the molecule is \( \text{MgO} \).<br /><br />So, the correct answer is:<br />\[ \boxed{\text{MgO}} \]