To determine the empirical formula of a compound containing 60.3% magnesium (Mg) and 39.7% oxygen (O), we follow these steps:
1. Assume a 100 g Sample:
- Magnesium: [tex]\( 60.3 \% \)[/tex] of [tex]\( 100 \)[/tex] g = [tex]\( 60.3 \)[/tex] g
- Oxygen: [tex]\( 39.7 \% \)[/tex] of [tex]\( 100 \)[/tex] g = [tex]\( 39.7 \)[/tex] g
2. Convert Masses to Moles:
- The molar mass of magnesium (Mg) is [tex]\( 24.305 \)[/tex] g/mol.
- The molar mass of oxygen (O) is [tex]\( 16.00 \)[/tex] g/mol.
- Moles of Mg:
[tex]\( \frac{60.3 \text{ g}}{24.305 \text{ g/mol}} \approx 2.481 \text{ moles} \)[/tex]
- Moles of O:
[tex]\( \frac{39.7 \text{ g}}{16.00 \text{ g/mol}} \approx 2.481 \text{ moles} \)[/tex]
3. Determine the Simplest Whole Number Ratio:
- Ratio of moles of Mg to moles of Mg:
[tex]\( \frac{2.481}{2.481} = 1.0 \)[/tex]
- Ratio of moles of O to moles of Mg:
[tex]\( \frac{2.481}{2.481} = 1.0 \)[/tex]
4. Simplify the Ratios to Smallest Whole Numbers:
The mole ratio [tex]\( 1.0: 1.0 \)[/tex] suggests that for every 1 mole of magnesium, there is 1 mole of oxygen. Thus, the simplest whole number ratio of Mg to O is [tex]\(1:1\)[/tex].
5. Write the Empirical Formula:
Based on the ratios, the empirical formula of the compound is [tex]\( \text{MgO} \)[/tex].
Therefore, the empirical formula of the compound is [tex]\( \text{MgO} \)[/tex].
