Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(42.1 \% \mathrm{Na}, 18.9 \% \mathrm{P}\), and \(39.0 \% \mathrm{O}\) (b) \(18.7 \% \mathrm{Li}, 16.3 \% \mathrm{C},\) and \(65.0 \% \mathrm{O}\) (c) \(60.0 \% \mathrm{C}, 4.4 \% \mathrm{H},\) and the remainder \(\mathrm{O}\)

For this compound, we are given the mass percentages as follows: 42.1% Na, 18.9% P, 39.0% O Assuming a 100g sample, we can convert these percentages to grams: 42.1g Na, 18.9g P, 39.0g O Now we can convert the grams to moles using the molar mass of each element: \(42.1 \, \text{g Na} \times \frac{1 \, \text{mol Na}}{22.99 \, \text{g Na}} = 1.83 \, \text{mol Na}\) \(18.9 \, \text{g P} \times \frac{1 \, \text{mol P}}{30.97 \, \text{g P}} = 0.610 \, \text{mol P}\) \(39.0 \, \text{g O} \times \frac{1 \, \text{mol O}}{16.00 \, \text{g O}} = 2.44 \, \text{mol O}\) Now we divide each mole value by the smallest mole value: \( \frac{1.83 \, \text{mol Na}}{0.610 \, \text{mol}} = 3\) \( \frac{0.610 \, \text{mol P}}{0.610 \, \text{mol}} = 1\) \( \frac{2.44 \, \text{mol O}}{0.610 \, \text{mol}} = 4\) The empirical formula is Na₃P₁O₄ or Na₃PO₄.

For this compound, we are given the mass percentages as follows: 18.7% Li, 16.3% C, 65.0% O Assuming a 100g sample, we convert these percentages to grams: 18.7g Li, 16.3g C, 65.0g O Now we convert the grams to moles using the molar mass of each element: \(18.7 \, \text{g Li} \times \frac{1 \, \text{mol Li}}{6.94 \, \text{g Li}} = 2.69 \, \text{mol Li}\) \(16.3 \, \text{g C} \times \frac{1 \, \text{mol C}}{12.01 \, \text{g C}} = 1.36 \, \text{mol C}\) \(65.0 \, \text{g O} \times \frac{1 \, \text{mol O}}{16.00 \, \text{g O}} = 4.06 \, \text{mol O}\) Now we divide each mole value by the smallest mole value: \( \frac{2.69 \, \text{mol Li}}{1.36 \, \text{mol}} = 2\) \( \frac{1.36 \, \text{mol C}}{1.36 \, \text{mol}} = 1\) \( \frac{4.06 \, \text{mol O}}{1.36 \, \text{mol}} = 3\) The empirical formula is Li₂C₁O₃ or Li₂CO₃.

For this compound, we are given the mass percentages as follows: 60.0% C, 4.4% H, The remainder is O, so 100 - 60 - 4.4 = 35.6% O Assuming a 100g sample, we convert these percentages to grams: 60.0g C, 4.4g H, 35.6g O Now we convert the grams to moles using the molar mass of each element: \(60.0 \, \text{g C} \times \frac{1 \, \text{mol C}}{12.01 \, \text{g C}} = 5.00 \, \text{mol C}\) \(4.4 \, \text{g H} \times \frac{1 \, \text{mol H}}{1.01 \, \text{g H}} = 4.36 \, \text{mol H}\) \(35.6 \, \text{g O} \times \frac{1 \, \text{mol O}}{16.00 \, \text{g O}} = 2.23 \, \text{mol O}\) Now we divide each mole value by the smallest mole value: \( \frac{5.00 \, \text{mol C}}{2.23 \, \text{mol}} = 2.24 ≈ 2\) \( \frac{4.36 \, \text{mol H}}{2.23 \, \text{mol}} = 1.96 ≈ 2\) \( \frac{2.23 \, \text{mol O}}{2.23 \, \text{mol}} = 1\) The empirical formula is C₂H₂O₁ or C₂H₂O.