Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(10.4 \% \mathrm{C}, 27.8 \% \mathrm{~S},\) and \(61.7 \% \mathrm{Cl}\) (b) \(21.7 \% \mathrm{C}, 9.6 \% \mathrm{O},\) and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al},\) and the remainder \(\mathrm{F}\)

: Assume a total mass of 100g for each compound, as it will make it easier to work with the given mass percentages. This means each mass percentage becomes the mass in grams for each element. (a) 10.4 g C, 27.8 g S, 61.7 g Cl (b) 21.7 g C, 9.6 g O, 68.7 g F (c) 32.79 g Na, 13.02 g Al, and the remainder is F

: Use the molar mass of each element to convert the grams to moles. Molar mass of C = 12.01 g/mol, S = 32.07 g/mol, Cl = 35.45 g/mol, Na = 22.99 g/mol, Al = 26.98 g/mol, O = 16.00 g/mol, F = 19.00 g/mol. (a) \( \frac{10.4\,g}{12.01\,g/mol} = 0.866\,mol\) C, \( \frac{27.8\,g}{32.07\,g/mol} = 0.867\,mol\) S, \( \frac{61.7\,g}{35.45\,g/mol} = 1.74\,mol\) Cl (b) \( \frac{21.7\,g}{12.01\,g/mol} = 1.81\,mol\) C, \( \frac{9.6\,g}{16.00\,g/mol} = 0.600\,mol\) O, \( \frac{68.7\,g}{19.00\,g/mol} = 3.62\,mol\) F (c) \( \frac{32.79\,g}{22.99\,g/mol} = 1.425\,mol\) Na, \( \frac{13.02\,g}{26.98\,g/mol} = 0.4828\,mol\) Al,_remaining mass = 100 g - 32.79 g - 13.02 g = 54.19 g F, \( \frac{54.19\,g}{19.00\,g/mol} = 2.852\,mol\) F

: Divide each of the moles calculated in step 2 by the smallest mole value in each compound. (a) \(\frac{0.866}{0.866} = 1\) C, \(\frac{0.867}{0.866} \approx 1\) S, \(\frac{1.74}{0.866} \approx 2\) Cl (b) \(\frac{1.81}{0.600} \approx 3\) C, \(\frac{0.600}{0.600} = 1\) O, \(\frac{3.62}{0.600} \approx 6\) F (c) \(\frac{1.425}{0.4828} \approx 3\) Na, \(\frac{0.4828}{0.4828} = 1\) Al, \(\frac{2.852}{0.4828} \approx 6\) F

: Combine the mole ratios obtained in step 3 to form the empirical formula for each compound. (a) CS₂ (b) C₃OF₆ (C) Na₃AlF₆ So, the empirical formulas for the compounds are (a) CS₂, (b) C₃OF₆, and (c) Na₃AlF₆.