Determine the empirical formula of (a) the rodenticide (rat killer) warfarin, which consists of \(74.01 \% \mathrm{C}\) \(5.23 \% \mathrm{H},\) and \(20.76 \%\) O, by mass; (b) the antibacterial agent sulfamethizole, which consists of \(39.98 \%\) C. \(3.73 \% \mathrm{H}, 20.73 \% \mathrm{N}, 11.84 \% \mathrm{O},\) and \(23.72 \% \mathrm{S},\) by mass.

First, convert the given percentages to grams (since it is a percentage, it would be percentage out of 100 g). This gives us \(74.01 \mathrm{g}\) C, \(5.23 \mathrm{g}\) H, and \(20.76 \mathrm{g}\) O. Next, convert these amounts into moles using atomic masses of the elements. For C, \( \frac{74.01 g}{12.01 g/mol} = 6.16 \mathrm{mol} \). For H, \( \frac{5.23 g}{1.008 g/mol} = 5.18 \mathrm{mol} \). For O, \( \frac{20.76 g}{16.00 g/mol} = 1.30 \mathrm{mol} \). Now we divide by the smallest number of moles to get the empirical formula.This leads to: C \((\frac{6.16}{1.30} = 4.74)\), H \((\frac{5.18}{1.30} = 3.98)\), and O \((\frac{1.30}{1.30} = 1)\). Since these are all close to whole numbers, the empirical formula is thus C9H4O.

Now, repeat the same process for Sulfamethizole. The given percentages convert to \(39.98 \mathrm{g}\) C, \(3.73 \mathrm{g}\) H, \(20.73 \mathrm{g}\) N, \(11.84 \mathrm{g}\) O, and \(23.72 \mathrm{g}\) S. Next, convert these amounts into moles using atomic masses of the elements. For C, \( \frac{39.98 g}{12.01 g/mol} = 3.33 \mathrm{mol} \). For H, \( \frac{3.73 g}{1.008 g/mol} = 3.70 \mathrm{mol} \). For N, \( \frac{20.73 g}{14.007 g/mol} = 1.48 \mathrm{mol} \). For O, \( \frac{11.84 g}{16.00 g/mol} = 0.74 \mathrm{mol} \). For S, \( \frac{23.72 g}{32.06 g/mol} = 0.74 \mathrm{mol} \). Now divide all the moles by the smallest number of moles (0.74), leading to C \((\frac{3.33}{0.74} = 4.5)\), H \((\frac{3.70}{0.74} = 5)\), N \((\frac{1.48}{0.74} = 2)\), O \((\frac{0.74}{0.74} = 1)\), and S \((\frac{0.74}{0.74} = 1)\). Since the ratio for C is not a whole number, double all of them to get the empirical formula C9H10N4O2S2.