Question

Determine the simplest formulas of the following compounds: (a) tetraethyl lead, the banned gasoline anti-knock additive, which is composed of \(29.71 \% \mathrm{C}, 6.234 \% \mathrm{H},\) and \(64.07 \% \mathrm{~Pb}\) (b) citric acid, present in most sour fruit, which is composed of \(37.51 \% \mathrm{C}, 4.20 \% \mathrm{H},\) and \(58.29 \% \mathrm{O}\) (c) cisplatin, a drug used in chemotherapy, which is composed of \(9.34 \% \mathrm{~N}, 2.02 \% \mathrm{H}, 23.36 \% \mathrm{Cl},\) and \(65.50 \% \mathrm{Pt}\)

Short answer

(a) Tetraethyl Lead: \(C_8 H_{20} Pb\) (b) Citric Acid: \(C_3 H_4 O_3\) (c) Cisplatin: \(N_2 H_6 Cl_2 Pt\)

Step-by-step solution

(a) Tetraethyl Lead

1. Assume a 100 g sample which means 29.71 g of Carbon (C), 6.234 g of Hydrogen (H), and 64.07 g of Lead (Pb) 2. Convert the mass into moles: moles of C = \(\frac{29.71}{12.01}\) = 2.475 moles moles of H = \(\frac{6.234}{1.008}\) = 6.185 moles moles of Pb = \(\frac{64.07}{207.2}\) = 0.309 moles 3. Divide each number of moles by the smallest moles (0.309): C ratio = \(\frac{2.475}{0.309}\) = 8 H ratio = \(\frac{6.185}{0.309}\) = 20 Pb ratio = \(\frac{0.309}{0.309}\) = 1 4. The simplest whole-number ratio is \(C_8 H_{20} Pb\)

(b) Citric Acid

1. Assume a 100 g sample which means 37.51 g of Carbon (C), 4.20 g of Hydrogen (H), and 58.29 g of Oxygen (O) 2. Convert the mass into moles: moles of C = \(\frac{37.51}{12.01}\) = 3.123 moles moles of H = \(\frac{4.20}{1.008}\) = 4.167 moles moles of O = \(\frac{58.29}{16.00}\) = 3.643 moles 3. Divide each number of moles by the smallest moles (3.123): C ratio = \(\frac{3.123}{3.123}\) = 1 H ratio = \(\frac{4.167}{3.123}\) = 1.33 ≈ 4/3 O ratio = \(\frac{3.643}{3.123}\) = 1.17 ≈ 1 4. Multiply each ratio by 3 to get the whole number ratio: \(C_3 H_4 O_3\)

(c) Cisplatin

1. Assume a 100 g sample which means 9.34 g of Nitrogen (N), 2.02 g of Hydrogen (H), 23.36 g of Chlorine (Cl), and 65.50 g of Platinum (Pt) 2. Convert the mass into moles: moles of N = \(\frac{9.34}{14.01}\) = 0.667 moles moles of H = \(\frac{2.02}{1.008}\) = 2.005 moles moles of Cl = \(\frac{23.36}{35.453}\) = 0.659 moles moles of Pt = \(\frac{65.50}{195.08}\) = 0.336 moles 3. Divide each number of moles by the smallest moles (0.336): N ratio = \(\frac{0.667}{0.336}\) = 1.98 ≈ 2 H ratio = \(\frac{2.005}{0.336}\) = 5.97 ≈ 6 Cl ratio = \(\frac{0.659}{0.336}\) = 1.96 ≈ 2 Pt ratio = \(\frac{0.336}{0.336}\) = 1 4. The simplest whole-number ratio is \(N_2 H_6 Cl_2 Pt\)