What amount (moles) is represented by each of these samples? a. 20.0 mg caffeine, \(C_{8} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{O}_{2}\) b. \(2.72 \times 10^{21}\) molecules of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) c. \(1.50 \mathrm{g}\) of dry ice, \(\mathrm{CO}_{2}\)

Use molecular weights for each element and add them to find the molar mass of each compound: Molar mass of caffeine: \(C_{8}H_{10}N_{4}O_{2} = 8 \times 12.01 + 10 \times 1.01 + 4 \times 14.01 + 2 \times 16.00 = 194.19 \,\mathrm{g/mol}\) Molar mass of ethanol: \(C_{2}H_{5}OH = 2 \times 12.01 + 6 \times 1.01 + 1 \times 16.00 = 46.07 \,\mathrm{g/mol}\) Molar mass of carbon dioxide: \(CO_{2} = 1 \times 12.01 + 2 \times 16.00 = 44.01 \,\mathrm{g/mol}\)

For each sample, divide the mass (or number of molecules for ethanol) by the corresponding molar mass or Avogadro's number: a. Moles of caffeine: \(20.0 \,\mathrm{mg} \, \times \, \frac{1 \,\mathrm{g}}{1000 \,\mathrm{mg}} \, \times \, \frac{1 \,-\! \mathrm{mol \, caffeine}}{194.19 \,\mathrm{g}} = 1.03 \times 10^{-4} \,\mathrm{mol \, caffeine}\) b. Moles of ethanol: \(\frac{2.72 \times 10^{21} \, \mathrm{molecules}}{6.022 \times 10^{23} \, \mathrm{molecules/mol}} = 4.52 \times 10^{-3} \,\mathrm{mol \, ethanol}\) c. Moles of dry ice: \(\frac{1.50 \,\mathrm{g}}{44.01 \,\mathrm{g/mol}} = 0.0341 \,\mathrm{mol \, CO_{2}}\) So, the amounts of each sample in moles are: a. \(1.03 \times 10^{-4} \,-\! \mathrm{mol \, caffeine}\) b. \(4.52 \times 10^{-3} \,-\! \mathrm{mol \, ethanol}\) c. \(0.0341 \,-\! \mathrm{mol \, CO_{2}}\)