Question

A sample of cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{O}_{4} \mathrm{~N},\) is diluted with sugar, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\) When a \(1.00-\mathrm{mg}\) sample of this mixture is burned, \(1.00 \mathrm{~mL}\) of carbon dioxide \((d=1.80 \mathrm{~g} / \mathrm{L})\) is formed. What is the percentage of cocaine in this mixture?

Short answer

Based on the information provided and the step-by-step solution, calculate the percentage of cocaine in a mixture. Answer: The percentage of cocaine in the mixture is 68.3%.

Step-by-step solution

Calculate the mass of carbon dioxide produced

We are given the volume of carbon dioxide produced (1.00 mL) and its density (1.80 g/L). To find the mass, we need to convert the volume to liters and multiply by the density: $$ \text{Mass of CO}_2 = 1.00 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \times 1.80 \frac{\text{g}}{\text{L}} = 0.0018\, \text{g} $$

Calculate the moles of carbon dioxide formed

Next, we'll find the moles of carbon dioxide formed. We'll need the molar mass of carbon dioxide (44.01 g/mol): $$ \text{Moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{0.0018 \, \text{g}}{44.01 \frac{\text{g}}{\text{mol}}} = 4.09 \times 10^{-5}\, \text{mol} $$

Calculate the moles of carbon in the original mixture

Since every mole of carbon dioxide has one mole of carbon, there were also 4.09 x 10^-5 mol of carbon in the mixture: $$\text{Moles of C in the mixture} = 4.09 \times 10^{-5}\, \text{mol} $$

Calculate the mass of carbon in the original mixture

To find the mass of carbon in the mixture, we multiply the moles of carbon by the molar mass of carbon (12.01 g/mol): $$ \text{Mass of C in the mixture} = 4.09 \times 10^{-5} \, \text{mol} \times 12.01\, \frac{\text{g}}{\text{mol}} = 4.91 \times 10^{-4}\, \text{g} $$

Calculate the mass of cocaine in the mixture and the percentage of cocaine

The mass of the carbon from cocaine must be equal to the mass of carbon in the mixture (4.91 x 10^-4 g). Since cocaine has a formula of \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{O}_{4} \mathrm{N}\), the percentage of carbon in cocaine can be calculated as follows: $$ \text{% C in cocaine} = \frac{17 \times 12.01}{(17 \times 12.01) + (21 \times 1.01) + (4 \times 16.00) + 14.01} \times 100 = 72.07\% $$ Now we can find the mass of cocaine in the mixture: $$ \text{Mass of cocaine} = \frac{\text{Mass of C in the mixture}}{\text{% C in cocaine}} = \frac{4.91 \times 10^{-4}\, \text{g}}{0.7207} = 6.83 \times 10^{-4}\, \text{g} $$ Finally, we can find the percentage of cocaine in the 1.00 mg mixture: $$ \text{Percentage of cocaine} = \frac{6.83 \times 10^{-4} \, \text{g}}{1.00 \times 10^{-3} \, \text{g}} \times 100 = 68.3\% $$ The percentage of cocaine in the mixture is 68.3%.