Question

Beta-blockers are a class of drug widely used to manage hypertension. Atenolol, a beta-blocker, is made up of carbon, hydrogen, oxygen, and nitrogen atoms. When a 5.000 -g sample is burned in oxygen, \(11.57 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(3.721 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are obtained. A separate experiment using the same mass of sample \((5.000 \mathrm{~g})\) shows that atenolol has \(10.52 \%\) nitrogen. What is the simplest formula of atenolol?

Short answer

Based on the given information and calculations, the simplest formula of atenolol is C\(_{51}\)H\(_{63}\)ON\(_6\).

Step-by-step solution

Determine the amount of carbon in the compound

First, let's determine the amount of carbon in atenolol based on the amount of \(\mathrm{CO}_{2}\) produced. From the balanced chemical equation of combustion, we know that for every mole of carbon in the compound, one mole of \(\mathrm{CO}_{2}\) is produced. Use the molar mass of \(\mathrm{CO}_2\) to calculate the amount of carbon present in the atenolol sample. $$ \text{Amount of carbon in atenolol} = 11.57 \mathrm{~g~CO_2} \times \frac{1 \mathrm{~mol~C}}{1 \mathrm{~mol~CO_2}} \times \frac{12.01 \mathrm{~g~C}}{44.01 \mathrm{~g~CO_2}} = 3.959 \mathrm{~g~C} $$

Determine the amount of hydrogen in the compound

Now, let's determine the amount of hydrogen in atenolol based on the amount of \(\mathrm{H}_{2} \mathrm{O}\) produced. From the balanced chemical equation of combustion, we know that for every two moles of hydrogen in the compound, one mole of \(\mathrm{H}_{2} \mathrm{O}\) is produced. Use the molar mass of \(\mathrm{H}_2\mathrm{O}\) to calculate the amount of hydrogen present in the atenolol sample. $$ \text{Amount of hydrogen in atenolol} = 3.721 \mathrm{~g~H_2O} \times \frac{2 \mathrm{~mol~H}}{1 \mathrm{~mol~H_2O}} \times \frac{1.01 \mathrm{~g~H}}{18.02 \mathrm{~g~H_2O}} = 0.412 \mathrm{~g~H} $$

Determine the amount of nitrogen in the compound

From the given information, we know that atenolol contains 10.52% nitrogen. Calculate the amount of nitrogen in the 5.000g atenolol sample. $$ \text{Amount of nitrogen in atenolol} = 5.000 \mathrm{~g~atenolol} \times 0.1052 = 0.526 \mathrm{~g~N} $$

Determine the amount of oxygen in the compound

To find the amount of oxygen in atenolol, subtract the amounts of carbon, hydrogen, and nitrogen from the total mass of atenolol. $$ \text{Amount of oxygen in atenolol} = 5.000 \mathrm{~g~atenolol} - (3.959 \mathrm{~g~C} + 0.412 \mathrm{~g~H} + 0.526 \mathrm{~g~N}) = 0.103 \mathrm{~g~O} $$

Convert the amounts of elements to moles

Use the molar masses of carbon, hydrogen, oxygen, and nitrogen to convert the amounts of elements to moles. $$ \text{Moles of C} = \frac{3.959 \mathrm{~g~C}}{12.01 \mathrm{~g/mol}} = 0.3298 \mathrm{~mol} $$ $$ \text{Moles of H} = \frac{0.412 \mathrm{~g~H}}{1.01 \mathrm{~g/mol}} = 0.408 \mathrm{~mol} $$ $$ \text{Moles of O} = \frac{0.103 \mathrm{~g~O}}{16.00 \mathrm{~g/mol}} = 0.00644 \mathrm{~mol} $$ $$ \text{Moles of N} = \frac{0.526 \mathrm{~g~N}}{14.01 \mathrm{~g/mol}} = 0.03757 \mathrm{~mol} $$

Determine the simplest formula

To determine the simplest formula, divide all the mole values by the smallest mole value. $$ \text{Ratio of C} = \frac{0.3298}{0.00644} \approx 51 $$ $$ \text{Ratio of H} = \frac{0.408}{0.00644} \approx 63 $$ $$ \text{Ratio of O} = \frac{0.00644}{0.00644} = 1 $$ $$ \text{Ratio of N} = \frac{0.03757}{0.00644} \approx 6 $$ The empirical formula is C51H63O1N6 which can be written as C\(_{51}\)H\(_{63}\)ON\(_6\).