Chapter 10, Problem 87A

Interpretation Introduction

Interpretation: The number of fluorine atoms in a 150 mg tablet of given antibacterial drug needs to be determined.

Concept Introduction: The relation between the number of moles, mass, and molar mass of a substance is as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Also, according to Avogadro’s law, in 1 mol of a substance, there are $6.022\times {10}^{23}$ units of that substance.

Explanation of Solution

The molecular formula of the given antibacterial drug is ${\text{C}}_{17}{\text{H}}_{18}{\text{FN}}_3{\text{O}}_3$ . The mass of the tablet of this drug is given 150 mg.

The molar mass of the drug can be calculated from the atomic masses of elements present in the formula. Here, the atomic number of C, H, F, N, and O is 12 g/mol, 1 g/mol, 19 g/mol, 14 g/mol, and 16 g/mol respectively.

Therefore, the molar mass of the drug will be:

  $\begin{array}{c}{\text{C}}_{17}{\text{H}}_{18}{\text{FN}}_3{\text{O}}_3=17\left(12\text{ g/mol}\right)+18\left(1\text{ g/mol}\right)+19\text{ g/mol+3}\left(14\text{ g/mol}\right)+3\left(16\text{ g/mol}\right)\\ =\left(204+18+19+42+48\right)\text{ g/mol}\\ =331\text{ g/mol}\end{array}$

Thus, the molar mass of the drug is $331\text{ g/mol}$ .

Now, the number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Substitute the values,

  $\begin{array}{c}n=\frac{150\text{ mg}\left(\frac{{10}^{-3}\text{ g}}{1\text{ mg}}\right)}{331\text{ g/mol}}\\ =4.53\times {10}^{-4}\text{ mol}\end{array}$

Therefore, the number of moles is $4.53\times {10}^{-4}\text{ mol}$ .

Now, in 1 mol of a drug, there is 1 mol of F; thus, the number of moles of F in the given tablet is $4.53\times {10}^{-4}\text{ mol}$ .

According to Avogadro’s number, 1 mol of a substance gives $6.022\times {10}^{23}$ units of that substance. Therefore, in $4.53\times {10}^{-4}\text{ mol}$ of drug or fluorine, the number of atoms of fluorine will be:

  $\begin{array}{c}n=\left(\text{4}{\text{.53×10}}^{\text{-4}}\text{ mol}\right)\left(\frac{6.022\times {10}^{23}}{1\text{ mol}}\right)\\ =27.28\times {10}^{-4+23}\\ =27.28\times {10}^{19}\\ =2.728\times {10}^{20}\end{array}$

Thus, the number of atoms of fluorine is $2.728\times {10}^{20}$ .