Question 4.22: Commercial pharmaceutical agents undergo extensive analysis ...
Commercial pharmaceutical agents undergo extensive analysis to establish how long they may be stored without degradation and loss of potency. A candidate drug has the following percent composition from combustion analysis: C, 62.50%; H, 4.20%. The drug is a diprotic acid. A standard tablet containing 325 mg of the drug is dissolved in 100.00 mL of water and titrated to the equivalence point with 16.45 mL of 0.2056 M NaOH(aq). After storage at 50°C under high humidity for one month, a second tablet, when dissolved in 50.00 mL of water, requires 10.10 mL of 0.1755 M NaOH(aq) to be completely neutralized.
a. What is the empirical formula of the drug?
b. What is its molar mass?
c. What is its molecular formula?
d. What is the percent of active drug substance remaining in the
tablet after storage?
Collect and Organize We are given the percent composition of the drug and information from titration experiments. From these data we can find the empirical formula, the molar mass, and the amount of active drug in the stored tablet.
Analyze The percent composition data from the combustion analysis do not add to 100%. We may assume that the amount missing is due to oxygen in the molecule. The drug is a diprotic acid, so neutralizing 1 mole of the drug requires 2 moles of NaOH. The titration of the drug requires about 20 mL of 0.2 M NaOH, which is about 0.004 mol OH^{-}. That means the pure sample is about 0.002 moles of the drug; if about 0.3 grams contains 0.002 moles, the drug must have a molar mass of about 150 g. If any of the drug decomposes upon storage, we should have less than 325 mg in the second sample.
Solve
a. The percent oxygen in the drug molecule is
100.00\% -(62.50\% + 4.20\% ) = 33.30 \%
Calculating the moles of C, H, and O in exactly 100 g of the drug:
\frac{62.50 g C}{12.01 g/mol}=5.20 mol C
\frac{4.20 g H}{1.008 g/mol}=4.17 mol H
\frac{33.30 g O}{16.00 g/mol}=2.08 mol O
Reducing these quantities to ratios of small whole numbers:
\frac{5.20 mol C}{2.08}=2.50 mol C
\frac{4.17 mol H}{2.08}=2.00 mol H
\frac{2.08 mol O}{2.08}=1.00 mol O
requires multiplying by 2, which gives us
5 mol C: 4 mol H:2 mol O
and the empirical formula: C_{5}H_{4}O_{2}.
b. To find the molar mass of the drug from the titration data, we first find the number of moles of drug in the sample:
0.01645 \sout{L NaOH} \times \frac{0.2056 \sout{mol NaOH}}{1 \sout{L NaOH}} \times \frac{1 mol drug}{2 \sout{mol NaOH}}
=0.001691 mol drug
If 0.325 g of drug is 0.001691 mol of drug, then the mass of 1 mole is
\frac{0.325 g}{0.001691 mol}=192.2 g/mol =\mathscr{M} _{drug}
c. The mass of 1 mole of empirical formula units (C_{5}H_{4}O_{2}) is
(5 × 12.01 g/mol) + (4 × 1.008 g/mol) + (2 × 16.00 g/mol)
= 96.08 g/mol
The number of moles of formula units per mole of molecules is
\frac{192.16 \frac{\sout{g}}{\sout{mol}}}{96.08 \frac{\sout{g}}{\sout{mol}}}=2
So the molecular formula of the drug is
(C_{5}H_{4}O_{2})_{2}=C_{10}H_{8}O_{4}
d. The stored tablet contains
0.01010 L NaOH \times \frac{0.1755 mol NaOH}{1 L NaOH} \times \frac{1 mol drug}{2 mol NaOH}
=0.0008863 mol drug = 8.863 \times 10^{-4} mol
Comparing this value with the original amount:
\frac{8.863 \times 10^{-4} mol}{1.691 \times 10^{-3} mol} \times 100 \% =52.41 \%
of the active drug is left in the tablet.
Think About It The molar mass we calculated is close to our estimated value, so it seems reasonable. If only half the active agent is present in the tablet after storage under these conditions, the manufacturer may have to reformulate the drug or package it differently to protect it from its surroundings.