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Chapter 14: Problem 158

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives $\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{N}_{2} \mathrm{O}_{10} \mathrm{S}$ as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise \(104 . )\) Why is codeine sulfate used instead of codeine?

Short Answer

Expert verified
Codeine sulfate has the molecular formula \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{N}_{2} \mathrm{O}_{10} \mathrm{S}\) and comprises approximately 67.21% carbon, 6.56% hydrogen, 3.63% nitrogen, 21.77% oxygen, and 0.83% sulfur by weight. It is used instead of codeine due to its enhanced stability and water solubility, which allows for better absorption in the body and improved shelf life.

Step by step solution

01

Calculate molar mass of codeine sulfate

To find the weight percentages of the elements, we first need to calculate the molar mass of codeine sulfate. To do this, we'll need the molar mass of each element (approximate values): C: 12.01 g/mol H: 1.01 g/mol N: 14.01 g/mol O: 16.00 g/mol S: 32.06 g/mol Now we can calculate the molar mass of codeine sulfate: \(Molar\,mass = 36 \cdot C + 44 \cdot H + 2 \cdot N + 10 \cdot O + S\) \(Molar\,mass = 36 \cdot 12.01 + 44 \cdot 1.01+ 2 \cdot 14.01 + 10 \cdot 16.00 + 32.06\)
02

Calculate the weight percentages of each element

Now that we know the molar mass of codeine sulfate, we can calculate the weight percentages of each element: Weight percentage of C: \[\frac{36 \cdot 12.01}{Molar\,mass} \times 100\%\] Weight percentage of H: \[\frac{44 \cdot 1.01}{Molar\,mass} \times 100\%\] Weight percentage of N: \[\frac{2 \cdot 14.01}{Molar\,mass} \times 100\%\] Weight percentage of O: \[\frac{10 \cdot 16.00}{Molar\,mass} \times 100\%\] Weight percentage of S: \[\frac{32.06}{Molar\,mass} \times 100\%\]
03

Discuss the advantages of codeine sulfate

Codeine sulfate is used instead of codeine due to its stability and water solubility. The sulfate form is more soluble in water, allowing it to be more readily absorbed in the body. Additionally, the sulfate salt offers improved stability, a desirable trait for pharmaceutical products.

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Most popular questions from this chapter

One mole of a weak acid HA was dissolved in 2.0 \(\mathrm{L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.45 \(\mathrm{M} .\) Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HA}\) .

At \(25^{\circ} \mathrm{C},\) a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

Consider 1000 . mL of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a $K_{\text { a value equal to } 1.00 \times 10^{-4} . \text { How much }}$ water was added or removed (by evaporation) so that a solution remains in which 25.0\(\%\) of \(\mathrm{HA}\) is dissociated at equilibrium? Assume that HA is nonvolatile.

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},$ assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

Calculate the pH of a \(0.010-M\) solution of iodic acid (HIO $_{3}, K_{\mathrm{a}}\( \)=0.17 )$
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