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Chapter 4: Problem 93

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{~N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, \(0.104 \mathrm{~g}\) silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

Short Answer

Expert verified
The mass percent of Chlorisondamine chloride in the medication is approximately 4.40%.

Step by step solution

01

Calculate the molar mass of Chlorisondamine chloride and Silver Chloride

Using the periodic table, we can find the molar mass of each element present in Chlorisondamine chloride and Silver Chloride. For Chlorisondamine chloride, \(\mathrm{C_{14}H_{20}Cl_{6}N_{2}}\), the molar mass is: \[ 14 \times \mathrm{M_C} + 20 \times \mathrm{M_H} + 6 \times \mathrm{M_{Cl}} + 2 \times \mathrm{M_N} = 14(12.01) + 20(1.01) + 6(35.45) + 2(14.01) \approx 465.49 \, \mathrm{g/mol} \]Where \(\mathrm{M_X}\) represents the molar mass of the element X. For Silver Chloride, \(\mathrm{AgCl}\), the molar mass is: \[ \mathrm{M_{Ag}} + \mathrm{M_{Cl}} = 107.87 + 35.45 \approx 143.32 \, \mathrm{g/mol} \]
02

Find the moles of chloride ions in the sample

From the data given, we have 0.104 g of silver chloride (AgCl) obtained after the reaction. To find the moles of chloride ions, we will first find the moles of silver chloride and then relate it back to the chloride ions. Moles of Silver Chloride = \(\frac{\text{Mass}}{\text{Molar Mass}}\) Moles of Silver Chloride = \(\frac{0.104 \, \mathrm{g}}{143.32 \, \mathrm{g/mol}} \approx 7.25 \times 10^{-4} \, \mathrm{mol}\) Since Silver Chloride has a 1:1 relationship with chloride ions, the moles of chloride ions will also be equal to the moles of Silver Chloride: Moles of Chloride ions = \(7.25 \times 10^{-4} \, \mathrm{mol}\)
03

Find the moles of Chlorisondamine chloride

Since there are 6 chloride ions in each Chlorisondamine chloride molecule, we can divide the moles of chloride ions by 6 to find the moles of Chlorisondamine chloride: Moles of Chlorisondamine chloride = \(\frac{\text{Moles of Chloride Ions}}{6}\) Moles of Chlorisondamine chloride = \(\frac{7.25 \times 10^{-4} \, \mathrm{mol}}{6} \approx 1.21 \times 10^{-4} \, \mathrm{mol}\)
04

Calculate the mass of Chlorisondamine chloride in the sample

Now, we will multiply the moles of Chlorisondamine chloride by its molar mass to get the mass of Chlorisondamine chloride present in the sample: Mass of Chlorisondamine chloride = \(\text{Moles} \times \text{Molar Mass}\) Mass of Chlorisondamine chloride = \(1.21 \times 10^{-4} \, \mathrm{mol} \times 465.49 \, \mathrm{g/mol} \approx 0.0563 \, \mathrm{g}\)
05

Calculate the mass percent of Chlorisondamine chloride in the medication

Finally, we can find the mass percent of Chlorisondamine chloride in the medication by using the following formula: Mass Percent = \(\frac{\text{Mass of Chlorisondamine chloride}}{\text{Mass of Medication Sample}} \times 100\) Mass Percent = \(\frac{0.0563 \, \mathrm{g}}{1.28 \, \mathrm{g}} \times 100 \approx 4.40 \% \) The mass percent of Chlorisondamine chloride in the medication is approximately 4.40%.

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Most popular questions from this chapter

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of \(3.0 \mathrm{M}\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{array}{l} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{array} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 M\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl}\) ?

An average human being has about \(5.0 \mathrm{~L}\) of blood in his or her body. If an average person were to eat \(32.0 \mathrm{~g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{~g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

The soda you are drinking contains \(0.5 \%\) by mass sodium benzoate as a preservative. What approximate mass of sodium benzoate is contained in \(1.00 \mathrm{~L}\) of the solution assuming that the density of the soda is \(1.00 \mathrm{~g} / \mathrm{mL}\) (the approximate density of water)?

A \(100.0-\mathrm{mL}\) aliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 M\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.
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