Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 91

Warfarin, \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4},\) is the active ingredient used in some anticoagulant medications. The amount of warfarin in a particular sample was determined as follows. A 13.96 g sample was first treated with an alkaline I_ solution to convert \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) to \(\mathrm{CHI}_{3}\). This treatment gives one mole of \(\mathrm{CHI}_{3}\) for every mole of \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) that was initially present in the sample. The iodine in \(\mathrm{CHI}_{3}\) is then precipitated as \(\mathrm{AgI}(\mathrm{s})\) by treatment with excess \(\mathrm{AgNO}_{3}(\mathrm{aq}):\) $$\begin{aligned} \mathrm{CHI}_{3}(\mathrm{aq})+3 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow & \longrightarrow 3 \mathrm{AgI}(\mathrm{s})+3 \mathrm{HNO}_{3}(\mathrm{aq}) &+\mathrm{CO}(\mathrm{g}) \end{aligned}$$ If \(0.1386 \mathrm{g}\) solid \(\mathrm{AgI}\) were obtained, then what is the percentage by mass of warfarin in the sample analyzed?

Short Answer

Expert verified
The percentage by mass of Warfarin in the analyzed sample is calculated following the four steps in the solution. Check calculations for the exact answer.

Step by step solution

01

Calculation of moles of AgI

Starting with the weight of the AgI solid obtained, 0.1386 g, calculate the number of moles it represents using the molecular weight of AgI (234.77 g/mol). The formula is: Mole = Mass/Molecular Weight.
02

Calculation of moles of Warfarin

Knowing the stoichiometric relationships between AgI and Warfarin, for every 3 moles of AgI formed, one mole of Warfarin was used. Thus, divide the moles of AgI by 3 to determine the moles of Warfarin.
03

Calculation of mass of Warfarin

Convert the moles of Warfarin back to grams using the molecular weight of Warfarin (308.33 g/mol). The formula used here is: Mass = Mole * Molecular Weight.
04

Calculation of percentage by mass of Warfarin

Finally, the percentage by mass of Warfarin in the sample can be calculated by dividing the mass of Warfarin by the mass of the sample and multiplying by 100.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) $\mathrm{Ca}^{2+}+2 \mathrm{I}^{-}+2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \longrightarrow$ (b) $\mathrm{Ba}^{2+}+\mathrm{S}^{2-}+2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow$

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) \(\mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}+\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow\) (b) \(\mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \longrightarrow\) (c) \(\operatorname{FeS}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow\) (d) \(\mathrm{K}^{+}+\mathrm{HCO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow\) (e) \(\mathrm{Mg}(\mathrm{s})+\mathrm{H}^{+} \longrightarrow\)

What volume of \(0.0962 \mathrm{M} \mathrm{NaOH}\) is required to exactly neutralize \(10.00 \mathrm{mL}\) of \(0.128 \mathrm{M} \mathrm{HCl} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free