Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

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

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is the heart of comparing quantities in chemical reactions. At its core is the concept of "mole ratios," which are derived from balanced chemical equations. When we look at the equation for forming AgI from CHI₃, stoichiometry tells us that for every mole of CHI₃ reacted, 3 moles of AgI are formed.
This relationship is crucial in solving problems such as determining the yield of a product or the consumption of a reactant. In the exercise, we rely on stoichiometry to convert between the moles of AgI and warfarin. Given the direct conversion shown in the exercise, we use the precise mole ratio from the balanced equation to translate moles obtained through measurement (like measuring the mass of AgI) back to the original moles of warfarin, allowing us to calculate the percentage mass.
  • Understand that stoichiometry is about the conversion of quantities between reactants and products.
  • Always use balanced equations to obtain the correct mole ratios.
  • This allows precise calculation from known masses to unknown quantities needed.
Chemical Reactions
Chemical reactions are processes where substances are transformed into different substances. In our exercise, the reaction involves converting warfarin to CHI₃ and subsequently to AgI. This specific transformation relies on a two-step mechanism involving conversion by an alkaline I₋ solution first and then treatment with AgNO₃.
The significance of understanding this reaction lies in understanding how the composition of substances changes and which substances are involved at each step. In such reactions:
  • Identify the reactants and products.
  • Understand the conditions required for the reaction, like alkaline solutions.
  • Recognize the intermediates, in this case, CHI₃, which serve as stepping stones to final products.
It's these transformations that allow us to understand how initial reactants, like warfarin, eventually form measurable, quantifiable outputs like AgI.
Mole Calculations
Mole calculations are an essential tool in chemistry for converting between the mass of a substance and the number of particles it represents, using the concept of the mole. In our task, we begin with the mass of AgI, which we convert into moles by dividing by AgI's molar mass.
This conversion is straightforward thanks to Avogadro's number, allowing large numbers of atoms to be expressed through mole quantities rather than individual counts. The importance of mole calculations is highlighted in several key steps:
  • Calculating moles of AgI: Mass divided by molar mass gives the number of moles.
  • Utilizing stoichiometry to assess moles of warfarin consumed in conversion.
  • Reversing the calculation to find mass from moles, using warfarin's molar mass.
Mole calculations, as shown, integrate seamlessly with stoichiometric analysis and chemical equations to reveal essential chemical insights and quantity relationships.

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

In this chapter, we described an acid as a substance capable of producing \(\mathrm{H}^{+}\) and a salt as the ionic compound formed by the neutralization of an acid by a base. Write ionic equations to show that sodium hydrogen sulfate has the characteristics of both a salt and an acid (sometimes called an acid salt).

Which solutions would you use to precipitate \(\mathrm{Mg}^{2+}\) from an aqueous solution of \(\mathrm{MgCl}_{2} ?\) Explain your choice. (a) \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (b) \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq});\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(\mathrm{aq})\).

A \(110.520 \mathrm{g}\) sample of mineral water is analyzed for its magnesium content. The \(\mathrm{Mg}^{2+}\) in the sample is first precipitated as \(\mathrm{MgNH}_{4} \mathrm{PO}_{4},\) and this precipitate is then converted to \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7},\) which is found to weigh 0.0549 g. Express the quantity of magnesium in the sample in parts per million (that is, in grams of \(\mathrm{Mg}\) per million grams of \(\mathrm{H}_{2} \mathrm{O}\) ).

Express the following data for cations in solution as molarities. (a) \(14.2 \mathrm{mg} \mathrm{Ca}^{2+} / \mathrm{L} ;\) (b) \(32.8 \mathrm{mg} \mathrm{K}^{+} / 100 \mathrm{mL};\) (c) \(225 \mu \mathrm{g} \mathrm{Zn}^{2+} / \mathrm{mL}\).

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) \(2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow\)

See all solutions

Recommended explanations on Chemistry Textbooks

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