/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 120 Saccharin \(\left(\mathrm{C}_{7}... [FREE SOLUTION] | 91影视

91影视

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of 0.5894 g were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO\(_{4}\) obtained was 0.5032 g. What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

Short Answer

Expert verified
The average mass of saccharin per tablet is approximately 0.03956 g or 39.56 mg, and the average mass percent of saccharin in the tablets is approximately 67.1%.

Step by step solution

01

Calculate the mass of saccharin in the experiment's initial state

To determine the mass of saccharin in the initial state of the experiment, we need to find the amount of sulfate ion (SO鈧劼测伝) present in the BaSO鈧 precipitate and use its stoichiometric relationship to saccharin. First, find the moles of BaSO鈧 produced using its formula mass (233.43 g/mol): Moles of BaSO鈧 = (mass of BaSO鈧) / (formula mass of BaSO鈧) = 0.5032 g / 233.43 g/mol 鈮 0.00216 mol Next, use the stoichiometric relationship between BaSO鈧 and saccharin to find the moles of saccharin, which is 1:1 due to the balanced reaction: Moles of saccharin = moles of BaSO鈧 = 0.00216 mol Now, find the mass of saccharin using its formula mass (183.18 g/mol): Mass of saccharin = (moles of saccharin) * (formula mass of saccharin) 鈮 0.00216 mol * 183.18 g/mol 鈮 0.3956 g
02

Calculate the average mass of saccharin per tablet

To calculate the average mass of saccharin per tablet, divide the total mass of saccharin by the number of tablets: Average mass of saccharin per tablet = mass of saccharin / number of tablets 鈮 0.3956 g / 10 tablets 鈮 0.03956 g/tablet
03

Calculate the mass percent of saccharin in the tablets

To calculate the mass percent of saccharin in the tablets, divide the mass of saccharin by the total mass of the tablets and then multiply by 100: Mass percent of saccharin = (mass of saccharin / total mass of tablets) * 100 鈮 (0.3956 g / 0.5894 g) * 100 鈮 67.1 % Thus, the average mass of saccharin per tablet is approximately 0.03956 g or 39.56 mg, and the average mass percent of saccharin in the tablets is approximately 67.1%.

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 91影视!

Key Concepts

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

Formula Mass
Formula mass, also known as molecular weight or molar mass, is the total mass of all the atoms in a chemical formula. To find the formula mass, sum up the atomic masses of each element in the compound using the periodic table.
For example, to calculate the formula mass of Saccharin (\(\text{C}_7\text{H}_5\text{NO}_3\text{S}\)), you:
  • Take 7 Carbon atoms, each with an atomic mass of approximately 12.01 u, totaling about 84.07 u.
  • Add 5 Hydrogen atoms at around 1.01 u each, totaling 5.05 u.
  • Include 1 Nitrogen atom with a mass of about 14.01 u.
  • Consider 3 Oxygen atoms, each weighing approximately 16.00 u, for a total of 48.00 u.
  • And finally, add 1 Sulfur atom with a mass of about 32.07 u.
Bringing these together, Saccharin's formula mass comes to approximately 183.18 u.
Understanding formula mass is crucial when calculating the amount of a substance needed or produced in a reaction, as it helps convert between the mass of a substance and the moles present.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced in a given reaction.
In the exercise, stoichiometry helps link the amount of sulfate ion formed to the amount of saccharin present. Once the barium sulfate (\(\text{BaSO}_4\)) forms from the reaction of barium chloride (\(\text{BaCl}_2\)) and sulfate ions, stoichiometry provides a 1:1 ratio of moles between BaSO鈧 and the sulfur in saccharin.
  • First, calculate the moles of \(\text{BaSO}_4\) from its mass by using its formula mass of 233.43 g/mol.
  • Next, because the ratio is 1:1, the moles of \(\text{BaSO}_4\) equal the moles of sulfur that originated from saccharin.
This concept is crucial for making efficient and accurate chemical calculations in laboratory settings and industrial processes.
Mass Percent
Mass percent describes the concentration of a particular component in a mixture, expressed as a percentage of the total mass. It is often used to determine how much of a specific substance is contained in a compound or mixture.
In this exercise, mass percent helps calculate the proportion of saccharin in the tablet mixture. To find this:
  • First, divide the mass of the saccharin (0.3956 g from the calculation) by the total mass of the tablets (0.5894 g).
  • Then, multiply the result by 100 to convert it to a percentage.
The calculated mass percent is approximately 67.1%, indicating that the saccharin constitutes a significant portion of the tablets. Understanding mass percent is essential not only in chemistry but also in areas like food nutrition and pharmaceuticals, as it helps ensure products meet specific concentration standards.
Precipitation Reaction
A precipitation reaction occurs when two dissolved substances react in solution to form an insoluble solid, known as a precipitate. These reactions are essential for isolating and removing specific ions or compounds in a mixture.
In the given exercise, the saccharin is dissolved in water, and the sulfur atoms in saccharin are converted into sulfate ions (\(\text{SO}_4^{2-}\)). By adding barium chloride to the solution, barium sulfate (\(\text{BaSO}_4\)) precipitates out because it is insoluble. This process helps isolate and quantify the amount of sulfate originally present in the saccharin.
Precipitation reactions are vital in fields like analytical chemistry for qualitative and quantitative analysis, as well as in environmental science where they help remove pollutants from water.

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

A \(10.00-\mathrm{mL}\) . sample of vinegar, an aqueous solution of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right),\) is titrated with \(0.5062 \mathrm{M} \mathrm{NaOH},\) and 16.58 mL is required to reach the equivalence point. a. What is the molarity of the acetic acid? b. If the density of the vinegar is \(1.006 \mathrm{g} / \mathrm{cm}^{3},\) what is the mass percent of acetic acid in the vinegar?

Consider the reaction of 19.0 g of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and 29.0 g of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$\mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }}$$ In the case of very dilute aqueous solutions, a concentration of 1.0 \(\mathrm{ppm}\) is equal to 1.0\(\mu \mathrm{g}\) of solute per 1.0 \(\mathrm{mL}\) , which equals 1.0 \(\mathrm{g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 \(\mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. 1.0 \(\mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. 10.0 \(\mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. 0.10 \(\mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{C} 1_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH? a. 0.100 \({M} {HCl} \quad \)c. 0.200\(M {HC}_{2} {H}_{3} {O}_{2}$$(1 \text { acidic hydrogen })\) b. 0.150 \({M} {HNO}_{3}\)

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.