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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 \(0.5894 \mathrm{~g}\) 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 \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). 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 \(0.0395 \: g\), and the average mass percent of saccharin in the tablets is \(67.0\% \).

Step by step solution

01

Write the balanced chemical equation for the oxidation of saccharin and precipitation of sulfate

The balanced chemical equation for the oxidation of saccharin (C7H5NO3S) and precipitation of sulfate (BaSO4) is: C7H5NO3S + 4O + BaCl2 -> BaSO4 + 7CO2 + 5H2O + N2 + 2HCl
02

Calculate the moles of BaSO4 formed

Given the mass of BaSO4 obtained, we can calculate the moles of BaSO4. The molar mass of BaSO4 is: \(M_{BaSO4} = 137.3 + 32.1 + (4 × 16.0) = 233.4 \: g/mol\) Now, calculate the number of moles: \(\text{moles of BaSO4} = \frac{\text{mass of BaSO4}}{M_{BaSO4}} = \frac{0.5032 \, g} {233.4 \, g/mol} = 2.156×10^{-3} \:mol \)
03

Calculate the moles of saccharin (C7H5NO3S) in the tablet sample

From the balanced chemical equation, 1 mole of C7H5NO3S produces 1 mole of BaSO4 So, the moles of C7H5NO3S are equal to the moles of BaSO4 formed: \(\text{moles of C}_{7}\text{H}_{5}\text{NO}_{3}\text{S} = 2.156×10^{-3} \: mol\)
04

Calculate the mass of saccharin (C7H5NO3S) in the tablet sample

The molar mass of C7H5NO3S is: \( M_{C7H5NO3S} = (7 × 12.0) + (5 × 1.0) + 14.0 + (3 × 16.0) + 32.1 = 183.2 \: g/mol \) Now, we can find the mass of saccharin: \(\text{mass of C}_{7}\text{H}_{5}\text{NO}_{3}\text{S} = \text{moles} × \text{molar mass} = 2.156×10^{-3} \: \text{mol} × 183.2 \: \text{g/mol} = 0.395 \: g \)
05

Calculate the average mass of saccharin per tablet

To find the average mass of saccharin per tablet, divide the total mass of saccharin by the number of tablets: \(\text{average mass of saccharin per tablet} = \frac{\text{total mass of saccharin}}{\text{number of tablets}} = \frac{0.395 \: g}{10} = 0.0395 \: g \)
06

Calculate the average mass percent of saccharin in the tablets

To find the average mass percent of saccharin in the tablets, divide the mass of saccharin by the total mass of the tablets and multiply by 100: \(\text{average mass \% of saccharin in the tablets} = \frac{\text{total mass of saccharin}}{\text{total mass of tablets}} × 100 = \frac{0.395 \: g}{0.5894 \: g} × 100 = 67.0 \% \) So, the average mass of saccharin per tablet is 0.0395 g, and the average mass percent of saccharin in the tablets is 67.0%.

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Most popular questions from this chapter

What mass of barium sulfate can be produced when \(100.0 \mathrm{~mL}\) of a \(0.100 M\) solution of barium chloride is mixed with \(100.0\) \(\mathrm{mL}\) of a \(0.100 \mathrm{M}\) solution of iron(III) sulfate?

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Commercial cold packs and hot packs are available for treating athletic injuries. Both types contain a pouch of water and a dry chemical. When the pack is struck, the pouch of water breaks, dissolving the chemical, and the solution becomes either hot or cold. Many hot packs use magnesium sulfate, and many cold packs use ammonium nitrate. Write reactions to show how these strong electrolytes break apart when they dissolve in water.

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