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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Saccharin, an artificial sweetener, has the formula C7H5NO3S. Suppose you have a sample of a saccharin-containing sweetener with a mass of 0.2140 g, After decomposition to free the sulfur and convert it to the SO42– ion, the sulfate ion is trapped as water-insoluble BaSO4 (Figure 4.4). The quantity of BaSO4 obtained is 0.2070 g. What is the mass percent of saccharin in the sample of sweetener?

Interpretation Introduction

Interpretation:

The mass percent of saccharin in the given sample has to be determined.

Concept introduction:

  • The relation between the number of moles and mass of the substance is ,

     Numberofmole=Mass in gramMolarmass

  Mass in gram of the substance =Numberofmole×Molarmass

  • The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
  • For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
  • Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
  • Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
  • Amountof substance = Concentration of substance × volumeofthesubstance
  • Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
Explanation

Here saccharin (C7H5NO3S ) decomposes to sulfur and converts to the SO42 ion.  This SO42 is trapped as water insoluble BaSO4.

Therefore the stoichiometric ratio between saccharin and the BaSO4 will be 1:1

Thus,

The amount of saccharin C7H5NO3S can be determined from the amount of BaSO4 as follows,

  amount of BaSO4 =0.2070gBaSO4233.38g/molBaSO4 = 0

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