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Chapter 4: Problem 116

The zinc in a \(1.343-g\) sample of a foot powder was precipitated as \(\mathrm{ZnNH}_{4} \mathrm{PO}_{4} .\) Strong heating of the precipitate yielded \(0.4089 \mathrm{~g} \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} .\) Calculate the mass percent of zinc in the sample of foot powder.

Short Answer

Expert verified
The mass percent of zinc in the sample of foot powder is approximately 12.97 %.

Step by step solution

01

Obtain the moles of zinc in Zn2P2O7

Using the molar mass, we will find the moles of Zn in the zinc pyrophosphate. The molar mass of Zn2P2O7 is calculated as follows: Molar mass of Zn2P2O7 = 2 * Molar mass of Zn + 2 * Molar mass of P + 7 * Molar mass of O Molar mass of Zn2P2O7 = 2(65.38 g/mol) + 2(30.97 g/mol) + 7(16.00 g/mol) Molar mass of Zn2P2O7 = 306.70 g/mol Now we can find the moles of Zn2P2O7: moles of Zn2P2O7 = mass of Zn2P2O7 / Molar mass of Zn2P2O7 moles of Zn2P2O7 = 0.4089 g / 306.70 g/mol moles of Zn2P2O7 = 0.001333 mol
02

Calculate the moles of Zn in Zn2P2O7

Since each molecule of Zn2P2O7 contains 2 Zn atoms, the moles of Zn can be calculated using the stoichiometric ratio of the moles of Zn2P2O7 and Zn: moles of Zn = 2 * moles of Zn2P2O7 moles of Zn = 2 * 0.001333 moles of Zn = 0.002666
03

Determine the mass of zinc in the sample

Now we will convert the moles of Zn to grams using the molar mass of Zn (65.38 g/mol): mass of Zn = moles of Zn * Molar mass of Zn mass of Zn = 0.002666 * 65.38 g/mol mass of Zn = 0.1742 g
04

Calculate the mass percent of zinc in the foot powder sample

Finally, we can determine the mass percent of Zn in the foot powder sample using the formula: mass percent of Zn = (mass of Zn in sample / mass of foot powder sample) * 100 mass percent of Zn = (0.1742 g / 1.343 g) * 100 mass percent of Zn = 12.97 % The mass percent of zinc in the sample of foot powder is approximately 12.97 %.

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Most popular questions from this chapter

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. \(\operatorname{copper}(\) II \()\) sulfate \((a q)+\) iron \((s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(II) sulfate \((a q)\) copper(II) sulfate \((a q)+\operatorname{iron}(s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{~mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{~g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate \(2.27 \mathrm{~g}\) of copper. Which equation above describes the reaction that occurred? Support your answer.

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