Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Chapter 4, Problem 127CP
Interpretation Introduction

Interpretation: The mass percentages in the sample have to be calculated.

Concept introduction: The mass percentage of a compound can be calculated with the calculated mass of the compound to the total mass of the compound.

Masspercent(in%)=Massofcompound(ing)Totalmass(ing)

Expert Solution & Answer
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Answer to Problem 127CP

Answer

The mass of SO42- was found to be 0.123g and its mass percent was 60.0% .  The mass percent of Na2SO4andK2SO4 in the sample are 39%Na2SO4and61%K2SO4

Explanation of Solution

Explanation

Record the given data

Mass of the sample mixture = 0.205g

Mass of the precipitate         = 0.298g

The mass of the sample mixture and mass of the precipitate are recorded as shown above.

To calculate the mass of SO42-

Molar mass of SO42- = 96.07g

Molar mass of BaSO4 = 233.4g

Mass of SO42- = 0.298gBaSO4×96.07gSO42-233.4gBaSO4

                       = 0.123gSO42-

The mass of SO42- is calculated by plugging in the values of mass of precipitate to the molar mass of SO42- and BaSO4 . The mass of SO42- was found to be 0.123g .

To calculate the mass percent of SO42-

Mass of SO42- = 0.123g

Mass percent of SO42- = 0.123g0.205g×100

                                    = 60.0%

The mass percent of SO42- is calculated by plugging in the values of mass of SO42- to the total mass of the sample. The mass percent of SO42- is found to be 60.0% .

To calculate the mass of Na2SO4andK2SO4

Molar mass of K+ = 39.10g

Molar mass of Na+ = 22.99g

Assume 100.0g of Na2SO4andK2SO4 mixture there are

60.0gofSO42-=1mol96.07g=0.625molSO42-

Hence,

2×0.625=1.25molof+1cations must be present to balance 2-chargeofSO42-

Let x= mass of K+

      y= mass of Na+

x+y=1.25......(1)

In 100.0g of mixture,

Total mass of K+ and Na+ = 40g. Then the equation can be given as,

xmolK+×39.10gmol+ymolNa+×22.99gmol=40.0g......(2)

Solving equation (1) and (2),

x+y=1.25g and xmolK+×39.10gmol+ymolNa+×22.99gmol=40.0g

x=1.25-ysubstitutethevalueofxin(2)39.10(1.25)y+(22.99)y=40.048.9-(39.10)y+(22.99)y=40.0-(16.11)y=-8.9y=0.55molNa+Therefore,x=1.25-0.55=0.70molK+

Therefore,

0.70molK+×1molK2SO42mol=0.35molK2SO4

0.35molK2SO4×174.27gmol=61gK2SO4

Massof Na2SO4=100-61=39g

MassofK2SO4=61gMassofNa2SO4=39g

The mass of Na2SO4andK2SO4 is calculated by plugging in the values of their respective masses to the total assumed mass. The moles of individual ions and moles of Na2SO4andK2SO4 are also calculated to get their respective masses. The masses of Na2SO4andK2SO4 are 61gand39g respectively.

To calculate the mass percent of Na2SO4andK2SO4

MassofK2SO4=61gMassofNa2SO4=39g

Assume total mass = 100g

MasspercentofK2SO4=61100×100 =61%

MasspercentofNa2SO4=39100×100=39%

The mass percent of Na2SO4andK2SO4 is calculated by plugging in the values of calculated mass of Na2SO4andK2SO4 to the total mass of the mixture. The mass percent of Na2SO4andK2SO4 is found to be 61%and39% .

Conclusion

Conclusion

The mass and mass percent of SO42- were calculated using the molar masses of SO42- , BaSO4 and mass of precipitate with the total mass of the sample. The mass percent of SO42- was found to be 60.0% and mass of SO42- was found to be 0.123g .

The mass percent of Na2SO4andK2SO4 was calculated by using the calculated mass of Na2SO4andK2SO4 to the total mass of the mixture. The mass percent of Na2SO4andK2SO4 was found to be 61%and39% .

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Chapter 4 Solutions

Chemistry

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