Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 2 - Chemical Formulas and Composition Stoichiometrhy - Exercises - Percent Purity - Page 77: 83


A sample with 110.5g of this ore contains 9.57g of lead.

Work Step by Step

- Purity of the ore: 100% ore has 10.0% $PbS$ by mass. - Molar mass for lead sulfide: $ 207.2* 1 + 32.07* 1 = 239.3g/mol (PbS)$ - Molar mass for lead: $207.2 g / mol (Pb)$ - Number of lead atoms in $PbS$: Each lead sulfide has 1 $Pb$ atom. The ratio is 1 mol(Pb) to 1 mol(PbS). Use these informations as conversion factors: $110.5 g (ore) \times \frac{10.0\%(PbS)}{100\%(ore)} \times \frac{1mol(PbS)}{239.3g (PbS)} \times \frac{1mol(Pb)}{1mol(PbS)} \times \frac{207.2g(Pb)}{1mol(Pb)} = 9.57g(Pb)$
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