Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 2 - Chemical Compounds - Exercise 2.16 - Grams, Moles, and the Avogadro Constant - Page 79: a


The iridium sample contains $3.133 \times 10^{22}$ atoms. The lithium sample contains $8.676 \times 10^{23}$ atoms, $8.363 \times 10^{23}$ more than the iridium sample.

Work Step by Step

For each sample, use the molar mass given in the periodic table to convert from grams to moles, then use Avogadro's number to convert from moles to the number of atoms. Lithium (6.941 grams per mole): $10.00g \times \frac{1 mol}{6.941 g} \times \frac{6.022\times10^{23}atoms}{1 mol}=8.676\times10^{23}$ atoms Iridium (192.217 grams per mole): $10.00g \times \frac{1 mol}{192.217g} \times \frac{6.022 \times 10^{23} atoms}{1 mol} = 3.133 \times 10^{22}$ atoms Round each to four significant figures, the least number of significant figures in any non-exact multiplicand. For the difference, simply subtract: $8.676 \times 10^{23} - 3.133 \times 10^{22} = 8.363 \times 10^{23}$
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