Chemistry: The Molecular Science (5th Edition)

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

Chapter 2 - Chemical Compounds - Problem Solving Practice 2.16 - Page 83: a


There are $1.60\times10^{23}$ O atoms in 12.0g asprin There are $1.86\times10^{23}$ O atoms in 12.0g Ca$_3$(PO$_4$)$_2$

Work Step by Step

Calculate the molar mass of each compound, and use this to determine how many moles there are in each sample. Then use Avogadro's number to determine how many formula units (molecules) there are of that compound. Finally, multiply by the number of Oxygen atoms in one formula unit of that compound. Apsrin, C$_9$H$_8$O$_4$: Molar mass is $12.0107 \times 9 + 1.0079 \times 8 + 15.9994 \times 4=180.1571$ g/mol $12.0g \times \frac{1 mol}{180.1571g} \times \frac{6.022\times10^{23} molecules}{1 mol}\times\frac{4\space O\space atoms}{1 molecule}=1.60\times10^{23}$ O atoms Ca$_3$(PO$_4$)$_2$: Expanding the parentheses there are 2 P atoms and 8 O atoms per formula unit. Molar mass is $40.078\times3+30.9738\times2+15.9994\times8=310.177$ g/mol $12.0g\times\frac{1mol}{310.177g}\times\frac{6.022\times10^{23}units}{1mol}\times\frac{8\space O\space atoms}{1 unit}=1.86\times10^{23}$ O atoms. Round each to 3 significant figures, the lowest number of significant figures in any non-exact number.
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