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 - The Mole Concept - Page 75: 38

Answer

a) $4.32\cdot10^{23}\ molecules$ b) $6.79\cdot10^{23}\ molecules$ c) $1.54\cdot10^{23}\ molecules$ d) $3.07\cdot10^{23}\ molecules$ e) Yes

Work Step by Step

a) Calculate the molar mass of CO2: Atomic weights: C - 12.011, O - 15.999 $M=12.011+2\cdot 15.999=44.009\ g/mol$ Calculate the number of moles: $n=\frac mM$ $n=\frac{31.6\ g}{44.009\ g/mol}=0.718\ mol$ Convert to the number of molecules using Avogadro's number: $n=\frac{N}{N_A}$ $N=0.718\ mol\cdot 6.022\cdot10^{23}=4.32\cdot10^{23}\ molecules$ b) Calculate the molar mass of N2: Atomic weights: N - 14.007 $M=2\cdot 14.007=28.014\ g/mol$ Calculate the number of moles: $n=\frac mM$ $n=\frac{31.6\ g}{28.014\ g/mol}=1.13\ mol$ Convert to the number of molecules using Avogadro's number: $n=\frac{N}{N_A}$ $N=1.13\ mol\cdot 6.022\cdot10^{23}=6.79\cdot10^{23}\ molecules$ c) Calculate the molar mass of P4: Atomic weights: P - 30.974 $M=4\cdot 30.974=123.896\ g/mol$ Calculate the number of moles: $n=\frac mM$ $n=\frac{31.6\ g}{123.896\ g/mol}=0.255\ mol$ Convert to the number of molecules using Avogadro's number: $n=\frac{N}{N_A}$ $N=0.255\ mol\cdot 6.022\cdot10^{23}=1.54\cdot10^{23}\ molecules$ d) Calculate the molar mass of P2: Atomic weights: P - 30.974 $M=2\cdot 30.974=61.948\ g/mol$ Calculate the number of moles: $n=\frac mM$ $n=\frac{31.6\ g}{61.948\ g/mol}=0.510\ mol$ Convert to the number of molecules using Avogadro's number: $n=\frac{N}{N_A}$ $N=0.510\ mol\cdot 6.022\cdot10^{23}=3.07\cdot10^{23}\ molecules$ e) Yes, part (d) has twice as many molecules, but each molecule in part (c) has twice as many phosphorus atoms, so the number of phosphorus atoms in parts (c) and (d) is the same.
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