## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95d: 77

#### Answer

$4\times10^{21}\ molecules$

#### Work Step by Step

Molecular Formula: $C_8H_9O_2N$ Atomic weights (g/mol): $C: 12.011,\ H: 1.008,\ O: 15.999,\ N: 14.007$ Molar mass: $8\times12.011+9\times1.008+2\times15.999+14.07=151.2\ g/mol$ Mass of the dose: $2\times500\ mg \times 10^{-3} mg/g = 1\ g$ Molecules in a dose: $1\ g\div151.2\ g/mol\times6.022\times10^{23}molecules/mol=4\times10^{21}\ molecules$

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