Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic: 69

Answer

$0.10mg\ C_{12}H_{22}O_{11}$

Work Step by Step

Convert molecules to moles by dividing by avagadro's number. Multiply by the molar mass of the compound ($342.30\frac{g}{mol}$ for sucrose) to get its mass in grams and then multiply by 1000 to convert to milligrams. $1.8\times10^{17}g\ C_{12}H_{22}O_{11}\times\frac{1\ mol}{6.022\times10^{23}}\times\frac{342.30g\ C_{12}H_{22}O_{11}}{1mol\ C_{12}H_{22}O_{11}}\times\frac{1000\ mg}{1\ g}= 0.10mg\ C_{12}H_{22}O_{11}$
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