Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 115: 3.40b


$ 3.40 \times 10^{-6} \space mole$ of aspartame.

Work Step by Step

1. Identify the conversion factors: - $mg$ to grams : $\frac{1 \space g}{1000 \space mg}$ - Grams (aspartame) to moles (aspartame): $\frac{1 \space mole \space (aspartame)}{294.3 \space g \space (aspartame)**}$ ** This was calculated in 3.40a 2. Calculate the number of moles in 1.00 mg of that substance. $1.00 \space mg \space (aspartame) \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mole \space (aspartame)}{294.3 \space g \space (aspartame) }=$ $ 3.40 \times 10^{-6} \space mole \space (aspartame)$
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