General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 7 - Chemical Reactions and Quantities - Challenge Questions - Page 285: 7.150

Answer

a. $6.80 \times 10^{-3} \space mole \space NaF$ b. $4.09 \times 10^{21} \space ions \space F^-$ c. $1.97 \times 10^{-3} \space g \space Na^+$ d. $7.4 \times 10^{20} \space molecules \space of \space triclosan$

Work Step by Step

a. - Calculate or find the molar mass for $ NaF $: $ NaF $ : ( 22.99 $\times$ 1 )+ ( 19.00 $\times$ 1 )= 41.99 g/mol - $$100 \space g \space toothpaste = 0.240 \space g \space NaF$$ $$ 119 \space g \space toothpaste \times \frac{0.240 \space g \space NaF}{100 \space g \space toothpaste} \times \frac{1 \space mole \space NaF}{ 41.99 \space g \space NaF} $$$$= 6.80 \times 10^{-3} \space mole \space NaF$$ b. - 1 mole NaF = 1 mole $F^{-}$ - 1 mole $F^-$ = $6.02 \times 10^{23}$ $F^-$ ions $$6.80 \times 10^{-3} \space mole \space NaF \times \frac{1 \space mole F^-}{1 \space mole \space NaF} \times \frac{6.02 \times 10^{23} \space ions \space F^{-}}{1 \space mole \space F^-}$$ $$= 4.09 \times 10^{21} \space ions \space F^-$$ c. $ NaF $ : 41.99 g/mol $Na^+$: 22.99 g/mol - 1 mole $Na^+$ = 1 mole $NaF$ $$ 1.50 \space g \space toothpaste \times \frac{0.240 \space g \space NaF}{100 \space g \space toothpaste} \times \frac{1 \space mole \space NaF}{ 41.99 \space g \space NaF} \times \frac{1 \space mole \space Na^+}{1 \space mole \space NaF} \times \frac{22.99 \space g \space Na^+}{1 \space mole \space Na^+} $$$$= 1.97 \times 10^{-3} \space g \space Na^+$$ d. - Calculate or find the molar mass for $ C_{12}H_7Cl_3O_2 $: $ C_{12}H_7Cl_3O_2 $ : ( 35.45 $\times$ 3 )+ ( 1.008 $\times$ 7 )+ ( 12.01 $\times$ 12 )+ ( 16.00 $\times$ 2 )= 289.53 g/mol $$100 \space g \space toothpaste = 0.30 \space g \space triclosan$$ $$ 119 \space g \space toothpaste \times \frac{0.30 \space g \space triclosan}{100 \space g \space toothpaste} \times \frac{1 \space mole \space triclosan}{ 289.53 \space g}\times \frac{6.02 \times 10^{23} \space molecules}{1 \space mole} $$ $$ = 7.4 \times 10^{20} \space molecules \space triclosan$$
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