## Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.

# Chapter 2 - Chemical Formulas and Composition Stoichiometrhy - Exercises - Percent Purity - Page 77: 85

#### Answer

(a) 99.7 g of $Sr$ (b) 31.9 g of $N$

#### Work Step by Step

(a) Sr: Purity of the sample: 90% $Sr(NO_3)_2$. Molar mass: $Sr(NO_3)_2$ = 87.62* 1 + ( 14.01* 1 + 16.00* 3 )*2 = 211.64g/mol $Sr(NO_3)_2$ Amount of $Sr$ moles in 1 mol of $Sr(NO_3)_2$: There is only 1 $Sr$ mol per mol of $Sr(NO_3)_2$ Molar mass: $(Sr) = 87.62g/mol(Sr)$ - Use these informations as conversion factors: $267.7g(Sample) \times \frac{90.0\%(Sr(NO_3)_2)}{100\%(Sr(NO_3)_2)} \times \frac{1mol(Sr(NO_3)_2)}{211.64g(Sr(NO_3)_2)} \times \frac{1mol(Sr)}{1mol(Sr(NO_3)_2)} \times \frac{87.62g(Sr)}{1mol(Sr)} = 99.7g(Sr)$ (b) N: Purity of the sample: 90% $Sr(NO_3)_2$. Molar mass: $Sr(NO_3)_2$ = 87.62* 1 + ( 14.01* 1 + 16.00* 3 )*2 = 211.64g/mol $Sr(NO_3)_2$ Amount of $N$ moles in 1 mol of $Sr(NO_3)_2$: There are 2 $N$ moles per mol of $Sr(NO_3)_2$ Molar mass: $(N) = 14.01g/mol(N)$ $267.7g(Sample) \times \frac{90.0\%(Sr(NO_3)_2)}{100\%(Sr(NO_3)_2)} \times \frac{1mol(Sr(NO_3)_2)}{211.64g(Sr(NO_3)_2)} \times \frac{2mol(N)}{1mol(Sr(NO_3)_2)} \times \frac{14.01g(N)}{1mol(N)} = 31.9g(N)$

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