Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 1 - Chemistry: The Study of Change - Questions & Problems - Page 33: 1.78

Answer

$$ 4.0 \times 10^4 \space g \space (O)$$ $$1.1 \times 10^4 \space g \space (C)$$ $$ 6.2 \times 10^3 \space g \space (H)$$ $$ 2 \times 10^3 \space g \space (N)$$ $$9.9 \times 10^2 \space g \space (Ca)$$ $$ 7.4 \times 10^2 \space g \space (P)$$ $$7.4 \times 10^2 \space g \space (Other \space elements)$$

Work Step by Step

$$62 \space kg \times \frac{65 \% (O) \space kg}{100\% \space kg} = 40 \space kg \space (O)$$ $$62 \space kg \times \frac{18 \% (C) \space kg}{100\% \space kg} = 11 \space kg \space (C)$$ $$62 \space kg \times \frac{10 \% (H) \space kg}{100\% \space kg} = 6.2 \space kg \space (H)$$ $$62 \space kg \times \frac{3 \% (N) \space kg}{100\% \space kg} = 2 \space kg \space (N)$$ $$62 \space kg \times \frac{1.6 \% (Ca) \space kg}{100\% \space kg} = 0.99 \space kg \space (Ca)$$ $$62 \space kg \times \frac{1.2 \% (P) \space kg}{100\% \space kg} = 0.74 \space kg \space (P)$$ $$62 \space kg \times \frac{1.2 \% (Other \space elements) \space kg}{100\% \space kg} = 0.74 \space kg \space (Other \space elements)$$ $$40 \space kg \space (O) = 40 \space (10^3)g \space (O) = 4.0 \times 10^4 \space g \space (O)$$ $$11 \space kg \space (C) = 11 \space (10^3)g \space (C) = 1.1 \times 10^4 \space g \space (C)$$ $$6.2 \space kg \space (H) = 6.2 \space (10^3)g \space (H) = 6.2 \times 10^3 \space g \space (H)$$ $$2 \space kg \space (N) = 2 \space (10^3)g \space (N) = 2 \times 10^3 \space g \space (N)$$ $$0.99 \space kg \space (Ca) = 0.99 \space (10^3)g \space (C) = 9.9 \times 10^2 \space g \space (Ca)$$ $$0.74 \space kg \space (P) = 0.74 \space (10^3)g \space (O) = 7.4 \times 10^2 \space g \space (P)$$ $$0.74 \space kg \space (Other \space elements) = 7.4 \times 10^2 \space g \space (Other \space elements)$$
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