Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 2 - Chemical Formulas and Composition Stoichiometrhy - Exercises - Composition Stoichiometry - Page 76: 57

Answer

The formula of bupropion is $C_{13}H_{18}ClNO$.

Work Step by Step

In a sample of 100 g of bupropion, there is 65.13 g of carbon, 7.57 g of hydrogen, 14.79 g of chlorine, 5.84 g of nitrogen, and 6.67 g of oxygen. Let's now calculate the moles of each element present in the sample. $n (C) = \frac{m(C)}{M(C)}=\frac{65.13}{12.01} $ mol $ =5.423 $ mol $n (H) = \frac{m(H)}{M(H)}=\frac{7.57}{1.008} $ mol $ =7.510 $ mol $n (Cl) = \frac{m(Cl)}{M(Cl)}=\frac{14.79}{35.45} $ mol $ =0.417 $ mol $n (N) = \frac{m(N)}{M(N)}=\frac{5.84}{14.01} $ mol $ =0.417 $ mol $n (O) = \frac{m(O)}{M(O)}=\frac{6.67}{16.00} $ mol $ =0.417 $ mol Divide those numbers by the smallest one of them and round the result to a whole number. $ N (C) = \frac{n(C)}{n(Cl)}=13$ $ N (H) = \frac{n(H)}{n(Cl)}=18$ $ N (Cl) = \frac{n(Cl)}{n(Cl)}=1$ $ N (N) = \frac{n(N)}{n(Cl)}=1$ $ N (O) = \frac{n(O)}{n(Cl)}=1$ The final formula is $C_{13}H_{18}ClNO$.
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