Answers and Explanations

REVIEW QUESTIONS

1 . D

In  order   to  answer  this    question,   the equation    must    first   be  balanced.   Starting    with    carbon, it

can be  seen    that    there   are six carbons on  the reactant    side    and only    one on  the product side,

so  a   coefficient of  six should  be  placed  in  front   of  the carbon  dioxide.    For the hydrogen,   there

are 12  atoms   on  the left    and only    two on  the right;  thus,   a   coefficient of  six should  go  in  front

of  water.  Now,    for oxygen, there   are eight   atoms   on  the left    and 18  on  the right.  In  order   to

balance the oxygen, ten more    atoms   of  oxygen  must    be  added   to  the left    side.   The best    way to

do  this    is  to  put a   coefficient of  six in  front   of  oxygen, since   putting a   stoichiometric  coefficient

in  front   of  the glucose molecule    would   unbalance   the equation    in  terms   of  carbon  and

hydrogen.   Therefore,  the final   balanced    equation    is:

C 6 H 12 O 6 + 6O 2 → 6CO 2 + 6H 2 O

Adding  up  the coefficients,   the result  is  (1  +   6   +   6   +   6)  =   19.

2 . C 10 H 14 N 2 , 74.1%   C,  8.6%    H,  17.3%   N

To  determine   the molecular   formula of  nicotine,   the empirical   weight  of  the compound    must

be  calculated.

The empirical   weight  (81 g/mol)  is  then    divided into    the molecular   weight  (162    g/mol)  to

determine   the number  by  which   each    subscript   in  the empirical   formula must    be  multiplied  to

obtain  the molecular   formula.