Page 127 - Theory and Problems of BEGINNING CHEMISTRY

116 FORMULA CALCULATIONS [CHAP. 7

Ans. (a)2,toget3:2 (b)4,toget5:4 (c)3,toget4:3 (d)3,toget5:3 (e)4,toget7:4

7.48. Which do we use to calculate the empirical formula of an oxide, the atomic mass of oxygen (16 amu) or
the molecular mass of oxygen (32 amu)?
Ans. The atomic mass. We are solving for a formula, which is a ratio of atoms. This type of problem has nothing
to do with oxygen gas, O 2 .

7.49. (a) Write a formula for a molecule with 4 phosphorus atoms and 6 oxygen atoms per molecule. (b) What
is the empirical formula of this compound?
Ans. (a)P 4 O 6 (b)P 2 O 3

7.50. Which of the formulas in Problem 5.2 obviously are not empirical formulas?
Ans. (a), (d), (e), and (g). The subscripts in (a), (d), (e), and (g) can be divided by a small integer to give a simpler
formula, so these cannot be empirical formulas. (They must have at least some covalent bonds.)

7.51. Calculate the empirical formula of a compound consisting of 92.26% C and 7.74% H.
Ans. Assume that 100.0 g of the compound is analyzed. Since the same percentages are present no matter what
the sample size, we can consider any size sample we wish, and considering 100 g makes the calculations
easier. The numbers of grams of the elements are then 92.26 g C and 7.74 g H.

1 mol C 1 mol H
92.26gC = 7.682 mol C 7.74gH = 7.68 mol H
12.01gC 1.008gH
The empirical formula (or any formula) must be in the ratio of small integers. Thus, we attempt to get the
ratio of moles of carbon to moles of hydrogen into an integer ratio; we divide all the numbers of moles by
the smallest number of moles:
7.682 mol C 7.68 mol H
= 1.00 mol C = 1.00 mol H
7.68 7.68
The ratio of moles of C to moles of H is1:1,sothe empirical formula is CH.
7.52. Calculate the empirical formula of a compound containing 69.94% Fe and the rest oxygen.
Ans. The oxygen must be 30.06%, to total 100.00%.

1 mol Fe 1 mol O
69.94gFe = 1.252 mol Fe 30.06gO = 1.879 mol O
55.85gFe 16.00gO
Dividing both of these numbers by the smaller yields
1.252 mol Fe 1.879 mol O
= 1.000 mol Fe = 1.501 mol O
1.252 1.252
This is still not a whole number ratio, since 1.501 is much too far from an integer to round. Since 1.501 is
1
about 1 , multiply both numbers of moles by 2:
2
2.000 mol Fe and 3.002 mol O
We can round when a value is within 1% or 2% of an integer, but not more. This is close enough to an integer
ratio, so the empirical formula is Fe 2 O 3 .
7.53. Determine the empirical formula of a compound which has a percent composition 23.3% Mg, 30.7% S,
and 46.0%, O.
Ans. In a 100-g sample, there are
1 mol Mg 1 mol O

23.3gMg = 0.959 mol Mg 46.0gO = 2.88 mol O
24.3gMg 16.0gO
1 mol S

30.7gS = 0.959 mol S
32.0gS