Calculation of Formulas from Chemical Analysis
149
3082 d of residue; the increase in weight was due to 3082 d - 1600 d = 1482 d of S
combining with Cr. Then use the relative atomic weight scale in d/atom to get
1600 d
atomsofCrin 1600d =
= 30.8 atoms Cr
52.0 d/atom
atoms of S in 1482 d = ,. . ,,
= 46.2 atoms S
32.1 d/atom
We could write the formula as Cr 3 o.8S 46 .2, but what we want is the simples! formula.
The proportions will be the same if we just divide each of the numbers (subscripts)
by the smallest, to get
Cr wjt S 4^2 — CrSi,5
10 8
in 8
This formula is unacceptable because it erroneously implies that we can split
atoms in chemical reactions. Therefore we multiply by 2 to get Cr 2 S 3 , the correct
empirical formula.
We could have thought of the experiment as being done with 0.1600 d of Cr and
0.1482 d of S; the final result would have been the same. By taking the larger
quantities we avoided the uneasy feeling you might have had when it looked as if
fractions of atoms were combining (0.00308 atoms of Cr and 0.00462 atoms of S).
PROBLEM:
A compound contains 90.6% Pb and 9.4% O by weight. Find the empirical
formula.
SOLUTION:
The results of analysis usually are given in percentages of the constituents, not in
terms of the amounts actually weighed (as in the last problem). This permits
comparison of results from different experiments. You recall that "percent"
means "per 100." We can say, then, that if we have 100 g of the compound, 90.6 g
is Pband9.4g is O. We can also say that if we have 100 d of compound, 90.6 d is Pb
and 9.4 d is O. Using Dalton atomic weights we can easily find the number of atoms
in 100 d of compound, as follows.
atoms of Pb in 90.6 d = -^r^T,
= °437 atoms pb
207.2 d/atom
9 4 d
atoms of O in 9.4 d =
= 0.587 atoms O
16.0 d/atom
As in the previous problem, we divide each of the numbers of atoms by the smaller
number to give
Pb^Cw = PbO,..,.,
We know that the subscripts must be simple whole numbers, a situation readily
obtained by multiplying through by 3 to give Pb 3 O 4 , the correct empirical formula.
Frequently the fractional numbers that must be multiplied through by simple
integers do not yield exactly whole numbers (for example, 3 x 1.34 = 4.02, not
149
3082 d of residue; the increase in weight was due to 3082 d - 1600 d = 1482 d of S
combining with Cr. Then use the relative atomic weight scale in d/atom to get
1600 d
atomsofCrin 1600d =
= 30.8 atoms Cr
52.0 d/atom
atoms of S in 1482 d = ,. . ,,
= 46.2 atoms S
32.1 d/atom
We could write the formula as Cr 3 o.8S 46 .2, but what we want is the simples! formula.
The proportions will be the same if we just divide each of the numbers (subscripts)
by the smallest, to get
Cr wjt S 4^2 — CrSi,5
10 8
in 8
This formula is unacceptable because it erroneously implies that we can split
atoms in chemical reactions. Therefore we multiply by 2 to get Cr 2 S 3 , the correct
empirical formula.
We could have thought of the experiment as being done with 0.1600 d of Cr and
0.1482 d of S; the final result would have been the same. By taking the larger
quantities we avoided the uneasy feeling you might have had when it looked as if
fractions of atoms were combining (0.00308 atoms of Cr and 0.00462 atoms of S).
PROBLEM:
A compound contains 90.6% Pb and 9.4% O by weight. Find the empirical
formula.
SOLUTION:
The results of analysis usually are given in percentages of the constituents, not in
terms of the amounts actually weighed (as in the last problem). This permits
comparison of results from different experiments. You recall that "percent"
means "per 100." We can say, then, that if we have 100 g of the compound, 90.6 g
is Pband9.4g is O. We can also say that if we have 100 d of compound, 90.6 d is Pb
and 9.4 d is O. Using Dalton atomic weights we can easily find the number of atoms
in 100 d of compound, as follows.
atoms of Pb in 90.6 d = -^r^T,
= °437 atoms pb
207.2 d/atom
9 4 d
atoms of O in 9.4 d =
= 0.587 atoms O
16.0 d/atom
As in the previous problem, we divide each of the numbers of atoms by the smaller
number to give
Pb^Cw = PbO,..,.,
We know that the subscripts must be simple whole numbers, a situation readily
obtained by multiplying through by 3 to give Pb 3 O 4 , the correct empirical formula.
Frequently the fractional numbers that must be multiplied through by simple
integers do not yield exactly whole numbers (for example, 3 x 1.34 = 4.02, not