the sample. The approach is to analyze as we did in Example 2-14 for all elements except
oxygen. Then we subtract the sum of their masses from the mass of the original sample
to obtain the mass of oxygen. The next example illustrates such a calculation.
EXAMPLE 2-15 Percent Composition
A 0.1014-g sample of purified glucose was burned in a C-H combustion train to produce
0.1486 g of CO 2 and 0.0609 g of H 2 O. An elemental analysis showed that glucose contains
only carbon, hydrogen, and oxygen. Determine the masses of C, H, and O in the sample and
the percentages of these elements in glucose.
Plan
Steps 1 and 2: We first calculate the masses of carbon and hydrogen as we did in Example
2-14.
Step 3: The rest of the sample must be oxygen because glucose has been shown to contain only
C, H, and O. So we subtract the masses of C and H from the total mass of sample.
Step 4: Then we calculate the percentage by mass for each element.
Solution
Step 1: _?_ g C0.1486 g CO 2 0.04055 g C
Step 2: _?_ g H0.0609 g H 2 O0.00681 g H
Step 3: _?_ g O0.1014 g sample[0.04055 g C0.00681 g H] 0.0540 g O
Step 4: Now we can calculate the percentages by mass for each element:
% C100% 39.99% C
% H100% 0 6.72% H
% O100% 53.2% O
Total 0 99.9%
You should now work Exercise 60.
DETERMINATION OF MOLECULAR FORMULAS
Percent composition data yield only simplest formulas. To determine the molecular for-
mula for a molecular compound, bothits simplest formula and its molecular weight must
be known. Some methods for experimental determination of molecular weights are in-
troduced in Chapters 12 and 14.
2-10
0.0540 g O
0.1014 g
0.00681 g H
0.1014 g
0.04055 g C
0.1014 g
2.016 g H
18.02 g H 2 O
12.01 g C
44.01 g CO 2
We say that the mass of O in the
sample is calculated by difference.
74 CHAPTER 2: Chemical Formulas and Composition Stoichiometry
Glucose, a simple sugar, is the main
component of intravenous feeding
liquids. Its common name is dex-
trose. It is also one of the products
of carbohydrate metabolism.