SAMPLE EXERCISE 3.15
Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g 
and 0.306 g 
Determine the empirical formula of isopropyl alcohol.
Solution
Analyze: We are given the quantities of 
and 
produced when a given quantity of isopropyl alcohol is combusted. We must use this information to determine the empirical formula for the isopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample.
Plan: We can use the mole concept to calculate the number of grams of C present in the
and the number of grams of H present in the 
These are the quantities of C and H present in the isopropyl alcohol before combustion. The number of grams of O in the compound equals the mass of the isopropyl alcohol minus the sum of the C and H masses. Once we have the number of grams of C, H, and O in the sample, we can then proceed as in Sample Exercise 3.13: Calculate the number of moles of each element, and determine the mole ratio, which gives the subscripts in the empirical formula.
Solve: To calculate the number of grams of C, we first use the molar mass of 
1 mol 
to convert grams of 
to moles of 
Because there is only one C atom in each 
molecule, there is one mole of C atoms per mole of 
molecules. This fact allows us to convert the moles of 
to moles of C. Finally, we use the molar mass of C, 1 mol 
to convert moles of C to grams of C. Combining the three conversion factors, we have
The calculation of the number of grams of H from the
grams of 
is similar, although we must remember that there are 2 mol of H atoms per 1 mol
of 
molecules:
The total mass of the sample, 0.255 g, is the sum of the masses of the C, H, and O. Thus we can calculate the mass of O as follows:
We then calculate the number of moles of C, H, and O
in the sample:
To find the empirical formula, we must compare the relative number of moles of each element in the sample. The relative number of moles of each element is found by dividing each number by the smallest number, 0.0043. The mole ratio of C
H
O so obtained is 2.98
7.91
1.00.
The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula 
PRACTICE EXERCISE
(a) Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g 
and 0.209 g 
What is the empirical formula of caproic acid? (b) Caproic acid has a molar
mass of 116 
What is its molecular formula?
Answers: (a) 
(b) 
indicates that water contains two H atoms for each O atom. This ratio also applies on the molar level; thus, 1 mol of 
contains 2 mol of H atoms and 1 mol of O atoms. Conversely, the ratio of the number of moles of each element in a compound gives the subscripts in a compound's empirical formula. Thus, the mole concept provides a way of calculating the empirical formulas of chemical substances, as shown in the following examples.
mercury and 
chlorine by mass. This means that if we had a 100.0-g sample of the solid, it would contain 73.9 g of mercury (Hg) and 26.1 g of chlorine (Cl). (Any size sample can be used in problems of this type, but we will generally use 100.0 g to make the calculation of mass from percentage easy.) Using the atomic weights of the elements to give us molar masses, we then calculate the number of moles of each element in the sample:
Hg mole ratio of 1.99
1:
This is the simplest, or empirical, formula because its subscripts are the smallest integers that express the ratios of atoms present in the compound. (Section 2.6) The general procedure for determining empirical formulas is outlined in Figure 3.11.
C, 
H, and 
O by mass. What is the empirical formula of ascorbic acid?
This suggests that if we multiply the ratio by 3, we will obtain whole numbers:
giving an empirical formula weight of 3(12.0 amu)
4(1.0 amu)
3(16.0 amu)
88.0 amu. The experimentally determined molecular weight is 176 amu. Thus, the molecule has twice the mass 
and must therefore have twice as many atoms of each kind as are given in the empirical formula. Consequently, the subscripts in the empirical formula must be multiplied
by 2 to obtain the molecular formula: 
The experimentally determined molecular weight of this substance is 121 amu. What
is the molecular formula of mesitylene?
C, 
H, and 
O by mass. Its molar mass is 62.1 
(a) What is the empirical formula of ethylene glycol? (b) What is its molecular formula?
(b) 
and all the hydrogen is converted to 
(Section
3.2) The amounts of 
and 
produced are determined by measuring the mass increase in the 
and 
absorbers. From the masses of 
and 
we can calculate the number of moles of C and H in the original compound and thereby the empirical formula. If a third element is present in the compound, its mass can be determined by subtracting the masses of C and H from the compound's original mass. Sample Exercise 3.15 shows how to determine the empirical formula of a compound containing C, H, and O.