The empirical formula for a substance tells us the relative number of atoms of each element it contains. Thus, the formula H_{2}O indicates that water contains two H atoms for each O atom. This ratio also applies on the molar level; thus, 1 mol of H_{2}O 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 forms a compound with chlorine that is 73.9 percent mercury and 26.1 percent chlorine by mass. This means that if we had a 100-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 use 100 g to make the calculation of mass from percentage easy.) We divide each of these masses by the appropriate atomic weight to obtain the number of moles of each element in 100 g:

We then divide the larger number of moles by the smaller to obtain the ratio 1.99 mol Cl/1 mol Hg. Because of experimental errors, the results may not lead to exact integers for the ratios of moles. The ratio obtained in this case is very close to 2, and we can confidently conclude that the formula for the compound is HgCl_{2}. This is the simplest, or empirical, formula because it uses as subscripts the smallest set of integers that express the correct ratios of atoms present. The general procedure for determining empirical formulas is outlined in Figure 3.12.

**Figure 3.12** Outline of the procedure used to calculate the empirical formula of a substance from its percentage composition. The procedure is also summarized as "percent to mass, mass to moles, divide by small, multiply 'til whole."

Ascorbic acid (vitamin C) contains 40.92 percent C, 4.58 percent H, and 54.50 percent O by mass. What is the empirical formula of ascorbic acid?

**SOLUTION ***Analyze:* We are given the mass percentages of the elements in ascorbic acid and asked for its empirical formula.

*Plan:* The strategy for determining the empirical formula of a substance from its elemental composition involves the four steps given in Figure 3.12.

*Solve:* We first assume that we have 100 g of material (although any number can be used). In 100 g of ascorbic acid, we will have 40.92 g C, 4.58 g H, and 54.50 g O.

Second, we calculate the number of moles of each element in 100 g of the compound:

Third, we determine the simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles, 3.406:

The ratio for H is too far from 1 to attribute the difference to experimental error; in fact, it is quite close to . This suggests that if we multiply the ratio by 3, we will obtain whole numbers:

The whole-number ratio gives us the subscripts for the empirical formula. Thus, the empirical formula is C_{3}H_{4}O_{3}.

*Check:* It is reassuring that the subscripts are moderately sized whole numbers. Otherwise, we have little by which to judge the reasonableness of our answer.

A 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance? ** Answer:** C

Remember that the formula obtained from percentage compositions is always the empirical formula. We can obtain the molecular formula from the empirical formula if we know the molecular weight of the compound. *The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula.* The multiple is found by comparing the formula weight of the empirical formula with the molecular weight. For example, in Sample Exercise 3.12 we saw that the empirical formula of ascorbic acid is C_{3}H_{4}O_{3}, giving a 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 (176/88.0 = 2.00) and must therefore have twice as many atoms as the empirical formula. The subscripts in the empirical formula must be multiplied by 2 to obtain the molecular formula: C_{6}H_{8}O_{6}.

Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C_{3}H_{4}. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene?

**SOLUTION** *Analyze:* We are given the empirical formula and molecular weight of mesitylene and asked to determine its molecular formula.

*Plan:* The subscripts in a molecular formula are whole-number multiples of the subscripts in its empirical formula. To find the appropriate multiple, we must compare the molecular weight with the formula weight of the empirical formula.

*Solve:* First we calculate the formula weight of C_{3}H_{4}:

Next we divide the molecular weight by the formula weight to obtain the factor used to multiply the subscripts in C_{3}H_{4}:

Only whole-number ratios make physical sense because we must be dealing with whole atoms. The 3.02 in this case results from a small experimental error in the molecular weight. We therefore multiply each subscript in the empirical formula by 3 to give the molecular formula: C_{9}H_{12}.

*Check:* We have confidence in the result because our division of the molecular weight by the formula weight yields nearly a whole number.

Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7 percent C, 9.7 percent H, and 51.6 percent O by mass. Its molar mass is 62.1 g/mol. **(a)** What is the empirical formula of ethylene glycol? **(b)** What is its molecular formula? *Answer: ***(a)** CH_{3}O; **(b)** C_{2}H_{6}O_{2}

The empirical formula of a compound is based on experiments that give the number of moles of each element in a sample of a substance. That is why we use the word "empirical," which means "based on observation and experiment." There are a number of different experimental techniques that chemists have devised to determine the empirical formulas of compounds. One of these is combustion analysis which is commonly used for compounds containing principally carbon and hydrogen as their component elements.

When a compound containing carbon and hydrogen is completely combusted in an apparatus such as that shown in Figure 3.13 , all the carbon of the compound is converted to CO_{2} and all the hydrogen to H_{2}O. The amounts of CO_{2} and H_{2}O produced can be measured by determining the mass increase in the CO_{2} and the H_{2}O absorbers. We can then use the masses of CO_{2} and H_{2}O to determine the number of moles of C and H in the original compound and thereby the empirical formula.

**Figure 3.13** Apparatus to determine percentages of carbon and hydrogen in a compound. Copper oxide helps to oxidize traces of carbon and carbon monoxide to carbon dioxide, and to oxidize hydrogen to water.

As an example, consider a sample of isopropyl alcohol, a substance sold as rubbing alcohol. The compound is known to contain only C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO_{2} and 0.306 g H_{2}O. From this experimental information we can use the mole concept to calculate the number of moles of C, H, and O in the sample.

To calculate the number of moles of C, we first convert the grams of CO_{2} to moles of CO_{2} and then convert moles of CO_{2} to moles of C. For the first conversion, we use the molar mass of CO_{2}, 1 mol CO_{2} = 44.0 g CO_{2}. For the second conversion, we use the fact that there is only one C atom in each CO_{2} molecule and therefore one mole of C atoms per mole of CO_{2} molecules. Using the two conversion factors, we have

The calculation of the number of moles of H from the grams of H_{2}O is similar, although we must remember that there are 2 mol of H atoms per mol of H_{2}O molecules:

The calculation of the number of moles of O is a bit more indirect. The total mass of the sample, 0.255 g, is the sum of the masses of the C, H, and O. If we know the mass of the sample, the mass of C, and the mass of H, we can calculate the mass of O. The masses of C and H are calculated from the number of moles of these elements:

We can now calculate the mass of O:

We then carry out a gram-to-moles conversion to determine the number of moles of 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, 0.0042. The ratio of C:H:O so obtained is 3.05:8.10:1. The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C_{3}H_{8}O.