3.5 Empirical Formulas from Analyses The empirical formula for a substance tells us the relative number of atoms of each element it contains. Thus, the formula 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 forms a compound with chlorine that is 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:  We then divide the larger number of moles (0.735) by the smaller (0.368) to obtain a Cl Hg mole ratio of 1.99 1: Because of experimental errors, the results may not lead to exact integers for the ratios of moles. The number 1.99 is very close to 2, and we can confidently conclude that the empirical formula for the compound is 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. Figure 3.11 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." Sample Exercise 3.13 Ascorbic acid (vitamin C) contains C, H, and 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 3.11. Solve: We first assume, for simplicity, that we have exactly 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 this sample:   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 mole ratio gives us the subscripts for the empirical formula. Thus, the empirical formula is 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. Practice Exercise 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: Molecular Formula from Empirical Formula 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. (Section 2.6) The multiple is found by comparing the empirical formula weight with the molecular weight. In Sample Exercise 3.13, for example, the empirical formula of ascorbic acid was determined to be 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: SAMPLE EXERCISE 3.14 Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of 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 the empirical formula,  Next we divide the molecular weight by the empirical formula weight to obtain the factor used to multiply the subscripts in  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: Check: We can have confidence in the result because dividing the molecular weight by the formula weight yields nearly a whole number. PRACTICE EXERCISE Ethylene glycol, the substance used in automobile antifreeze, is composed of 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? Answer: (a) (b) Combustion Analysis The empirical formula of a compound is based on experiments that give the number of moles of each element in a sample of the compound. That is why we use the word "empirical," which means "based on observation and experiment." Chemists have devised a number of different experimental techniques 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 3.12, all the carbon in the compound is converted to 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. Figure 3.12 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. 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) 