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Elemental Analysis

Introduction

One of the first things that chemists want to know when they isolate or synthesize a new compound is its composition. What elements are present? What are their relative amounts? One way to answer questions like these involves combustion analysis. Before we get into the details of this method, let's consider the relationship between an empirical formula and the percentage composition of the elements in that formula. We'll use methane, CH4, as an example. The formula weight of methane is [12.01 + (4 x 1.008)] = 16.04. In terms of mass, methane is (4.032/16.04) x 100 = 25.13% hydrogen and (12.01/16.04) x 100 = 74.87% carbon. N.B. In this example the numerical values of the atomic masses are given to 4 significant figures. The values of the per-cent composition were obtained by rounding off the output of the Rothenberger Calculator.

Exercise 1 Calculate the formula weight and the per-cent composition of water. Express your answer to 4 significant figures.

The formula weight of water is. Water contains per-cent hydrogen and per-cent oxygen by weight.

Exercise 2 Calculate the formula weight and the per-cent composition of carbon dioxide. Express your answer to 4 significant figures.

The formula weight of carbon dioxide is. Carbon dioxide contains per-cent carbon and per-cent oxygen by weight.

In order to calculate the per-cent composition of a compound, you need to know its empirical formula and the atomic mass of each of the elements in the molecule. Conversely, if you know the per-cent composition of a compound and the atomic masses of each of the elements in it, you can calculate its empirical formula. For example, suppose you know that a compound contains 79.86% C and 20.14% H. Since these values sum to 100%, you know that this compound is a hydrocarbon. To figure out the relative proportions of C and H, divide the percentage of each element by its atomic mass: 79.86/12.01 = 6.649, 20.14/1.008 = 19.98. This says that the formula of the compound is C6.649H19.98. This is non-sensical since the numbers of carbon and hydrogen atoms must have integer values. If you divide each non-integer value by 6.649, you generate a new formula in which the C/H ratio is the same, namely C1.000H3.005. Rounding off gives CH3 as the empirical formula of the compound. The simplest molecular formula that is consistent with this empirical formula is C2H6, i.e. ethane.
Exercise 3

a. Is C3H9 a valid molecular formula for the empirical formula CH3? yes no

b. Is C4H12 a valid molecular formula for the empirical formula CH3? yes no

Exercise 4 Calculate the empirical formula for each of the following compositions.

a. C, 83.21; H, 16.79.

b. C, 92.24; H, 7.76.

Knowing the composition of a compound can be very useful first step in determining its structure. How is combustion analysis performed experimentally? Generally a sample is sent to an analytical laboratory where a carefully weighed amount is completely burned in an oxygen atmosphere. The carbon dioxide and water that are formed are collected and weighed. Knowing the mass of the initial sample and the masses of CO2 and H2O formed allows the percentage composition of the sample to be determined in a straightforward manner.

We'll use the combustion of methane, as shown in Equation 1, to illustrate.

Equation 1 is a balanced equation; one mole of methane reacts with two moles of dioxygen to produce one mole of carbon dioxide and two moles of water. So if we start with 16.04 g of methane and react it with 64.00 grams of dioxygen, we should form 44.01 grams of carbon dioxide and 36.03 grams of water. (Note that the sum of the masses of the reactants, 80.04 grams, equals the sum of the masses of the products.) Of the 44.01 grams of carbon dioxide produced in reaction 1, 12.01 grams is carbon and 32.00 grams is oxygen. Since all of the carbon in the carbon dioxide originated from the methane, we can state with certainty that 12.01 of the 16.04 grams of methane was carbon. In other words, carbon constitutes 74.87% of the mass of methane. By the same token, 4.04 of the 36.04 grams of water formed in reaction 1 was hydrogen. Since all of the hydrogen in the water was originally in part of the methane, we can calculate that hydrogen constituted 25.13% of the mass of the methane. We already knew that.

Exercise 5 What mass of carbon dioxide would be formed upon complete combustion of 0.1028 g of methane? Express your answer to 4 significant figures. g What mass of water would be formed? g
Exercise 6 Complete combustion of a sample of methane produced 0.1735 g of CO2. What was the mass of the sample of methane? Express your answer to 4 significant figures. g
Exercise 7 A chemist found an unlabeled bottle of liquid in the stockroom. The boiling point of the compound was 78oC. Combustion analysis indicated the composition to be 52.13% C, 13.15% H, and 34.72% O. What is the simplest molecular formula for this compound? How many unique Lewis structures are there for this formula? What is the name of the compound?
Elemental analysis is just a small part of what's required to establish the identity of a compound. But it's an important part, and it's generally included, along with supporting data, in the experimental section of any publication of the results of research. Figure 1 presents a typical summary of the physical data for a compound named (4S,6R)-4-hydroxy-2,2,6-trimethylcyclohexanone, an intermediate that was prepared during the synthesis of a natural product called (-)-boscialin. The structures of these compounds are given in Figure 2. The work was reported in 1998 on pages 591-597 of volume 61, number 5, of the Journal of Natural Products. (This type of citation is abbreviated J. Nat. Prod., 61 (5), 591-597, 1998.) We will examine all of the data in Figure 1 as we progress, but for now focus your attention on the line in bold-face type that begins with the abbreviation Anal. This line compares the experimentally determined composition of the sample to the theoretical values; combustion analysis indicates that the compound contains 69.15% carbon vs. the theoretical value of 69.19%. The theoretical and experimental values for hydrogen match exactly.

Figure 1

An Experimental Section from a Scientific Publication

(4S,6R)-4-Hydroxy-2,2,6-trimethylcyclohexanone: colorless needles (Et2O-hexane); mp 51.9-52.6C (lit.13 51.5-52.5 C); [a]25D -52.5 (c 0.75, MeOH; ee ca. 50%) [Lit.13 -105 (c 0.8, MeOH)]; IR (KBr) max 3700-3100 (OH), 2966, 2929 (C-H), 1698 (C=O) cm-1; 1H NMR (300 MHz, CDCl3) 4.32 (1H, m, H-4), 2.72 (1H, m, H-6), 2.28 (1H, m, Heq-5), 2.21 (1H, s, OH), 2.07 (1H, m, Heq-3), 1.58 (1H, m, Hax-3), 1.41 (1H, m, Hax-5), 1.21 (3H, s, CH3-2), 1.07 (3H, s, CH3-2), 1.03 (3H, d, J = 6.6 Hz, CH3-6); 13C NMR (75 MHz, CDCl3) 215.8 (C-1); 65.6 (C-4); 49.4 (C-3); 44.6 (C-2, C-5); 37.9 (C-6); 26.4 (CH3eq-C-2); 25.8 (CH3ax-C-2); 14.8 (CH3-C-6); EIMS m/z 156 [M]+ (13), 138 [M-H2O]+ (6), 110 (5), 95 (11), 83 (100), 69 (51), 57 (64), 41 (60).

Anal. C, 69.15%; H, 10.32%; O, 20.52%; calcd for C9H16O2: C, 69.19%; H, 10.32%; O, 20.48%.

Figure 2

The Structures of (4S,6R)-4-Hydroxy-2,2,6-trimethylcyclohexanone and (-)-boscialin

Additional Exercises

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