| % C = | 2 moles C (12.01 g/mole C) 46.07 g ethanol |
100 % = 52.14 % |
The composition of a compound is often expressed in terms of the weight percent of each element in the compound.
For example, consider ethanol, which has the formula C2H6O. One mole of ethanol has a mass of 46.07 g. The elemental formula indicates that one mole of ethanol contains two moles of carbon, six moles of hydrogen, and one mole of oxygen. Thus the composition of the compound by mass is
| % C = | 2 moles C (12.01 g/mole C) 46.07 g ethanol |
100 % = 52.14 % |
Similarly the weight percents (actually mass percents) of hydrogen and oxygen in ethanol are
| % H = | 6 moles H (1.008 g/mole H) 46.07 g ethanol |
100 % = 13.13 % |
| % O = | 1 mole O (16.00 g/mole O) 46.07 g ethanol |
100 % = 34.73 % |
Notice that the sum of the weight percents of all the elements in a compound must equal 100 %.
Elemental analysis is an experiment that determines the amount (typically a weight percent) of an element in a compound. Just as there are many different elements, there are many different experiments for determining elemental composition. The most common type of elemental analysis is for carbon, hydrogen, and nitrogen (CHN analysis). This type of analysis is especially useful for organic compounds (compounds containing carbon-carbon bonds).
The elemental analysis of a compound is particularly useful in determining the empirical formula of the compound. The empirical formula is the formula for a compound that contains the smallest set integer ratios for the elements in the compound that gives the correct elemental composition by mass.
Consider the compound octane, whose molecular formula is C8H18. The ratio of carbon to hydrogen atoms in octane is 8:18. This ratio has a value of 0.444444, which can also be expressed by ratios such as 16:36 and 1:2.25. The smallest ratio of integers that gives this ratio is 4:9 and therefore the empirical formula of octane is C4H9.
Note that the molecular formula is an integer (n) multiple of the empirical formula. For octane the empirical formula is C4H9; thus the molecular formula is C4nH9n. In this case n = 2, which produces C8H18.
If one prepares or isolates a new compound, the determination of the molecular formula for the compound is an important goal. This determination typically involves two steps: (1) elemental analysis of the compound and (2) determination of the molecular mass. The elemental analysis permits determination of the empirical formula, and the molecular weight and elemental analysis permit determination of the molecular formula.
To illustrate this process, consider an unknown compound containing only carbon and hydrogen whose elemental composition is 84.72% carbon and 15.41% hydrogen. (Notice that the experimental error has produced values whose sum is not exactly 100.00 %.) Suppose one had a 100.0 g sample of this compound. (Any amount could be chosen; 100.0 g just happens to be a convenient number to use.) The 100.0 g sample of the compound contains 100.0 g (0.8472) = 84.72 g carbon and 100.0 g (0.1541) = 15.41 g hydrogen.
These amounts can be expressed in terms of moles of each element.
| 84.72 g C | 1 mole C 12.01 g C |
= 7.054 mole C |
| 15.41 g H | 1 mole H 1.008 g H |
= 15.29 mole H |
The atom ratio C:H is thus 7.054:15.29 or C7.054H15.29. Of course formulas are generally expressed in terms of integers, and the empirical formula is the smallest ratio of integers that yields the correct atom ratio. There are several ways to arrive at the correct set of integers. If the molecule is reasonably small, trial and error is frequently the most direct approach.
First divide each term in the ratio by the smallest value (7.054 in this case).
7.054 : 15.29 = 2.168 = 1.000 : 2.168
In this case, the integer for C cannot be 1, because this integer does not lead to an integer value for H. Keep multiplying this ratio by small integers (2, 3, 4, ...) until a ratio of integers is obtained.
7.054 : 15.39 = 2.182 = 2.000 : 4.336
7.054 : 15.39 = 2.182 = 3.000 : 6.504
7.054 : 15.39 = 2.182 = 4.000 : 8.672
7.054 : 15.39 = 2.182 = 5.000 : 10.84
7.054 : 15.39 = 2.182 = 6.000 : 13.01
In this case, the smallest ratio of integers that produces 2.182 for C:H is 6:13 and thus the empirical formula is C6H13.
Be aware that the elemental analysis is not perfectly accurate. The experimental error will generally produces atom ratios that are not perfect integers but are close to integers.
A separate measurement of the molecular mass is required to determine the molecular formula. For this compound mass spectrometry gave a molecular mass of 170.
The molecular formula is C6nH13n, thus the molecular mass is
The molecular formula is thus C12H26.
Objectives
Carbon-Hydrogen (CH) analysis is performed by burning the unknown sample. As a result of the complete combustion of the compound, all of the carbon in the compound is converted to carbon dioxide gas and all of the hydrogen in the compound is converted to water vapor.
x CO2 (g) + y/2 H2O (g) + A
The combustion is conducted in the presence of excess oxygen ( z >> x + y/2 ). The symbol A represent other elements in the system, which burn to produce either a solid residue or an unreactive gas (unreactive for the purposes of this analysis).
The sample is typically wrapped in a tin capsule and inserted into a furnace held at 1200oC. A steady stream of an unreactive gas (e.g., helium) is passed through the furnace and a burst of pure oxygen gas is added as the sample capsule is inserted into the oven. As the sample burns, the temperature rises to approximately 1700oC. The furnace contains a catalyst (Cr2O3) that facilitates the complete combustion of the sample. Copper turnings are also present to ensure that any nitrogen in the sample is reduced to N2, which is unreactive.
As the gas leaves the furnace, the carbon dioxide and water vapor produced by the combustion reaction are carried out of the furnace. The gas stream first passes through a desiccant, typically Mg(ClO4)2, that removes all moisture from the gas stream. Next the gas stream passes through a tube containing NaOH, which removes all carbon dioxide from the gas stream (forming NaHCO3).
By weighting the tubes containing Mg(ClO4)2 and NaOH before and after the combustion reaction, it is possible to determine the amounts of water and carbon dioxide produced by the combustion reaction and thus the amounts of hydrogen and carbon in the unknown compound.
The experimental apparatus is shown in the image below. The furnace lies at the left side of the image. The tubes containing Mg(ClO4)2 (center of the image) and NaOH (right side of the image) rest on balances, which permit their masses to be determined. (The gas line connecting the two tubes is very light and flexible. For the purposes of this simulation, it is assumed they do not perturb the measurements of the masses of the tubes.)
The experimental procedure is as follows.
1. Select an unknown compound (A, B, C, D, or Random).
2. Zero the readings on the balances.
3. Initiate the combustion reaction by selecting the "Burn Sample" button.
4. Wait a few seconds for the combustion reaction to occur, the water and carbon dioxide to be absorbed, and the mass readings to stabilize.
5. Record the masses of water and carbon dioxide produced by the combustion of the sample.
6. Calculate the empirical formula for the unknown compound. The molecular mass of the compound is provided; use this value and the empirical formula to determine the molecular formula for the unknown compound.
Furnace
H2O Trap
CO2 Trap |