2.5B Analysis of Mixtures and Combustion Analysis

Gravimetric Analysis of Mixtures

The logic used in the preceding section can be extended to the analysis of mixtures. Some mixtures can be separated by simple physical processes. Let’s say we had a mixture of sand (SiO₂) and salt (NaCl). The latter is soluble in water but the former is not. You could weigh a sample of the mixture, wash it with water, dry the remaining sand and weigh it. The “missing” mass can be assumed to be due to the salt that dissolved and was rinsed away. You can now state what fraction of the mixture is sand and what is salt.

Now let’s make the above scenario a bit less straightforward. What if the mixture consisted of sodium chloride (NaCl) and sodium carbonate (Na₂CO₃). They are both white solids that can dissolve in water, making a simple physical separation based on solubility very difficult, if not impossible. But these two compounds behave very differently when subjected to high temperatures. NaCl is exceptionally thermally stable and can be heated via over a Bunsen burner to no effect. Na₂CO₃, on the other hand, decomposes at high temperature according to the following reaction:

\[ \ce{Na2CO3 (s) ->[heat] Na2O (s) + CO2(g)} \nonumber \]

This equation explicitly shows the states of the reactants and products: (s) refers to solids and (g) indicates a gas; this notation is often used when the states of the various reaction components need to be specified or is otherwise helpful in a particular context. More importantly for our current purpose is the fact that the above reaction has a well-defined stoichiometry. If we know how much CO₂ is produced, we can calculate the amount of Na₂CO₃ must have been present to produce it. The logic of the analysis now becomes evident: the amount of Na₂CO₃ is found by measuring the mass of the sample before and after heating; the amount of mass lost can be used to find the moles of CO₂ produced, which according to the above equation, is equal to the moles of Na₂CO₃ present in the original mixture. An example of this calculation is shown below.

Analysis of a Mixture: NaCl and Na2CO3

Problem 2.8. A mixture containing calcium carbonate CaCO3 (the main component of limestone and seashells) and sand (SiO2) as a mass of 10,195 g. To find the amount of calcium carbonate, the mixture is heated to decompose the calcium carbonate according to the following equation:

After heating, the mass of the remaining mixture is 9.610 g. What is the percentage of calcium carbonate in the original mixture?

Combustion Analysis

We started this chapter by briefly mentioning the value of knowing the mass of a particular sample of matter. If the matter is pure, that is, either an element or a compound, then if you know its mass you can calculate the quantity of its component particles. This is the value of the concept of the mole.

"So what?" you may ask. Granted, knowing the number of sucrose molecules in a heaping teaspoon may not seem particularly important and, for the most part, it isn't. But the underlying principles can be applied in different contexts to reveal the elemental composition of compounds, the first step in unraveling the nature of their component molecules.  To illustrate, let's say you are interested in plants and you want to know what compounds give rise to the aroma of lilies. You collect the oils from the flowers and separate them into several pure compounds, but you don't know anything about them other than their physical properties, such as their density, solubility, etc. What do you do? Perhaps surprisingly, a common first step is to burn it. Literally. Why? Because complete combustion of if you carefully measure the masses of the water and carbon dioxide that are generated, you can determine the relative amounts of carbon and hydrogen in the original compound, and that brings you very close to having the molecular formula.

Let's say you have isolated 0.5292 g of one of the pretty smelling compounds found in lilies. You burn it and find 1.7095 g and 0.5598 g of CO₂ and H₂O are produced, respectively. Converting each of these to moles allows us to find the moles of carbon atoms and hydrogen atoms in the original sample, as shown below.

\[ \text{moles C }= (1.7095 \text{ g } \ce{CO2}) (\dfrac{1 \text{ mol }\ce{CO2}}{44.01 \text{ g }\ce {CO2}}) (\dfrac{1 \text{ mol C}}{1 \text{ mol} \ce{CO2}}) = 0.03884 \text{ mol C} \nonumber \] 

\[ \text{moles H }= (0.5598 \text{ g } (\dfrac{1 \text{ mol }\ce{H2O}}{18.02 \text{ g }\ce {H2O}}) (\dfrac{2 \text{ mol H}}{1 \text{ mol} \ce{H2O}}) = 0.06217 \text{ mol H} \nonumber \]

Before proceeding we need to pause a moment to determine if this compound is a hydrocarbon or not. If no products other than H₂O and CO₂ are formed then there are only two possibilities: 1) the compound is composed solely of carbon and hydrogen, or 2) the compounds is composed of carbon, hydrogen and oxygen. The presence of oxygen atoms does not lead to any other combustion product so we need to determine from the masses of CO₂ and H₂O if any oxygen was part of the original structure. How? We use the moles of carbon and hydrogen to calculate their respective masses: if they sum to the mass combusted then we can be sure there was no oxygen in the compound. If, on the other hand, the sum of those two masses is less than that combusted, the "missing" mass is assumed to be oxygen. In the above case, we find:

\[ \text{mass carbon} = (0.03884 \text{mol C})(\dfrac{12.011 \text{g C}}{1 \text{mol C}}) = 0.4665 \text{g C} \nonumber \]

\[ \text{mass hydrogen} = (0.06217 \text{mol H})(\dfrac{1.0079 \text{g H}}{1 \text{mol H}}) = 0.0627 \text{g H} \nonumber \]

These add up to 0.5292 g, which matches the initial mass of the unknown sample. We conclude, therefore, the there was no oxygen in the compound and it solely consists of carbon and hydrogen. We can now proceed with the molecular formula determination.

The molar ratio of hydrogen:carbon is therefore 0.06217:0.03884, which can be written as a fraction as follows:

\[ \frac{\text{moles H}}{\text{moles C}} = \frac{0.06217}{0.03884} = 1.601 \sim \frac{8}{5} \nonumber \]

Because the molar ratio of H:C is 8:5, we know that the ratio of atoms in the molecule must also be 8:5. At this point we have determined that the empirical formula of the unknown compound is C₅H₈. The empirical formula is the simplest whole number ratio of atoms in a compound. It is important to note that the empirical formula may or may not be the same as the molecular formula. For example, compounds having the molecular formula of C₅H₈, C₁₀H₁₆, and C₁₅H₂₄ all have the same empirical formula: C₅H₈. 

To recap, we know the empirical formula is C₅H₈; because the molecular formula must be equal to a multiple of this expression, it can be written as (C₅H₈)_x, where x is an integer (usally a small one). To determine the value of x, we need some additional information about the molecular weight of the compound and that must be obtained via additional experiments [2]. Given an estimate of the molecular weight, it is simply a matter of determining how many C₅H₈ units (known generally as empirical mass units) are required to correspond to an equal mass. In this case, let's assume that another experiment revealed that the molecular weight was somewhere in the range of 130 - 140 g/mol. You can often determine by inspection the value of x. The mass of C5H8 is 68 g/mol, so we would need 2 such empirical units to have a mass that is in the stated range of possible molecular weights. Ths the molecular formula is C10H16. If you want to see the algebraic logic, it is provided below, using the midpoint of the stated molecular weight range. Note that in this case, the calculated value may be differ from an integer by a few percent; unlike in the previous step where rounding could cause serious errors, the typical uncertainty in molecular weight measurements make it less likely that the number of empirical mass units will match the estimated molecular weight perfectly. 

\[ x \text{ empirical mass units} = \dfrac{135 \frac{\text {g}}{\text{mol}}}{68 \frac{\text {g}}{\text{empirical mass unit}}} \sim 2 \frac{\text {empirical mass units}}{\text{mol}} \nonumber \] 

Footnotes and References

[1]. Why is water sometimes incorporated into crystals? As we will see, water molecules have a high affinity for ions because of attractive electrostatic forces. A magnesium ion, which as a +2 charge, is usually surrounded by a "shell" of six water molecules which more or less act as a single unit as it condenses into a crystalline structure with the nagatively charged sulfate ion (SO₄^2-). 

[2] Why refer to a "unit" of MgSO₄ instead of a molecule. As we discuss in Chapter 4, many compounds between metallic and nonmetallic elements don’t exist as molecules, but pairs of oppositely charged ions. There are no “molecules” of magnesium sulfate in the compound but, rather, discrete ions – Mg²⁺ and SO₄^2- – in a 1:1 ratio. 

[1]. Easier said than done. Although the concepts behind combustion analyses are not complicated, special equipment is needed to efficiently trap the water and carbon dioxide.

[2] The molecular weight of compounds can be obtained by several different techniques, ranging from classic "wet" methods to those based on modern instrumentation. These include:

• mass spectrometry: an instrumental method that measures the charge/mass ratio of ionized forms of molecules; it is quite sensitive and can make highly accurate measurements;
•  colligative effects: the addition of "particles" (read, molecules) to a solvent will increase its boiling point and decrease its freezing point by an amount that is proportional to the number of particles added; by measuring either the freezing point depression or boiling point elevation you can calculate the the number of particles added. The molecular weight is simply the ratio of the mass of particles added to the number of them.
• size-exclusion chromatography: this technique can separate molecules on the basis of size; the time it takes a given compound to move through the column (the retention time) is a function of its size and by comparing it to the response of a series of known standards, an estimate of the molecular weight can be obtained.
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https://openverse.org/image/ce8dc963...=epsom%20salts 

https://www.flickr.com/photos/109785...00/28844946711 "Pure Epsom Salts and Carters Senna Pods - The General Store - The Village Row - Gressenhall Farm & Workhouse, Dereham, Norfolk" by Glen Bowman is licensed under CC BY 2.0.