WebLecture

Chemical analysis is like a complex logic problem. First, we need to isolate the compound we are interested in from other compounds in a mixture or solution. Sometimes the only way to measure this amount is to force the compound of interest to react with some other chemical, and measure the output, then calculate the moles of the compound that participated in the reaction. Then we look at the molar ratios to determine the empirical or simplest formula for the compound of interest.

For example, let's assume that we have a compound which we know consists only of hydrocarbons (hydrogens bound to carbon atoms which are themselves linked in a chain). We want to determine the formula for the hydrocarbon. When we burn a sample in air, it is completely consumed and yields 8.8 grams of CO_{2} and 7.2 grams of water.

We know all the oxygen involved in the reaction came from the (for our purposes) unlimited supply in the atmosphere, but all the carbon and hydrogen in the products came from the sample. All the carbon went into the CO_{2} and all the hydrogen went into the H_{2}O.

The molecular mass of CO_{2} is 12.011 + 2 * 15.9994 = 44.01 g/mole. If we have 8.8 grams of CO_{2}, we have 8.8/44.01 = .200 moles of CO_{2}, and therefore .200 moles of C to start with.

The molecular mass of H_{2}O is 2 * 1.0079 + 1 * 15.9994 = 18.02. If we have 7.20 g of water as output, we have 7.20/18.02 = .400 moles of water, and (since we have 2 hydrogens for each molecule of water) .800 moles of hydrogen.

Our ratio of hydrogen to carbon is .8/.2 or 4:1, and the simplest formula or empirical formula for the gas we burned in air is therefore CH_{4}. The actual molar formula can be any integer multiple of this (CH_{4}, C_{2}H_{8}, C_{3}H_{12}, etc.); we cannot determine the molar formula without further experimental data.

Of course, the whole process is more complicated because our reaction is never 100% efficient, but we ignore that for the moment.

Suppose that we combust 0.290 grams of some carbohydrate. We know from our previous discussion that all carbohydrates dissociate to water and carbon dioxide, and in this particular experiment, we measure 0.440 grams of CO_{2} and 0.090 grams of water in these two products.

To determine the empirical or simplest formula from this experimental data, we follow these steps:

- Identify the products and reactants.
C

_{x}H_{y}O_{z}--> CO_{2}+ H_{2}O - Convert mass data to moles from the data collected in the lab for the known masses.
0.440g CO

_{2}/44.01 g/mole CO_{2}= 0.0100 moles CO_{2}

0.090g H_{2}O/18.02 g/mole H_{2}O = 0.0050 moles H_{2}O - Determine the moles reacting.
Since all the C comes from the carbohydrate, if we get 0.0100 moles of CO

_{2}, we started with 0.100 moles of C.

Since all the H comes from the carbohydrate, if we get 0.0050 moles of H_{2}O, we started with 2*0.0050 moles of hydrogen, or 0.0100 moles H. So there was an equal number of moles of carbon and hydrogen in the original compound. - Determine the masses for the known components. The products contain
0.0100 moles H, for a mass of 0.0100 * 1.0079 g/mole = 0.010g

0.0100 moles C, for a mass of 0.0100 * 12.01 g/mole = 0.120g

The total mass of the products is the sum of these two: 0.130g. - Determine the remaining mass and moles of the final component.
Since we burned 0.290g of the carbohydrate to start with, and we've accounted for 0.130g, the oxygen must be responsible for the remaining mass of 0.290 - 0.130 = 0.160g.

0.160g /16.0g/mole = 0.0100 moles of oxygen - Determine the ratios of the moles of components.
0.0100 moles C : 0.0100 moles H : 0.0100 moles O.

- Determine the simplest formula.
x = y = z = 1. The carbohydrate we started with must have an empirical formula CHO.

- The molecular formula will be any multiple of this ratio: CnHnOn.

- Why doesn't the empirical formula equal the molecular formula in all cases?
- What is the equivalence point of a titration? Why do we care?
- How can we use indicators?

- Practice with more chemical analysis at the University of Wisconsin Stoichiometry site.
- A very brief introduction to the basic methods of chemical analysis from the University of Vermont Chemistry department gives you a better idea of the complexity of this part of chemistry.

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