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# Chemical Analysis in Solutions I

## Chemical Analysis Basics

#### Chemical Analysis

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.

#### Using Stoichiometry in Chemical Analysis

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 CO2 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 CO2 and all the hydrogen went into the H2O.

The molecular mass of CO2 is 12.011 + 2 * 15.9994 = 44.01 g/mole. If we have 8.8 grams of CO2, we have 8.8/44.01 = .200 moles of CO2, and therefore .200 moles of C to start with.

The molecular mass of H2O 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 CH4. The actual molar formula can be any integer multiple of this (CH4, C2H8, C3H12, 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 CO2 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:

1. Identify the products and reactants.

CxHyOz --> CO2 + H2O

2. Convert mass data to moles from the data collected in the lab for the known masses.

0.440g CO2/44.01 g/mole CO2 = 0.0100 moles CO2
0.090g H2O/18.02 g/mole H2O = 0.0050 moles H2O

3. Determine the moles reacting.

Since all the C comes from the carbohydrate, if we get 0.0100 moles of CO2, we started with 0.100 moles of C.
Since all the H comes from the carbohydrate, if we get 0.0050 moles of H2O, 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.

4. 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.

5. 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

6. Determine the ratios of the moles of components.

0.0100 moles C : 0.0100 moles H : 0.0100 moles O.

7. Determine the simplest formula.

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

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

### Practice with the Concepts

##### Determining the empirical formula

Test your understanding of each step by working through the following "lab" exercise.

0.105g of YACH (Yet another carbohydrate) is analyzed by combustion. In this case, the resulting products are 0.257 g carbon dioxide and 0.0350 g water.

How many moles of carbon and hydrogen are involved?

How much mass is due to each of the elements carbon and hydrogen?

How much mass is due to oxygen? How many moles of oxygen were involved?

What are the ratios of carbon to oxygen and hydrogen to oxygen?

What is the empirical formula for the original carbohydrate?

### Discussion Questions

• 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?