A hundred years ago, chemists didn’t have much in the way of instruments that could analyze substances.    They had a balance and a Bunsen burner.   So the natural thing to do was to weigh something first, then burn it and then weigh the ashes.

If you started with plant material, the ashes would be minerals such as calcium carbonate, sodium chloride, potassium carbonate and more.   Interestingly, the name potassium came from potash which is potassium carbonate. Water was poured over ashes and the salts that in the ashes were dissolved. That water was placed in iron pots and the water boiled away. The white residue was high in potassium carbonate know as "potash".

So let's say you had 100 grams of dry plant material and burned it, ant its ashes weighed 5 grams. So you can say that the plant material was 5% minerals.
You might learn that different kind of plants have a different percentage of minerals in them. So this kind of analysis could help you identify an unknown plant material. The tools to do this were simple—a balance and a fire.

Let's say we weighed out 100.0 grams of glucose. The weight of the anhydrous MgSO4 powder before the glucose was burned was 110.0 grams. Afterwards, the powder weighed 169.3 grams. So that's an increase of 59.3 grams, which is from the absorbed water. We can turn the 59.3 grams of water to moles of water with our usual grams to mole factor.
59.3 grams H₂O x 1 mole H₂O = 3.29 moles H₂O
                              18.02 grams

This counts the water molecules for us. For H₂O, we see there are 2 hydrogen atoms for every H₂O. So that means we have 2 x 3.29= 6.58 moles of hydrogen atoms. So we now know that 100 grams of glucose has 6.58 moles of hydrogen atoms in it, but this isn't enough to know its formula because we also need to know the number of carbon atoms and oxygen atoms. It's not easy to count oxygen atoms because when we burn the glucose, we are pumping more oxygen atoms into the chamber. However, we could measure the carbon atoms by measuring the CO₂ that is released.

You may remember the Apollo 13 problem where you calculated how much CO₂ two pounds of lithium hydroxide could absorb. Most hydroxides can absorb carbon dioxide. NASA likes lithium hydroxide because lithium is a very light metal. A cheaper hydroxide is lye (sodium hydroxide/NaOH). It's sold as a drain opener and also sold to those who make soap and biodiesel. Here's the equation for absorbing CO₂:
2NaOH + CO₂ --> Na₂CO₃ + H₂O

The NaOH turns into sodium carbonate (washing soda). This process also produces water, which raises a problem. If we absorb the CO₂ first, there will be water produced that will throw off our measurement of water in the canister of MgSO₄ that absorbs water. So we have to place the NaOH canister after the MgSO₄ canister.

Here's our setup. We have two canisters. The first is to capture the H₂O coming from the hydrogen atoms in the glucose. The second canister is to capture the CO₂ which is coming from the carbon atoms in the glucose. Our plan is to burn 100.0g of glucose again, but first we have to know the starting weights of the MgSO₄ and the NaOH. It doesn't matter if we know the starting weight of these powders or the starting weights of the canisters that hold these powders. We are only interested in the difference in weight from before and after the glucose is burned. So let's say the MgSO₄ canister starts off weighing 223.4 grams and after glucose is burned it weighs 282.8 grams. That's a difference of 59.4 grams (similar as to before). The NaOH canister started with a weight of 155.2 grams and then after the burn, it weighed 241.8 grams. That's a difference of 86.56 grams.

C_?H_?O_?

Using the above measurements and the power of math, we will come up with a formula for glucose. As usual we can't count atoms, but we can measure their mass. Here we have measured the mass of water absorbed by measuring the before and after weights of the MgSO₄ canister. Then we can convert that to a count of hydrogen atoms:

The increase of mass in MgSO₄ canister is just from H₂O. However, the NaOH canister's increase in weight is from CO₂ absoption but some mass from the original NaOH is lost as water vapor.
2NaOH + CO₂ --> Na₂CO₃ + H₂O

The canister's start weight was from the canister itself plus an unknown mass of the NaOH inside. The end weight was the canister itself plus an unknown amount of NaOH and an unknown amount of Na₂CO₃. Even though in chemistry we don't have to use algebra that often; this is one of those times that we do. (The below algebra is not something you will have to do in this class, but it shows the usefulness of algebra).

Let N be the starting grams of NaOH. Let C be the canister's weight. So the beginning canister's weights is C+N. Let S be the grams of Na2CO3 created. Let M be the grams of NaOH that got converted.

Canister after glucose burned:  241.8 g = C + N + S - M
Canister with NaOH start weight:  155.2 g = C + N
Subtracting the the start weight from the end weight:  86.6 g=(C+N+S-M) - (C+N) = S - M
This shows that the difference in weight is the grams of Na₂CO₃ created minus the grams of NaOH converted.
The number of grams of Na₂CO₃ created is the number of moles of Na₂CO₃ created times the 106 grams per mole that it weighs.
Let Y = moles of Na₂CO₃ created.
S=Y moles of Na₂CO₃ x 106 g/mol
From our equation: 2NaOH + CO₂ --> Na₂CO₃ + H₂O the moles of NaOH is twice that of the moles of Na₂CO₃ created. So moles of NaOH consumed is 2Y. So the grams of NaOH consumed is as follows.
M= 2Y moles of NaOH x 40.0 g/mol
Earlier we showed the the 86.6grams = grams Na₂CO₃ created minus the grams NaOH converted. We now have another way to write that:
86.6 g =Y moles of Na₂CO₃ x 106 g/mol - 2Y moles of NaOH x 40.0 g/mol
Moles cancels and the "2Y x 40" becomes 80Y, so the above simplifies to
86.6 g=106Y-80Y
86.6 g=26Y
       Solving for Y
86.6 g =  26Y
  26          26

Y=3.33 moles of Na₂CO₃
There is one carbon for each molecule of Na₂CO₃, so that means there are also 3.33 moles of carbon in this 100.0g sample of glucose. Earlier we showed that there was 6.60 moles of hydrogen in the glucose sample. Glucose also has oxygen in it, so how is that measured? We don't have to. We can subtract the grams of carbon and hydrogen from the 100.0 grams, and oxygen will be the remaining grams.
3.33 moles of carbon x 12 g/mol = 39.96 grams.   6.60 moles of hydrogen x 1.01= 6.666 g. So these total 46.6 grams. So that means that oxygen in the 100.0grams of glucose must be 100.0g-46.6g or 53.4grams.
Now we can get a count of the oxygen atoms.
53.4 g x 1 mole = 3.34 moles oxygen
               16 g

Now we are almost there. We now have a count of all 3 elements. We can write our formula as:

C_3.33H_6.60O_3.34

Normally, we don't see a mixed fractions in the subscript. So the way to fix that is to divide by the smallest number. The smallest is 3.33. So dividing that into all of these, we get these numbers: 1, 1.98, 1.003. Atoms don't combine as fractions. So the slight departure from a whole number is just the error in measurement of the weights of the canisters or not capturing 100% of the elements. So the final formula would be:

CH₂O

A weakness of this technique of finding the formula is that we don't know if the molecule of glucose is just the 4 atoms in CH₂O or is a multiple of this formula. In other words, it could be C₂H₄O₂ or C₃H₆O₃, C₄H8O₄, C₅H₁₀O₅, C₆H₁₂O₆, and so forth. The formula CH₂O is called an empirical formula because it is supported by experiment data (empirical data) but it hasn't been proven as the formula for the real molecule of glucose. Other types of experiments are needed to show that its actual formula is 6 times the empirical formula of CH₂O which is C₆H₁₂O.

Many compounds also have nitrogen in them, such as amino acids, proteins, drugs, plant extracts, and more. So another trap is added to capture any nitrogen that is in the molecule. Again, because the sample is burned, the nitrogen is in the form of nitrogen oxide (NO). The same technique is used to measure the before and after weights of the canisters. With three canisters, empirical formulas for compounds that contain carbon, hydrogen, nitrogen, and oxygen can be determined.

Instruments that capture C, H, and N are often referred to as a Combustion CHN Elemental Analyzers. Remember they also find oxygen by subtracting the mass of C, H, and N from the sample mass to find the oxygen's mass. Instruments like these make the job much easier.

Here's another CHN analyzer. On the top is an automatic sample feeder. It can do a hundred or more analyses one after another. From the calculations you saw above, these analyses can be tedious, but instruments like this can make it a lot easier.

These instrument can analyze a very small sample. They say about 1 to 3 milligrams. That's hardly visible.

Here's a third type of CHN analyzer. This one can capture sulfur dioxide, so it can be used with compounds that have sulfur in them. Some elemental analyzers can also analyze for halogens (F, Cl, I, Br). The old elemental analysis methods required weighing the before and after weights of the canisters manually. The new ones do this automatically and then give a readout of percent of each element by weight. To get the actual molecular formula, other experiments need to be done to find the molar mass (the mass of one mole of the substance).

If the compound can be turned into a vapor, the volume, mass, and temperature of the vapor can be measured and from that the molar mass can be determined. That coupled with the elemental analysis can let you turn the empirical formula into its true molecular formula.

So let's do some problems that find the molecular formula using this type of data.