# Empirical Formulas

## Chm 1311 Lecture for 17 June 1999 (cont.)

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### AceticAcid

We've seen HC2H3O2 as the expression that Brady & Holum prefer for acetic acid. They do that since the boldface H is the acidic proton in this molecule, and it makes the formula follow the common practice for acids, HCl, H2SO4, etc.
In other contexts, you'll have to put up with reading acetic acid into other formulae like CH3CO2H which comes close to the molecule's structural formula (about which more in later chapters): ```   H    O
|   //
H--C--C          Acetic Acid or Ethanoic Acid
|   \
H    O--H
```
For the purposes of counting atoms, however, we'll just use the obvious:

C2H4O2

### PercentageComposition

When analyzing molecules, their atomic makeup is often given in terms of their percentage composition. This is an experimental result of elemental analyses. In this case, it would be expressed as x% C, y% H, and z% O. If obtained by weighing, these will be weight percents.
Of course, if we know the molecular formula (C2H4O2), these percentages are a snap! We just compare the weights of each kind of atom (AW) in the molecule to its overall molecular weight (MW).

• MW of C2H4O2 = 2×AW C + 4×AW H + 2×AW O
= 2(12.011 g/mol) + 4(1.008 g/mol) + 2(15.999 g/mol)
= 24.022 g/mol + 4.032 g/mol + 31.998 g/mol
= 60.052 g acetic acid / mol acetic acid
• wt % C = (24.022 g / 60.052 g) × 100% = 40.002% by weight
• wt % H = (4.032 g / 60.052 g) × 100% = 6.714% by weight
• wt % O = (31.998 g / 60.052 g) × 100% = 53.284% by weight
But this is doing the actual experimental problem exactly backwards! Here we're deriving weight % from the molecular formula. In the real world, we're trying to obtain the molecular formula from weighing the decomposed components of a molecule.

Sounds unappetizing, doesn't it?

Fortunately, molecular corpses aren't as messy as those of higher organisms. However, they aren't quite as clean as I've led you to believe. Rarely does a decomposition yield pure elements to be weighed and quoted as above. A more realistic example is of elemental analysis by combustion. Most things burn, if suitably encouraged. (Indeed, that's how Lavoisier proved that diamond was naught but a form of carbon...he burned a diamond and recovered only CO2.)

We won't be so profligate. Combustion analysis is wonderfully suited to organic compounds; they're often flammable and often consist, like acetic acid, of just carbon, hydrogen, and oxygen. They burn to CO2, H2, and ... er ... well, the oxygens don't burn but we can figure out how much of the weight to assign the oxygen by subtracting what we've found of the carbon and hydrogen. So oxygen gets done by difference.

### CombustionAnalysis

In practice, a combustible sample is placed in an oven under an oxygen stream (no need to worry about the added oxygen since we won't anaylze it anyway) and heated to burn completely.
The H2O in the effluent gas stream is trapped in a dessicant (like P4O10, which folks still call "phosphorous pentoxide" for reasons explained below), the CO2 is neutralized in a base (like NaOH), and the excess O2 just vents off. Weighing the traps before and after the combustion gives one the weight of the evolved CO2 and H2O.

Imagine that a 5.0000 g sample of some unknown (actually acetic acid) is subjected to such a combustion analysis. The traps yield 7.3285 g CO2 and 3.000 g H2O. (Too neat. Too neat. I've not introduced the requisite experimental error here; fuzz it up a bit if you like.)

But CO2 isn't C, and H2O isn't H. We have to find out how much hydrogen those 3 g of water are worth. Ditto the carbon. Do do that, we'll need the molecular weights of CO2 and water so that we can compare mols to mols, the only valid comparison:

• MW CO2 = (AW C) + 2×(AW O)
= 12.011 g/mol + 2(15.999 g/mol) = 44.009 g CO2 / mol CO2
```                    1 mol CO2     1 mol C  12.011 g C
wt C = 7.3285 g CO2 ------------ --------- ----------
44.009 g CO2 1 mol CO2 1 mol C
= 2.000 g C
```
• MW H2O = (AW O) + 2×(AW H)
= 15.999 g/mol + 2(1.008 g/mol) = 18.015 g H2O / 1 mol H2O
```                     1 mol H2O     2 mol H   1.008 g H
wt H = 3.000 g H2O  ------------ ----------- ---------
18.015 g H2O  1 mol H2O   1 mol H
= 0.3357 g H
```

So a total of 2.000+0.3357=2.336 g of the unknown is C and H. The rest, 5.0000-2.336=2.664 g, must be O (with confirmatory experimental evidence that there are no other kinds of atoms in the unknown). Now we could quote the unknown's composition by weight percent:
• wt % C = (2.000 g C/5.0000 g X) × 100% = 40.00% (surprise²)
• wt % H = (0.3357 g H / 5.0000 g X) × 100% = 6.714%
• wt % O = (2.664 g O / 5.0000 g X) × 100% = 53.28%
But this time derived from "experimental" data.

### Empirical Formula

If we're aiming at the formula subscripts, we must compare atom counts to atom counts not weights to weights. So we convert these percentages into moles (atom count). Thus, in 100 g (for convenience) of unknown X:
• # mol C = 40.00 g C [ 1 mol C/12.011 g C ] = 3.330 mol C
• # mol H = 6.714 g H [ 1 mol H/1.008 g H ] = 6.661 mol H
• # mol O = 53.28 g O [ 1 mol O/15.999 g O ] = 3.330 mol O
And we're ready to call it C3.330 H6.661 O3.330 ?!?

No, of course not. All we wanted was the ratio, and that ratio is terribly, suspiciously, mathematically like 1:2:1. We could verify that by dividing each of the subscripts by 3.330, and hydrogen's would work out 2.0003, a bit too good! But then we've been playing with exact (not real experimental) weights for this example. Had you fuzzed it up back there when I suggested it, this might've come out 2.016 which I would still have insisted is 2!

So the molecule is CH2O, and what happened to the acetic acid? (The salad and salad oil is premature at this point.) What we have discovered is the empirical formula (with the simplest subscript ratio). That happened to P4O10 and we thought it was P2O5 for the longest time; thus "phosphorus pentoxide".

How do we disabuse ourselves of this too simple formula? We need one additional piece of information: the molecular weight of the unknown! But wait, you say, isn't this circular? Don't we need the molecular formula to calculate the molecular weight? True, but we can determine the MW experimentally too! There are many physical properties that depend upon the number of molecules (or moles) of a substance and NOT on what exactly the substance is. The easiest one to understand is the volume of a vapor of the substance; we'll see in chapter 10 that (Avogadro's Law) at a fixed temperature and pressure, the volume of a vapor is directly proportional to the number of molecules (mols) in it.

So this empirical formula gives a formula weight of

(AW C) + 2×(AW H) + (AW O) = 12.011 + 2.016 + 15.999 = 30.03 g/mol

From the vapor volume of a known weight of the substance X, we find its molecular weight to be (surprise²) 60.052 g/mol. So our empirical formula weight was obviously a factor of 2 low. We conclude then that the molecular formula of X must be C2H4O2.

### Isomers

iso = same
meros = share
Perhaps. But perhaps not. It certainly has that formula, but does anything else?

This combustion analysis is very powerful for getting empirical formulae of organic compounds, but it turns out that their ubiquitous carbon atoms are too versatile in their bonding!
Isomer:
Greek for
"Sharing the
same"
atoms.
This is too early in the course to introduce the complication of isomers, but it'll emphasize the richness of organic bonding. So just ponder the following with your "suspension of disbelief" on high for a moment. Isomers are bond rearrangements of the same atoms; they still form one molecule, but the rearrangment gives a new compound with unique properties. We can test for those properties to see which isomer we've found. (Don't confuse these with isotopes, which are atoms with differing neutron counts. Those guys are chemically identical!)

We can rearrange 2 carbons, 2 oxygens, and 4 hydrogens in more than one way. The following are two perfectly reasonable examples. By the end of the course, you'll be able to knock out examples for yourselves.
```   H    O              H    H    O
|   //               \   |   //
H--C--C          or      O--C--C
|   \                    |   \
H    O--H                H    H
```
which have very different chemistries. For one thing, only acetic acid's bold H is acidic. None of the hydrogens on the other molecule (hydroxyethanal) are acidic. So it'd be easy to tell which one we had. Drop it in water and measure the acidity; if it's significant, then we had acetic acid all along!

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