Empirical and Molecular Formulas Empirical determined by observation

Empirical and Molecular Formulas Empirical: determined by observation or measurement. Not related to “empire, ” surprisingly.

Going Back to a Worksheet Problem How much does 1 mole of sodium oxide weigh? 1 mole Na 2 O = 23*2 + 16 = 62 grams How much of that is the sodium? 2 moles Na+ = 46 grams How much of that is the oxide? 1 mole O 2 - = 16 grams

Turn Those Into Percents 46 g Na+/62 g total = 74. 2% Na+ 16 g O 2 -/62 g total = 25. 8% O 2 - You can do this same calculation for any compound, given it’s formula: Fe 2(CO 3)3: 111. 7 g Fe 3+/291. 7 g total = 38. 3% iron 36 g C/291. 7 g total = 12. 3% carbon 144 g O/291. 7 g total = 49. 4% oxygen On its own, this is actually pretty useless. (much like this cat)

However…. There are various gizmos* out there that will take an unknown sample for you, and measure what the percents of each element are. *Atomic absorption spectrometers are the most common, but not the only method. Search for “elemental analysis”.

So We Find Some White Powder* Given some unknown compound, you can send it through the instrument (or, more typically, pay somebody else to do it for you). and get some results, for example: 25. 43% Na 39. 19% Cl 35. 39% O *About 95% of compounds on earth are white powders; not just illicit substances.

So We Find Some White Powder* Given some unknown compound, you can send it through the instrument (or, more typically, pay somebody else to do it for you). and get some results, for example: 25. 43% Na 39. 19% Cl 35. 39% O Because we already know something about chemistry, we can narrow things down a lot: • It’s ionic (there’s sodium) • Cl and O combine to make one of four different anions: Cl. O-, Cl. O 2 -, Cl. O 3 -, Cl. O 4 • Because charges must balance, there are only four possible formulas: Na. Cl. O 2 Na. Cl. O 3 Na. Cl. O 4 You could find the percentages for each of those formulas, but… *About 95% of compounds on earth are white powders; not just illicit substances.

That’s Too Much Work Also, if your compound had been covalent, there would have been nearly infinite possible combinations…

How to go Backwards? 25. 43% Na 39. 19% Cl 35. 39% O Start with an initial assumption: we have 100 g of our unknown. (This number doesn’t matter, but if you’re going to use percents, why make it hard? ) Therefore we have: 25. 43 g Na 39. 19 g Cl 35. 39 g O

How to go Backwards? 25. 43% Na 39. 19% Cl 35. 39% O Start with an initial assumption: we have 100 g of our unknown. (This number doesn’t matter, but if you’re going to use percents, why make it hard? ) Therefore we have: 25. 43 g Na 39. 19 g Cl 35. 39 g O Convert each of these values to moles (using atomic masses): 1. 105 moles Na 1. 104 moles Cl 2. 212 moles O

How to go Backwards? 25. 43% Na 39. 19% Cl 35. 39% O Start with an initial assumption: we have 100 g of our unknown. (This number doesn’t matter, but if you’re going to use percents, why make it hard? ) Therefore we have: 25. 43 g Na 39. 19 g Cl 35. 39 g O Convert each of these values to moles (using atomic masses): 1. 105 moles Na 1. 104 moles Cl 2. 212 moles O *IMPORTANT*: you cannot just round these numbers and reliably get the formula, you need to find the ratios between them. This is most easily done by dividing all of them by the smallest one: 1 Na 1 Cl 2 O = Na. Cl. O 2

A Few Notes • If, after your final step (dividing by the smallest), you get something like 1. 003 or 2. 985, it’s ok to round at that point. • Sometimes you will get things like: 1 C 1. 5 H = CH 1. 5 ? !? If that happens, double everything: 2 C 3 H = C 2 H 3 • On the other hand, if you get: This is more likely an error than: • Don’t round atomic masses. 1 C 1. 125 H 8 C 9 H = C 8 H 9

52. 14% C An Example 13. 13% H 34. 73% O

52. 14% C 1. Assume 100 g. 52. 14 g C An Example 13. 13% H 34. 73% O 13. 13 g H 34. 73 g O

52. 14% C 1. Assume 100 g. 52. 14 g C 2. Convert to moles 4. 34 moles C An Example 13. 13% H 34. 73% O 13. 13 g H 34. 73 g O 13. 02 moles H 2. 17 moles O

52. 14% C 1. Assume 100 g. 52. 14 g C 2. Convert to moles 4. 34 moles C An Example 13. 13% H 34. 73% O 13. 13 g H 34. 73 g O 13. 02 moles H 3. Divide each by smallest (2. 17) to get ratios 2 C 6 H 4. Formula is C 2 H 6 O 2. 17 moles O 1 O

Empirical vs Molecular Formula C 2 H 6 O is your empirical formula. If were an ionic compound, it is the formula of your compound, but not necessarily for a covalent compound. The actual formula (molecular formula) could be any multiple of that: If you do out the percents for C 4 H 12 O 2, you get… 52. 14% C 13. 13% H 34. 73% O Same for C 6 H 18 O 3, C 8 H 24 O 4, etc. We need more information to figure what the real formula is. Question to think about: WHY do covalent compounds have a molecular formula that might be different than the empirical formula, but ionic compounds don’t?

Need More Information Take our C 2 H 6 O. Let’s say we find some way to measure that the molecular weight of our compound is 138. 2 g/mole* C 2 H 6 O = 46. 07 g/mole (empirical formula) 138. 2 / 46. 07 = 3 So our molecular formula is three times heavier than our empirical formula, so our formula must be three times as big: C 6 H 18 O 3 (molecular formula) If you don’t get this information, the best you can do is find the empirical formula. * Typically mass spectrometry

Summary • Percents can be found from formulas just using mass • Working backwards from percents to formula is more difficult • Steps are: 1. Assume some amount (typically 100 g) 2. Convert masses to moles 3. Divide by smallest to get ratios 4. If they’re still not whole numbers, double or triple everything 5. If you have a molecular weight, determine what multiple of the empirical formula the molecular formula is.