Empirical Formulae and Limiting Reagents and Some Other Stuff

Okay, the final post for unit 2A is just adding on to that Stoichiometry stuff we started earlier.

I'm going to start off with the "some other stuff" part. First up on this course outline is mass to volume and volume to moles calculations... wait, haven't we already done that? I guess maybe it's time for a recap... First you need to remember this handy little formula:

n = (V/22.71) where n = number of moles and V = volume of gas at STP

When doing volume to moles calculations, just whack the volume into the equation and BAM! Conversion done.

Mass to volume is slightly harder. Convert the mass to moles, and then rearrange the above equation to convert moles to volume. Look, I'll even rearrange the equation for you so you don't have to think about it.

V = 22.71n

Limiting reagents is just the next step to the stuff covered in Simple Calculations Involving Moles. Basically, in limiting reagent questions, you get told how much you have of more than one of the reactants. You have to figure out which one is the limiting reagent (i.e. which one will run out first). To do this, first you convert the masses or volumes or whatever of all of the reagents given to moles. Then you look at the ratio between the number of moles and the coefficients to see if they line up.

Let's just say that one substance (let's call it Substance A) in the equation doesn't have a coefficient before its formula, while another substance (let's call it Substance B) has a coefficient of 2 before its formula. Since coefficients tell you the ratio between number of moles of substances in the equation, these coefficients are telling you that the ratio between substances A and B is 1:2 (i.e. Substance B has twice the number of moles of Substance A). If you have the number of moles of both substances, you can then check to see if they fit the ratio.

For example, if you have 3 moles of Substance A, and 7 moles of Substance B, the ratio doesn't quite fit the 1:2 ratio as above. You have more Substance B than is required; therefore, Substance A is the limiting reagent. Similarly, if you have 3 moles of Substance A but only 4 moles of Substance B, you don't have enough Substance B (you need 6 moles), so Substance B is the limiting reagent. One easy way to work out which substance is the limiting reagent is by dividing the number of moles of each substance by their respective coefficients. Unfortunately Chem markers nowadays prefer that you write "if/then" statements rather than pure mathematical equations (which is a bummer because I reckon mathematical equations explain stuff better). For example, you could write something along the lines of "If I have 3 moles of Substance A, then I need 6 moles of Substance B, but I have 7 moles, which is more than I need, so Substance A must be the limiting reagent" or "If I have 3 moles of Substance A, then I need 6 moles of Substance B, but I only have 4 moles; therefore, Substance B is the limiting reagent."

Once you find your limiting reagent, you can do all of the other calculations that the question requires, basing your calculations off the number of moles of the limiting reagent.

The last thing I need to cover is Empirical Formulae. I've left this until last because I think I really do need to use an example to explain how to do these (mind you, some of the topics I've covered before probably also need examples but I'm too lazy to provide them). An empirical formula is basically a formula showing all of the atoms in the simplest ratio. For example, propane's molecular formula is C3H6, but its empirical formula is CH2 (just divide both subscripts by the lowest common denominator). They're pretty easy to work out though.

1. Work out the mass of each element in the sample. This is the hardest step and one day I'll have to provide you with a more diverse range of examples to show you all the different ways you might have to work out the mass of each element in the sample. However, sometimes you might get lucky and have the masses spoonfed to you- always nice!
2. Work out the number of moles of each element.
3. Divide the number of moles of each element by the number of moles of the element with the smallest number (sounds confusing, but it might make more sense when I provide an example).
4. If each number of moles is within 0.1 of a whole number (like 1.05 or 2.01), then round each to the nearest whole number to get your simplest ratio. If not, you might have to multiply each number of moles by 2 or 3 or whatever in order to make each number of moles within 0.1 of a whole number and get a simplest ratio.
5. Write down the elements with the ratio numbers as subscripts. This is your empirical formula.

Now here's an example!

An oxide of iron was reduced to metallic iron by reaction with hydrogen gas. 11.2g of iron oxide formed 7.83g of metallic iron. Calculate the empirical formula of the iron oxide.

Now, an oxide of iron is basically a compound containing iron and oxygen.

This question doesn't provide the numbers straight away, so I guess we'll have to work them out.

Assuming none of the other products of the reaction contained iron, the 7.83g must be the mass of all of the iron in the original compound (remember, mass doesn't increase or decrease as the result of a reaction). To work out the mass of the oxygen, therefore, just subtract 7.83 from 11.2 to get 3.37g.

Now we can work out the empirical formula! I've put my working in a table because I find that it makes my working out neater and easier to follow.

Yay, pretty colours.

The next step is working out the molecular formula, if the question asks you to do so. You can only work out the molecular formula if you're given the molar mass of the molecular formula.

Basically, there's only two steps involved. Work out the molar mass of the empirical formula by adding together the mass numbers of all of the elements, then see how many times the molar mass of the empirical formula of the mass number fits into the empirical formula of the molecular formula (i.e. divide the molecular formula's molar mass by the empirical formula's molar mass). Multiply the result by the coefficients of the empirical formula.

The above question has a part b) involving molecular formulae. Part b) is "[The oxide of iron's] molar mass was 159.6g/mol. Determine its molecular formula."

First we get the molar mass of the empirical formula:

(55.85 x 2) + (16 x 3) = 159.7g/mol

And then you divide the molecular formula molar mass by the empirical formula molar mass, rounding if necessary:

(159.6/159.7) = close to 1

Well, that makes things easy. 2 x 1 = 2, and 3 x 1 = 3. Therefore the molecular formula is the same as the empirical formula- Fe2O3 which is iron (III) oxide. (Yeah, I can't be bothered with subscripts any more, I'm just typing in the subscripts using a smaller front size.)

One last thing. Sometimes, instead of given the mass of each element, you're given the percentage by mass of the compound. This makes things easy. Percentage by mass is basically how many grams of each element you have in 100g of the compound (per cent = out of 100), so just pretend you have 100g of the compound and go from there.

So that's all from 2A Chemistry, unless you're desperate for me to explain the Applied Chemistry bits (describe and explain examples where rates of reaction have been altered in and around the home). 2B just has acids and bases, organic chemistry and oxidation and reduction (a.k.a. redox), but I hate redox, not because I'm terrible at it, but because I found it a rather annoying topic for some reason. ANYWAY, that's it from me for this post. I'll be back soon to talk about acids and bases!