empirical formula. + titrations.

okay, so, my standard grade chemistry teacher isn't good at all! and i'm not the best at maths anyway :wink:

So could someone explain, with examples? In complete dumbdumb terms, the empirical formula, and titrations?

Cheers very muchly.

:yep:

Reply 1

Kaeroll

Empirical formula: lowest ratio of integers that describe the elemental composition of a compound.

Take glucose as an example - you can find the structure on Wikipedia or Google. Its formula is C6H12O6; this is the true number of atoms in one molecule of glucose. The empirical formula is CH2O - the lowest ratio of integers that describes the relative numbers of atoms. Multiply the empirical formula by six and you get the molecular formula.

NB- this may seem obvious, but not every compound has the empirical and molecular formulae related by a factor of six. Carbon dioxide, for example, has identical molecular and empirical formulae - CO2.

Typically you'll be asked to work out, from the percentage composition (by mass) of a compound, its empirical formula, and you'll be given its molecular mass. By simple division you can then work out the molecular formula. So for ethane, you'd work out that its empirical formula is CH3 (mass = 15). If you were told that ethane's molecular mass is 30, you'd be able to see quite clearly that its molecular formula is C2H6. Do you see how that works?

Titrations are a bit more complicated. In a titration, you have a known volume of a solution but don't know its concentration. It's usually an acid or base, or a redox active compound. The idea is to take a known solution - as in known volume and concentration - and react it with the unknown.

It's probably easiest to explain by an example. Say I have a solution of NaOH and don't know its concentration. I titrate 100 mL of NaOH against a 1 molar solution of strong acid, say HCl. Let's say it takes 50 mL of my HCl to neutralise the base (you'd monitor this with a pH indicator - the choice of which is a discussion in itself).

(It's nice to tabulate this kind of thing to see what information we have. Unfortunately that doesn't seem to be possible on this forum... if anyone knows how please let me know.)

First thing to do is quickly work out the number of moles of HCl needed to neutralise the NaOH.

mol = 0.05 \;m^{3}\times1\;mol\;dm^{-3} = 0.05\;mol

As this reaction proceeds in a 1:1 ratio, 100 mL of NaOH solution must therefore contain 0.5 mol. It's then a simple division:

conc (NaOH) = \frac{0.05\;mol}{0.1\;dm^{3}} = 0.5 \; mol \; dm^{-3}

Do you follow this? (I'm aware I'm not the best communicator on earth...). This is a very simple example. A2 exams often feature potassium permanganate reacting with something (I forget), which goes in some crazy ratio like 5:2. The principles still hold:

1. Carry out a stoichiometric reaction such as a neutralisation, stopping at the precise moment of neutralisation (or as close as possible).
2. Figure out how many moles you needed of the known solution.
3. Figure out how many moles of unknown that equals.
4. Divide by volume of unknown.

Key tip... keep an eye on your units. It's easy to forget that we usually deal in mL of solution but calculate by the litre (one dm3). Extra steps may be added to the calculation such as having you make your own standard solution - dissolving a given mass in a given volume, and you need to work out the concentration. The use of weak acids and bases, which are not fully dissociated in solution, or unusual stoichiometries is often done to make you work that bit harder.

If you were looking for practical tips on carrying out a titration... whoops, sorry. Hopefully someone else will find this useful. I remember seeing a very smooth demo carried out by a member of staff at Liverpool uni, who caught a neutralisation to the drop. Very smooth indeed.

Hope this helps.

Kaeroll

Reply 2

seejayybee

Kaeroll

If you were looking for practical tips on carrying out a titration... whoops, sorry. Hopefully someone else will find this useful. I remember seeing a very smooth demo carried out by a member of staff at Liverpool uni, who caught a neutralisation to the drop. Very smooth indeed.

Thanks! I'll be reading over this a good few times to make sure I def. know it!
I've carried out a titration before, you were right with giving me the calculation, i'd like to see someone doing it right, I keep faffing it up!!!

Thanks a lot!! :woo:

Reply 3

Kaeroll

You're welcome.

It's something that comes with practice. I've not done one in a good two years so I'd probably be very rusty if I had to carry one out tomorrow. I used to label everything meticulously (always a good idea, to be honest), and be very patient - if you rush, you'll just end up with poor results. Running a quick eyeballed run to get a rough idea of where the volume is can be handy if you want to speed things up.

Kaeroll

empirical formula. + titrations.

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