(In most chemistry questions, you can take the Avagadro constant as being 6 x 1023, and forget about the .02 on the end!)
This number is special because it is the number of carbon atoms that makes exactly 12 grammes of carbon, i.e. if you had 6.02 x 1023 carbon atoms on a scale, it would read exactly 12 grammes. Although the Avagadro constant is officially defined in terms of carbon, it is probably easier to think of it in terms of the simplest atom, hydrogen.
6.02 x 1023 hydrogen atoms have a mass of 1 gramme. This means that the mass of one hydrogen atom must be 1/(6.02 x 1023) of a gramme. We refer to 6.02 x 1023 hydrogen atoms as a mole of hydrogen atoms. Similarly, if you had 2 grammes of hydrogen, that would be two "Avagadro constant's worth", in other words 2 moles of hydrogen atoms.
Similarly, 3 grammes of hydrogen is 3 moles of hydrogen atoms and 17.361 grammes of hydrogen is 17.361 moles of hydrogen atoms. It gives you a convenient way of counting atoms.
Relative Atomic MassEvery atom has a relative atomic mass (R.A.M.) which tells you how heavy the atom is when compared to hydrogen. For instance, every helium atom is 4 times as heavy as a hydrogen atom. Helium has a R.A.M. of 4.So how much would a mole of helium atoms weigh? |
 4 hydrogen atoms weigh the same as 1 helium atom |
Well, if a mole of hydrogen atoms has a mass of 1g, and helium atoms are 4 times as heavy as hydrogen atoms, then a mole of helium atoms must be 4g (4 times as much as for hydrogen). Similarly, 2 moles of helium atoms would be 8g (twice 4g), 10 moles of helium atoms would be 40g (ten times 4g) etc.
What about other elements? Take oxygen for example. This has a Relative Atomic Mass of 16, meaning that each oxygen atom is 16 times as heavy as a hydrogen atom. One mole of oxygen must be 16g. 3 moles of oxygen must be 3 x 16 = 48g etc.
In short, if you know what the R.A.M. of an element is, then you can find the mass of one mole of the element is by slapping 'grammes' on the end of that figure. If you want the mass of a different number of moles (16 moles, 23.5 moles etc.) then multiply the mass of one mole by that figure.
Suppose you have the number of grammes of an element and you want the number of moles that are present. In this case, you do the opposite - you divide by the R.A.M., so 48 grammes of oxygen contains 48 / 16 = 3 moles of oxygen. Since the R.A.M. of mercury is 200.6, then 500 grammes of mercury must contain 500 / 200.6 = 2.492 moles of mercury. ####
The basic formula is:
Of course, the formula can be turned round:
| Relative Atomic Mass = | |
or even:
| Number of moles = | |
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Here's a "magic triangle" that might help you when you come to moles calculations. Simply cover up the quantity that you want to find in the triangle and the other two quantities appear in the right relationship for the calculation. For example, to calculate the number of moles, just cover up the number of moles in the triangle and you end up with the mass of the element over Relative Atomic Mass. |  |
One molecule of iron sulphide consists of one atom of iron chemically bonded with one atom of sulphur - they join on a one-to-one basis. It therefore follows that one mole of iron sulphide must contain one mole of iron atoms joined to one mole of sulphur atoms.
One mole of iron atoms = 56 grammes (as the R.A.M. of iron is 56).
One mole of sulphur atoms = 32 grammes (as the R.A.M. of sulphur is 32), so ...
One mole of iron sulphide = 56 + 32 = 88 grammes. Similarly, two moles of iron sulphide are contained in 2 x 88 = 176 grammes, and 4 moles in 4 x 88 grammes etc. After that it is exactly the same as what you met up above.
So, to find the Relative Molecular Mass (not Atomic - we are dealing with compounds here), just add the R.A.M. values for the atoms that make up the molecule. You must be careful to add them in the correct proportion - there may be more than one atom of an element in a compound.
For instance, glucose has a chemical formula of C6H12O6, where carbon has a R.A.M. of 12, oxygen has an R.A.M. of 16 and hydrogen has an R.A.M. of 1. To find the R.M.M. of glucose, you don't just do 12 + 16 + 1, because there are 6 carbons in there, 12 hydrogens and 6 oxygens. The final calculation is:
The R.M.M. of glucose is 72 + 12 + 96 = 180. This means that one mole of glucose is 180 grammes, half a mole would be half of 180 = 90 grammes, eleven moles would be 11 x 180 = 1980 grammes (or 1.98 kilogrammes) etc.
Carbon (symbol C) = 12, Mercury (symbol Hg) = 200.6, Hydrogen (symbol H) = 1, Oxygen (symbol O) = 16, Sodium (symbol Na) = 23, Sulphur (symbol S) = 32, Chlorine (symbol Cl) = 35.5.
Perhaps an example would clear this up. Suppose we had exactly 29.25g of sodium chloride. If you do the calculations you will see that this is exactly half a mole. We then dissolve this sodium chloride in 200cm3 of a litre of distilled (totally pure) water. This means that we have half a mole of salt dissolved in 0.2 of a litre, which is the same concentration as if we had 2.5 moles dissolved in 1 litre. We therefore say that the concentration of the solution is 2.5 Molar (or 2.5M), even though we only have one fifth of a litre.
To calculate the concentration of any solution, divide the number of moles of substance by the volume of water in litres (dm3). Here is another example:
Dissolve 34.11g of sodium hydroxide (R.M.M. = 40) in 30cm3 of distilled water. 34.11g is 34.11 / 40 = 0.85275 moles of NaOH dissolved in 0.03 litres of water, so the concentration is 0.85275 / 0.03 = 28.425 Molar. That's very concentrated! Typically, you won't come across concentrations much higher than 2M.
The basic formula is:
Of course, the formula can be turned round:
| Concentration (Molar) = | |
or even:
| Volume (litres) = | |
|
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Here's another "magic triangle" that might help you, this time when you come to molarity concentrations. It works in the same way as the previous one - simply cover up the quantity that you want to calculate and the other two will appear in the right relationship. |  |
So what's titration then? It is the process of working out the concentration of a solution by reacting it with a solution of known concentration. This is often (almost always) done with acids and alkalis. Let me explain with an example:
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Suppose we have a sample of sulphuric acid (H2SO4) of unknown concentration in a flask, but whose volume is known accurately. We set up a burette (a long thin graduated tube with a tap at the base) above it and fill it to a certain graduated mark with sodium hydroxide solution of concentration 0.5M, as shown in the diagram on the right.
Then we put some litmus solution in the acid - it goes red, of course. This litmus will tell us when enough of the sodium hydroxide has reacted to neutralise the acid. We gradually let the sodium hydroxide run into the acid by opening and closing the tap on the burette. The flask is gently swirled after each part of the hydroxide has been let in, and the colour carefully noted. The swirling is necessary to make sure that the solutions are thoroughly mixed - you will find a dribble of blue appears as you let the hydroxide in, but that the litmus goes back to red as you swirl. Eventually, even with swirling, you notice that the colour is starting to change. This is the point to let the hydroxide in one drop at a time, until you are sure that all the red colour has disappeared. At this point, the reading on burette is noted again. |
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It would help if we put some figures in this example:
| Volume of acid in flask: | 120cm3 = 0.12 litres |
| Initial burette reading: | 100cm3 |
| Final burette reading: | 35cm3 |
| Volume of hydroxide used: | 100cm3 - 35cm3 = 65cm3 |
| Concentration of hydroxide: | 0.5M |
| Number of moles of hydroxide: | 0.5M x 0.065 litres = 0.0325 moles |
The balanced chemical equation for the neutralisation of sodium hydroxide with sulphuric acid is as follows:
2 H2O (l) + Na2SO4 (aq)
The important thing to note here is the left side of the equation: It takes 2 moles of sodium hydroxide to neutralise 1 mole of sulphuric acid. However, we don't have 2 moles of sodium hydroxide, we only have 0.0325 moles. This will neutralise half as many moles of acid:
| Number of moles of acid: | 0.0325 / 2 = 0.01625 moles |
| Concentration of acid: | = moles / volume (in litres) |
| = 0.01625 / 0.12 | |
| = 0.13542 Molar |
We have finally calculated the concentration of the acid.
| + | H2O + NaCl |
||
| (One molecule of hydrochloric acid) |
(One molecule of sodium hydroxide) |
Similarly, 0.1 moles of hydrochloric acid neutralise 0.1 moles of sodium hydroxide, 237.4 moles of acid neutralise 237.4 moles of hydroxide etc. They always react in a 1 : 1 ratio. N.B. The same isn't true of volumes! You can't assume that 1 litre of hydrochloric acid neutralises 1 litre of hydroxide - that depends on the concentration of the chemicals as well!
| + | 3 H2O + Na3PO4 |
||
| (One molecule of phosphoric acid) |
(Three molecules of sodium hydroxide) |
In this case, it takes 3 molecules of sodium hydroxide to neutralise every molecule of phosphoric acid. This means that you would need 3 moles of hydroxide to neutralise 1 mole of the acid, or 0.9 moles to neutralise 0.3 moles of acid etc. In this case, the chemicals react in a 3 : 1 ratio.
For each reaction, you should write a balanced equation. Then decide on the ratio of the reactants that neutralises them exactly.
