Empirical Formula
The empirical formula just shows the simplest ratio of the atoms present. For example:
3g of ethane contains 2.4g of carbon (Ar = 12.0) and 0.6g of hydrogen (Ar = 1.0).
What is the empirical formula?
\(Number\;of\;moles\;of\;carbon = \;\frac{{2.4}}{{12.0}} = 0.2\;mol\;of\;carbon\)
\(Number\;of\;moles\;of\;hydrogen = \;\frac{{0.6}}{{1.0}} = 0.6\;mol\;of\;hydrogen\)
Then you need to divide each number by the SMALLEST number (which in this case is 0.2):
\(\frac{{0.2}}{{0.2}} = 1\)
\(\frac{{0.6}}{{0.2}} = 3\)
The ratio is 1:3 and therefore the empirical formula for ethane is CH3.
Molecular Formula
The molecular formula shows the numbers of each atom in the molecule. It is found from:
- The empirical formula
- The relative molecular mass of the empirical formula
- The relative molecular mass of the molecule
For example: The empirical formula of ethane is CH3 and this group has a molecular mass of 15.0. The molecular mass of ethane is 30.0, which is
\(2 \times 15.0 = 30.0\)
Therefore there must be two units to the molecular formula in every molecule of ethane. The molecular formula is therefore C2H6
Percentage Yield
The yield of a chemical reaction is the amount of product formed. First you need to work out the theoretical yield. This is the maximum possible amount of product from the reactants used. In practice you are unlikely to get the theoretical yield. The actual yield is usually much less than the theoretical yield and there are several reasons for this, including:
- the reactants may be impure
- the reaction may not go to completion
- some of the product may be left in the container
- it may be difficult to purify the product.
In a laboratory situation this might just be annoying, as you will have to use more of each reactant than you calculated. But in an industrial situation more raw materials and energy will be used, more waste will be produced, and it could cost a lot of money.
Calculating percentage yield
The percentage yield tells you how close to the theoretical yield you have got. The higher the percentage yield, the closer you are to the theoretical yield. Here is the equation for calculating percentage yield:
\(Percentage\;Yield = \;\frac{{Actual\;Yield}}{{Theoretical\;Yield}}\; \times 100\)
The percentage yield would be 100% if all the reactants were converted into products, and there were no losses during processes such as pouring and filtering.
The theoretical yield for a certain method to obtain copper (II) sulphate crystals is 2.0g. The actual yield obtained was 1.8g. What was the percentage yield?
\(Percentage\;Yield = \;\frac{{1.8g}}{{2.0g}}\; \times 100 = 90\% \;\)
A common percentage yield question revolves around the Haber process.
Example: In the Haber process Ammonia is manufactured from nitrogen and hydrogen. Over one hundred million tonnes of ammonia is manufactured worldwide each year. About 85% of this is used to make fertilizers. Without the Haber process it would be very difficult to produce enough food for everyone. Yet under typical conditions the percentage yield of ammonia is only about 15%. The reaction is reversible and does not go to completion:
N2(g) + 3H2(g) --> 2NH3(g)
If the percentage yield of ammonia in the Haber process was 15%, what mass of ammonia is produced from six tonnes of hydrogen?
Mass of hydrogen = 6x106g
Ar of hydrogen = 2.0
\(\frac{{Mass}}{{Ar}} = Moles\)
\(\frac{{{\rm{6}} \times {{10}^6}}}{{2.0}} = 3 \times {10^3}moles\)
Look at the mole ratio (from the original equation) and you will see that for every 3 moles of hydrogen you would get 2 moles of ammonia, so:
\(\frac{{3 \times {{10}^3}}}{3}\; \times 2 = 2 \times {10^3}\;moles\;of\;ammonia\)
Now convert the moles back into grams:
\(Moles = Mass \times Mr\)
\(2 \times {10^3} \times 17.0 = 3.4 \times {10^7}g\)
Now you can work out the percentage yield:
\(Percentage\;Yield = \;\frac{{Actual\;Yield}}{{Theoretical\;Yield}}\; \times 100\)
\(\frac{{Percentage\;Yield \times Theoretical\;Yield}}{{100}} = Actual\;Yield\)
\(\frac{{15 \times 3.4 \times {{10}^7}}}{{100}} = 5.1 \times {10^6}g\;or\;5.1tonnes\)