GCSE OCR Gateway Chemistry C5.1g-h Percentage and theoretical yield

GCSE OCR Gateway Chemistry C5.1g-h

Percentage and theoretical yield

C5.1g  

To be able to calculate the theoretical amount of a product from a given amount of reactant

C5.1h  

To be able to calculate the percentage yield of a reaction product from the actual yield of a reaction

Percentage and theoretical yield

So we talk a lot these days about atom economy trying not to waste any reactant molecules and convert all of them in a reaction into useful product molecules.

But what if we do not convert all the reactants into useful molecules.  Then we have waste molecules.  We also have a measure of just how efficient the reaction is in converting reactants into useful products. 

That's what we have always called percentage yield. 

Lets suppose our reaction between sodium carbonate and hydrochloric acid involves 25g of sodium carbonate reacting with 50 cm³ of 2M acid. 

What is the percentage yield of sodium chloride?

First, we have to see which of the two reactants limits the yield of sodium chloride.  In other words, is the sodium carbonate in excess or is the acid in excess (by which we mean there is more of one than the other reactant.)

Finding the limiting reactant.

To find out which reactant is in excess we have to work out how many moles there are of each:

Here is the equation with the molar masses of the reactants and products.

Na₂CO₃   +   2HCl      ⟶    2NaCl     +   H₂O     +   CO₂

106                 73                   117                 18            44

So

amount sodium carbonate n (mol)  =   mass (g) /molar mass (g.mol^—1)

Therefore     n = 25g/106g.mol^—1   =  0.236 mol sodium carbonate

Next

amount hydrochloric acid  n (mol)  =  c (mol.dm^—3)   ⨉   V (dm³)

Therefore     n =  2 mol.dm^—3   ⨉   50 ⨉ 10^—3 dm³  =  0.100mol acid

You can see from the equation:

Na₂CO₃   +   2HCl      ⟶    2NaCl     +   H₂O     +   CO₂

that  2moles hydrochloric acid reacts with 1 mole of sodium carbonate so we would need 0.236  ⨉  2 =  0.472mol of acid to completely react with the sodium carbonate.

That means that the acid is the limiting reactant.  Its amount determines by the equation how many moles of salt will be produced. 

If we start with 0.1 mole  hydrochloric acid then we end up with 0.1 mole of salt.

0.1 mole of salt or 5.85g of salt is the theoretical yield. 

But suppose we carry out this salt preparation and find we only produce 4.67g of salt. 

4.67g of salt is called the actual yield. 

How efficient have we been in producing salt?

We measure the efficiency of salt production if we calculate the percentage yield

Comparing percentage yield with atom economy:

At this point, we could also calculate the atom economy of the sodium chloride:

So in our equation above, 65% of the reactant atoms and molecules are converted into useful salt but in the experiment the practical efficiency of changing one into the other comes in at about 4 in every 5 particles or 80% in our case. 

Here is a table summarising the different properties of the reaction which we have just calculated:

Reaction	Atom economy of NaCl(s)	Theoretical yield of NaCl(s)	Actual yield of NaCl(s)	Percentage yield of NaCl(s)
0.1 mol HCl + excess Na2CO3	65%	0.1mole or 5.85g	4.67g	79.8%