Percent Yield & Percent Purity

PERCENT YIELD

Sometimes in a chemical process, not all the products are used up. Percent yield is there to help us calculate the % of the product recovered.

% Yield =   .      grams of actual product recovered             X 100%

                    grams of product expected from stoichiometry

The percent yield should never be higher than 100%. Where did all that extra stuff come from anyway?

Let’s do an example:

Consider this equation: 1 CrCl2 + 2 Ca à 2 CaCl + 1 Cr

If 15.0g of CrCl2 is reacted with excess Ca and 4.10g of Cr was produced, what is the % yield?

STEP 1: Check to see if the equation is balanced. If not, balance it.

It is balanced in this case.

STEP 2: Find the amount of Cr produced when 15.0g of CrCl2 reacts

15.0g CrCl2 X  1mol CrCl2  = 0.12195mol CrCl2    ßconvert to moles

                          123g CrCl2

0.12195mol CrCl2 X  1mol Cr .  = 0.12195mol Cr   ßuse mole ratio

                                   1mol CrCl2

0.12195mol Cr  X 52.0g Cr  = 6.34g Cr   ßconvert to grams

                                1mol Cr

STEP 3: Find % yield

% yield =  4.10g Cr obtained .  X 100%  = 64.7% yield

                   6.34g Cr expected

For every 100.0g of product predicted, only 64.7g of product is actually formed.

 

PERCENT PURITY

Sometimes the reactants that were used are not pure, meaning there are bits of other elements in the main element. % purity calculates how much is actually there.

% Purity =  mass of pure substance  X 100%

                      mass of impure sample

Again, the % purity should never go above 100%.

Let’s start with an easy example:

If a 10.0g sample of zinc ore contains 7.3g of zinc metal, what is the percent purity?

Find the percent purity

% Purity =  7.3g Zn metal (pure substance)  X 100%  = 73.0% purity

                      10.0g Zn ore (impure sample)

Here’s a harder example:

Consider this equation: Ca + H2O à CaOH + H2

If 6.2g of impure calcium, that is 82% pure, is reacted, how many moles of hydrogen gas is produced?

STEP 1: balance the equation

2 Ca + 2 H2O à 2 CaOH + 1 H2

STEP 2: calculate the mass of Ca

% Purity X mass of impure Ca = mass of pure Ca

82.0%  X 6.2g impure Ca  = 5.084g pure Ca

­100%

STEP 3: calculate moles of H2 produced by 5.1g Ca

5.1g Ca X  1mol Ca  =  0.1268mol Ca   ßconvert to moles

                   40.1g Ca

0.1268mol Ca X 1mol H2  = 0.063mol H2   ßuse mole ratio

                            2mol Ca

6.3 x 10^-2 mol of H2 is produced.