Percent Yield and Percent Purity

Today we learned the last things for the test.

1) Percent Yield: it is the amount of actual product obtained as a percentage of the theoretical amount of product.

It is found using the formula:

Actual mass produced(g) / theoretical mass produced(g)

Lets try some problems!!!!

Example :

The reaction SiO2(s) + 4 HF(g) -----> SiF4(g) + 2H2O(l) produces 3.0 grams of H2O when 13.30 g of SiO2 is treated with small excess of HF.

What mass of SiF4 is formed?

Looking back at previous lessons this is fairly straightforward.


Mass of SiF4 formed:

3.0 g H20 x 1mole/ 18.0 x 1SiF4/2H20 x 104.1/1 mole=8.7 g of SiF4


What mass of SiO2 is left unreacted?

Mass of SiO2 used - 3.0 g H2O x 1mole/18.0 x 1 SiO2/2H2O x 60.1/1mole=5.0 g SiO2

13.30-5.0 g = 8.3 g SiO2 in excess

What is percent yield of reaction?

amount of SiF4 expected:  13.30 g SiO2 x 1 mole/ 60.1 x 1 SiF4/1SiO2 x 104.1/1mole = 23.0 g SiF4

Now remember Percent Yield is amount obtained/amount expected x 100

8.7/23 x 100 = 37 .8% yield.


Percent Purity:

(Mass of pure substance/mass of impure substance) x100

If the mass of silver ore is 66.4 grams and contains 12.0 g of pure silver what is percent purity?

12.0/66.4

x 100 = 18.1 % purity





http://www.youtube.com/watch?v=LicEaaXhlEY