Percent Yield and Percent Purity

So today we learned the awesome wonders of the "percent," i had no idea that we could relate percentage to chemistry! Just keep making our lives hard O CHEMISTRY GODS! Anywho I must get this done! So here we go.......



Percent Yield!
-the percent yield is calculated because sometimes not all of the reactants are used up, nor is it possible to recover all of the product.

%yield : (grams of product recovered /grams of product expected from stoichiometry) X 100

Heres an example video on how to calculate percent yield
http://www.youtube.com/watch?v=TKNxdL7DN1I

Example:

Given that the chemical formula for salicylic acid is C₇H₆O₃ and the chemical formula for aspirin is C₉H₈O₄.
In an experiment, 100.0 grams of salicylic acid gave 121.2 grams of aspirin. What was the percent yield?

Solution:

Step 1: Calculate the M_r (relative molecular mass) of the substances.

A_r : C = 12, H = 1, O = 16
So, M_r : salicylic acid = 138, aspirin = 180.

Step 2: Change the grams to moles for salicylic acid

138 g of salicylic acid = 1 mole
So, 100 g = 100 ÷ 138 mole = 0.725 moles

Step 3: Work out the calculated mass of the aspirin.

1 mole of salicylic acid gives 1 mole of aspirin
So, 0.725 moles gives 0.725 moles of aspirin
0.725 moles of aspirin = 0.725 × 180 g = 130.5 g
So, the calculated mass of the reaction is 130.5 g

Step 4: Calculate the percent yield.

The actual mass obtained is 121.2 g
So, the percent yield = 121.2 ÷ 130.5 × 100% = 92.9%

Percent Purity!
-reactants that are used in the equations and or experiments arent always pure, so thats why you must calcute the amount of pure substance

%purity:  (mass of pure substance/mass of impure substance) X 100

Example:

Chalk is almost pure calcium carbonate. We can work out its purity by measuring how much carbon dioxide is given off. 10 g of chalk was reacted with an excess of dilute hydrochloric acid. 2.128 liters of carbon dioxide gas was collected at standard temperature and pressure (STP).
The equation for the reaction is
CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Solution:

Step 1: Calculate the M_r of calcium carbonate

A_r: Ca = 40, C = 12, O = 16)
M_r of CaCO₃ = 100

Step 2: Calculate the grams from the volume

1 mole of CaCO₃ gives 1 mole of CO₂
1 mole of gas has a volume of 22.4 liters at STP.
22.4 liters of gas of gas is produced by 100 g of calcium carbonate
and 2.128 liters is produced by 2.128 ÷ 22.4 × 100 = 9.5 g

Step 3: Calculate the percent purity

There is 9.5 g of calcium carbonate in the 10 g of chalk.
Percent purity = 9.5 ÷ 10 × 100% = 95%

 Heres a link, it contains numerous practise sheets! enjoy
http://misterguch.brinkster.net/pra_equationworksheets.html