1. Percent Yield How efficient is a chemical reaction?

2. Theoretical Yield- calculated maximum yield ( grams 1 → moles 1 → moles 2 → grams 2 ) Theoretical Yield – maximum amount of product that could be produced based on calculation and assuming perfect lab conditions. Actual Yield – amount actually obtained under laboratory conditions. Note: Actual yield is almost always less than theoretical yield and cannot exceed theoretical yield.

3. Why does total mass of reactant not always = to isolated product in the lab?

4. Why is perfection (Actual Yield = Theoretical Yield) rarely achieved? Lab technique – spills, incomplete transfers(some material left behind), etc. Reactions do not always run to completion i) Chemical equilibrium requires some reactant to remain unreacted ii) Side reactions (competing reactions may siphon) Rxn of interest: A + B → Product If another chemical reacts with A or B to form a differ product, less desired product is formed. Ex: B + X → Y, less B available to react with A.

5. K 2 CO 3 + 2 HCl → 2 KCl + H 2 O + CO 2 Calculate the theoretical yield of KCl solid if 45.8 g of K2CO3 are reacted. 45.8 g K 2 CO 3 ( 138.2 g K 2 CO 3 ) ( 2 mole KCl ) 1 mole KCl ( 74.6 g KCl ) = 49.4 g KCl 45.8 g? g 1 mole K 2 C O 3

6. Percent Yield Formula If actual yield of KCl recovered in lab is 46.3 g, calculate the % yield. % yield = actual yield % yield = 46.3 g/ 49.4 g = 93.7 % Theoretical yield x 100 %

7. A student in the lab calculated a yield over 100%. What is the most likely explanation for this apparent creation of matter? Most likely explanation: Product is not pure. For example, if a solid product is suspended in water and isolated by filtration, it is likely that the product is still wet and that water is adding extra mass.

8. Importance of Percent Yield Chemists generally are trying to synthesize a target amount of a product. Example: Goal to produce: 100 grams of product R → P ; 50 % yield Problem 50% yield means will only actually obtain 50 g of product. Chemists use % yield to adjust amounts of starting material to account for material “lost” during reaction. Solution: Double starting amount of R to amount needed to produce 200 g of product; 50% of 200 g = 100 g, target amount. B