Theoretical Yield

The amount of product(s) calculated using stoichiometry problems

Actual Yield

The amount of product actually recovered when the reaction is done

in the lab

Percent Yield

(Actual yield/Theoretical Yield) x 100

A reaction between solid sulfur and oxygen produces sulfur dioxide.

1S6 (s) + 6O2 (g) 6SO2 (g)

The reaction started with 384 grams of S6 (s). Assume an unlimited supply of oxygen. What is the theoretical (predicted) yield and the percent yield if only 680 grams of sulfur dioxide are produced?

Find moles of S6

384 g S6 ( 1 moles S6/ 192 g S6) = 2 moles S6

Find moles of SO2

2 moles S6 (6 moles SO2 /1 mole S6) = 12 moles SO2

Find theoretical grams SO2

12 moles SO2 (64 g SO2/1mole) = 768 g SO2

Find Percent Yield

(680 g / 768g) x100 = 89%

Zinc reacts with copper sulfate in a single replacement reaction as follows

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)

• 50.00 grams of zinc metal were added to excess copper sulfate dissolved in a water solution. 42.50 grams of copper were recovered. Calculate the theoretical yield of copper in this experiment

Step 1: Find the theoretical mass of Cu

• Mass of Cu = – 50.00g Zn(mol/65.38gZn) = 0.7648 mol Zn– 0.7648 mol Zn (1 molCu/1 mol Zn) = 0.7648 mol

Cu– 0.7648 mol Cu (63.55g Cu/mol) = 48.60 g Cu

Step 2: Find Percent Yield

• % Yield =

(42.50 g Cu/48.60 g Cu)x 100= 87.44%

Silver can produced by reacting silver nitrate with magnesium in the following

reactionMg(s) + 2 AgNO3 (aq) Mg(NO3)2

(aq) + 2 Ag (s)

• How much Silver can be recovered by reacting a silver nitrate solution with 50.00 grams of powdered magnesium.

• 50.00g Mg(mol/24.31 g Mg) = 2.057 mol Mg• 2.057 mol Mg ( 2mol Ag/ 1 mol Mg) = 4.114

mol Ag• 4.114 mol Ag (107.9 g/mol) =443.8 g Ag

Assume that 95% of the silver can be recovered.

• 443.8 g theoretically produced.• 443.8 g (0.95) = 421.58 g Ag actually produced

MOLARITY

Concentration

Solutions

• Solute: the substance being dissolved• Solvent: the substance doing the dissolving

Units of Concentration

• Molarity

– Ratio of – “Amount” of Solute/ “Amount” of Solution

Molarity

• Moles of solute/L of solution• M

Prepare 500mL of a 1M sucrose solution

Prepare 250 mL of a 6M solution of dye

• 4 drops = 1 mole