April 7, 2014

• Today: Stoichiometry and % Yield

Percent Yield

• Remember, stoichiometry is used to tell you how much product you can form from X amount of product without doing the reaction

• Percent yield tells you how much product you actually got in the lab compared to how much you could have got

Theoretical yield

• The maximum amount of product that can be formed from a given amount of reactant.

– This is a value you calculate on paper

• In other words: (Write your own definition of Theoretical yield here)

Actual yield

• The measured amount of a product obtained from a reaction

– This is a measurement of a product formed in an actual chemical reaction

• In other words: (Write your own definition of Actual yield here)

Percent Yield Formula

Actual yield x 100 = Theoretical yield

Example 1Mg(s) + 2H2O(g) Mg(OH)2(s) + H2(g)

A. If 16.2 g Mg are heated with excess H2O how many grams of hydrogen gas could theoretically be formed?

Known: mass MgUnknown: mass H2

Plan: g Mg mol Mg mol H2 g H2

Relationships: molar mass Mg = 24.31 g/mol, molar mass H2 = 2.02 g/mol, mole ratio = 1 mol Mg : 1 mol H2

16.2 g Mg | 1 mol Mg | 1 mol H2 | 2.02 g H2 = 16.2 x 2.02 = 1.35 g

24.31 g 1 mol Mg 1 mol H2 24.31

Example 1 B. If only 0.905 g H2 are actually formed, what is the percent

yield of this reaction?

Actual yield x 100 = % yieldTheoretical yield

0.905 g H2 x 100 = 67.0%

1.35 g H2

(This means that you produced or collected only 67% of the product that it was possible to form with the amount of reactant you started with.)

Example 2CO(g) + 2H2(g) CH3OH(l)

If 11.0 g H2 reacts with CO to produce 68.4 g CH3OH, what is the percentage yield of CH3OH?

We need to determine the theoretical yield. Known: mass H2

Unknown: mass CH3OH

Plan: g H2 mol Mg mol H2 g H2

Relationships: molar mass H2 = 2.02 g/mol

molar mass CH3OH = 32.04 g/mol

mole ratio = 2 mol H2 : 1 mol CH3OH

Example 2CO(g) + 2H2(g) CH3OH(l)

If 11.0 g H2 reacts with CO to produce 68.4 g CH3OH, what is the percentage yield of CH3OH?

Known: mass H2

Unknown: mass CH3OH

Plan: g H2 mol Mg mol H2 g H2

Relationships: molar mass H2 = 2.02 g/mol, molar mass CH3OH = 32.04 g/mol, mole ratio = 2 mol H2 : 1 mol CH3OH

11.0 g H2 | 1 mol H2 | 1 mol CH3OH | 32.04 g CH3OH = 11.0 x 32.04 = 87.2

2.02 H2 2 mol H2 1 mol CH3OH 2.02 x 2 g

Example 2

CO(g) + 2H2(g) CH3OH(l)

If 11.0 g H2 reacts with CO to produce 68.4 g CH3OH, what is the percentage yield of CH3OH?

Actual yield x 100 = % yieldTheoretical yield

68.4 g CH3OH x 100 = 78.4%

87.2 g CH3OH