Chapter 9

Mass – Mass Calculations

2Al + 6HCl → 2AlCl3 + 3H2

• How many mol Al will react with 6.37 mol HCl?

2.12 mol Al

C6H12O6 + 6O2 → 6CO2 + 6H2O

• How many grams of O2 are needed to react with 12.5 g of C6H12O6?

• How is this problem different than the previous problem?

• This is an example of a mass – mass calculation.

Mass of substance

StoichiometricfactorMoles of

substanceMoles ofDifferentSubstance

Mass of DifferentSubstance

General Plan for Mass – Mass Calculations

C6H12O6 + 6O2 → 6CO2 + 6H2O

• How many grams of O2 are needed to react with 12.5 g of C6H12O6?

13.3g O2

2KClO3 → 2KCl + 3O2

• If 80.5 g O2 are produced how many grams of KCl will be produced?

125g KCl

2Al + 6HCl → 2AlCl3 + 3H2

• 9.23g of Al will produce how many grams of H2?

1.03g H2

Percent Yield

Theoretical Yield vs. Actual Yield• Theoretical yield is the amount of product we should produce from

the complete conversion of a given amount of reactant to product – must be calculated or given.

• Actual yield is the amount of product that is actually produced during a reaction. – it is always less than theoretical yield because of incomplete reactions,

impure reactants, side reactions, etc– must be given within the problem or measured during an experiment.

• Theoretical yield is calculated from the amount of limiting reactant.

Making Popcorn

• If we started with 80 popcorn kernels and found only 72 of them popped.

• What is the percent yield of popcorn?

Percent Yield

actual yield% yield = x 100%

theoretical yield

A reaction was supposed to make 2.50 grams of Zn. You actually made 2.32 grams. What is the percent yield?

92.8%

2Al + 3CuCl2 → 3Cu + 2AlCl3

• A chemistry student is trying to prepare copper metal by the reaction of aluminum with copper (II) chloride.

→

2Al + 3CuCl2 → 3Cu + 2AlCl3

• If the student is able to prepare 3.67g of copper metal by the reaction of 1.27g of aluminum with excess copper (II) chloride.

• How do I know that the aluminum gets used up and the copper (II) chloride does not?

This is the concept of the “limiting reactant”

2Al + 3CuCl2 → 3Cu + 2AlCl3

• If the student is able to prepare 3.67g of copper metal by the reaction of 1.27g of aluminum with excess copper (II) chloride. What is the percent yield?

Chromium (III) hydroxide will dissolve in concentrated sodium hydroxide according to the following equation:

NaOH + Cr(OH)3 NaCr(OH)4

If you begin with 66.0g of Cr(OH)3 and obtain 38.4g of NaCr(OH)4, what is the percentage yield?

41.9%

Homework

• “Mass-Mass Problems” Worksheet

Two other types of percent yield problems

• In both of these types of problems you will be given the percent yield.

1. Given the percent yield determine the amount of a reactant you need.

– You use more.

2. Given the percent yield determine the amount of a product you can produce.

– You make less.

You use more.• A bakery produces cookies at

a percent yield of 92%. A customer places an order for 2150 cookies. The bakers will need enough ingredients for how many cookies.

2150 0.92

= 2337 cookies

Divide by percent yield. “You use more”

You make less• A bakery produces

cookies at a percent yield of 92%. The bakers add enough ingredients to make 1525 cookies. How many cookies do they produce?

1525 (0.92) = 1403 cookies

Multiply by the percent yield. “You make Less”

A chemical engineer needs to produce 212 g of zinc oxide by the process below:

2ZnS + 3O2 2ZnO + 2SO2

By doing trials runs he expects the process to run at a 78% yield. How much zinc sulfide should he use to prepare the 212 g of zinc oxide?

326g ZnS

Ammonia is synthesized from hydrogen and nitrogen according to the following equation:

N2 + 3H2 2NH3

If an excess of nitrogen is reacted with 3.41g of hydrogen, how many grams of ammonia can be produced assuming the reaction has a percentage yield of 48.8%?

9.42g NH3

Homework

• Lab Summary: Percent Yield Lab

• Percentage Yield Worksheet (front and back).