# Stoichiometry Stoichiometry

Moles, masses, reactions, oh my!

Stoikheion + metria

From the Greek for “element” and

“the process of measuring”

Quantitative analysis of reactants and products

Moles

Molar masses

Balanced equations

Grinch Sandwiches

Ingredients per sandwich

◦ Sauerkraut (10 g)

◦ Arsenic sauce (5 g)

Balanced Equations

2H

2

+ O

2

 2H

2

0

◦ Coefficients = moles

◦ 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water

◦ What if you only had 1 mole of hydrogen?

◦ What if you had 10 moles of both hydrogen and oxygen?

◦ What if you needed to make 5 moles of water?

Law of Conservation of Mass

Convert each of the moles in the previous equation to grams

Compare to the product’s mass.

Be in awe!

Mole Ratios

Equation must be balanced

Relate reactants to each other, products to each other, or reactants to products

Molar Masses

If mass is mentioned, find that substance’s molar mass

Making Stoichiometry Work

Dimensional Analysis

◦ Number, unit, substance…always!

Problem #1

When 2.50 g of hydrogen gas are mixed with an excess of oxygen gas, how many grams of water should be produced?

Problem #1—step 1

Write a balanced equation.

◦ 2H

2

+ O

2

 2H

2

0

Problem #1—step 2

Begin with your given information and make a plan of conversions.

◦ Grams of hydrogen  moles of hydrogen  moles of water  grams of water

Write it as dimensional analysis.

g H

2 x mol H

2 g H

2 x mol H

2

O x g H

2

O = mol H

2 mol H

2

0

Problem #1—step 3

2.5 g H

2 x 1 mol H

2

18.02 g H

1 mol H

2

2

0

2.02 g H

2

O = x 2 mol H

2

O x

2 mol H

2

• Cross out what divides out.

Problem #1—step 4

2.5 g H

2 x 1 mol H

2

2.02 g H

2 x 2 mol H

2

O x

2 mol H

2

18.02g H

2

O = 22.3 g H

2

O

1 mol H

2

0

• How much oxygen gas would be required?

Problem #2

The thermite reaction: aluminum foil reacts with ferric oxide

Problem #2

Wow! Now let’s assume that 0.010 g of aluminum reacted with the excess ferric oxide on the ball bearing.

What mass of aluminum oxide should be produced?

Problem #2—Step 1

Write a balanced equation.

2Al + Fe

2

O

3

 Al

2

O

3

+ 2Fe

Problem #2—Step 2

g Al x mol Al x mol Al

2

O g Al

3 x g Al

2

O mol Al mol Al

3

2

O

3

=

Problem #2—Step 3

Place the proper numbers and calculate.

0.010 g Al x 1 mol Al x 1 mol Al

2

O

26.98 g Al 2 mol Al

3

101.96 g Al

1 mol Al

2

O

2

3

O

3

= x

0.0189 g of Al

2

O

3 should be formed

Problem #3

What if only 0.0150 g of Al

2

O

3 were actually produced in the previous problem? What is the percent yield?

So, you can calculate the % yield

◦ % yield = actual amt x 100 theoretical amt

Problem #3

Thus, the % yield calculation

0.0150 g x 100 = 79.4 % yield

0.0189 g

Limiting Reagent

Ideally, the perfect amount of each reactant

Often one reactant limits how much product can be formed—limiting reagent (LR)

Often the other reactant(s) is(are) in excess (ER)

You will be given at least 2 quantities of reactants

Grinch Sandwiches

Ingredients per sandwich

◦ Sauerkraut (10 g)

◦ Arsenic sauce (5 g)

Problem #4

0.500 g of silver nitrate in solution reacts with 0.750 g of tin (IV) chloride in solution. If 0.400 g of solid silver chloride are retrieved, what is the percent yield?

Problem #4—Step 1

Write a balanced equation.

4AgNO

3

A

+ SnCl

4

B

 4AgCl + Sn(NO

3

C D

)

4

Write shorthand, to save time.

You’ll write the real stuff in the answer.

Problem #4—Step 2

Ask yourself, when I have 0.500 g of

AgNO

3

, how many grams of SnCl

4

I really need? This will determine do the LR.

Set up the plan g A x mol A x mol B x g B = g A mol A mol B

Problem #4—Step 3

Plug in the numbers.

0.5 g A x 1mol A x 1 mol B

169.88g A 4 mol A x 260.50 g B = 0.192 g SnCl

4

1 mol A

Problem #4—Step 4

Analyze and determine the LR

◦ 0.750 g of SnCl

4 are available, and only

0.192 g of it need to be used to completely react with the AgNO

3

• Which substance will be in excess?

• Which substance will you run out of first?

Problem #4—Step 5

• Now, use the given amount of the

LR to find how much product you should be able to make.

g A x mol A x mol C x g C = g A mol A mol C

Problem #4—Step 6

• Plug in your numbers and calculate.

0.5 g A x 1 mol A x 4 mol C x

169.88 g A 4 mol A

143.32 g C = 0.422 g AgCl should be

1 mol C formed

Problem #4—Step 7

• Look at the actual amount of product that was formed, and determine percent yield using the theoretical you just calculated.

0.400 g AgCl x 100 = 94.8% yield

0.422 g AgCl

Practice

Sit down with some bread, sauerkraut, toadstools, and arsenic sauce, and work some stoichiometry problems just for fun.