Chocolate Chip Cookies• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 1 bag chocolate chips

Chocolate Chip Cookies• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 1 bag chocolate chips

How much?

What units?

Of what?

Chocolate Chip Cookies• 2.25 flour• 8 butter• 0.5 shortening• 0.75 sugar• 0.75 brown sugar• 1 salt• 1 baking soda• 1 vanilla• 0.5 Egg Beaters• 1 chocolate chips

How much?

Of what?

Chocolate Chip Cookies• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 1 bag chocolate chips

How much?

What units?

Of what?

Get on with it!

What does this have to do with CHEMISTRY?

2.25 cups flour + 8 Tbsp butter + 0.5 cups shortening +

0.75 cups sugar + 0.75 cups brown sugar + 1 tsp salt +

1 tsp baking soda + 1 tsp vanilla + 0.5 cups Egg Beaters + chips

(a synthesis reaction)

(177ºC)

1 batch of chocolate chip cookies!

coefficient

unitsubstance

Welcome to STOICHIOMETRY

What is stoichiometry?

• Composition stoichiometry deals with the mass relationships of elements in compounds.

• Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.

• All reaction stoichiometry calculations start with a balanced chemical equation.

Let’s Revisit the Cookies…

• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters

For 1 batch: The Egg Beaters I have are close to expiring! I’d like to use the rest of them in this recipe. I have 1.5 cups of Egg Beaters.

How many batches of cookies can I make with that many Egg Beaters?

Let’s Revisit the Cookies…

• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters

For 1 batch: I have 1.5 cups of Egg Beaters.

How many batches of cookies can I make with that many Egg Beaters?

1.5 cups E.B.x

1 batch cookies

0.5 cups E.B.=

3.0 batches of cookies

Let’s Revisit the Cookies…

• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters

For 1 batch: I have 1.5 cups of Egg Beaters.

How much butter do I need to deplete (use up) the Egg Beaters?

1.5 cups E.B.x

8 Tbsp butter

0.5 cups E.B.=

24 Tablespoons of butter

… Back to Chemistry• There are three types of stoichiometry

problems:– Mole-Mole problems (1 conversion)– Mass-Mole problems (2 conversions)– Mass-Mass problems (3 conversions)

given required

Stoichiometry Problems

• Stoichiometric problems are solved by using ratios from balanced chemical equations to convert the given quantity.

• A mole ratio is a conversion factor that relates the amount in moles of any two substances involved in a chemical reaction.

• This information is obtained from the balanced chemical equation.

Mole Ratios

• Example:

• The relationships between product and reactants or reactants can be expressed in the following mole ratios:

2 H2 + O22 H2O

Mole Ratios

2 H2 + O2 2 H2O

2

2

2

2

H mol 2

OH mol 2or

OH mol 2

H mol 2

2

2

2

2

O mol 1

OH mol 2or

OH mol 2

O mol 1

2

2

2

2

H mol 2

O mol 1or

O mol 1

H mol 2

Practice

• For each reaction, write all possible mole ratios.

)g(O3)s(Al4)l(OAl2 232

)g(O)l(Hg2)s(HgO2 2

32

32

OAl mol 2

Al mol 4or

Al mol 4

OAl mol 2

32

2

2

32

OAl mol 2

O mol 3or

O mol 3

OAl mol 2

Al mol 4

O mol 3or

O mol 3

Al mol 4 2

2

HgO mol 2

Hg mol 2or

Hg mol 2

HgO mol 2

HgO mol 2

O mol 1or

O mol 1

HgO mol 2 2

2 2

2

O mol 1

Hg mol 2or

Hg mol 2

O mol 1

Mole-Mole ProblemsExample:

2 H2 + O22 H2O

How many moles of water can be formed from 0.5 mol H2?

0.5 mol H2 x

2 mol H2

2 mol H2O=

0.5 mol H2O

Mole-Mole Practice

3CuSO4 + 2 Al Al2(SO4)3 + 3 Cu

1. Convert 0.5 mol CuSO4 to mol Cu

0.5 mol CuSO4 x 3 mol Cu

3 mol CuSO4

= 0.5 mol Cu

2. Convert 0.5 mol Al to mol CuSO4

0.5 mol Al x 3 mol CuSO4

2 mol Al= 0.8 mol CuSO4

Mass – Mole Problems

• Step 1: Write a BALANCED EQUATION.• Step 2: Calculate the molar mass of your

given substance and convert from mass to moles.

• Step 3: Determine the mole ratio from the coefficients in the balanced equation.

• Step 4: Set up the conversion and solve.

Mass-Mole ProblemsExample:

2 H2 + O22 H2O

How many moles of water can be formed from 48.0 g O2?48.0 g O2 x

1 mol O2

2 mol H2O=

3.00 mol H2O

32.00 g O2

1 mol O2x

Mass-Mole PracticeCuSO4 Al Al2(SO4)3 Cu3 2 3+ +

1. a. 13.5 g Al 1 mol Al

26.98 g Al=x 0.751 mol

CuSO4

b.

c.

13.5 g Al 1 mol Al2(SO4)3

2 mol Al=x 0.250 mol

Al2(SO4)3

13.5 g Al 3 mol Cu

2 mol Al=x 0.751 mol

Cu

Mole ratio

x3 mol CuSO4

2 mol Al

1 mol Al

26.98 g Al

x

1 mol Al

26.98 g Alx

Mole – Mass Practice

Ca + AlCl3 CaCl2 + Al3 2 3 2

0.095 mol AlCl3 x 3 mol Ca

2 mol AlCl3

x 40.08 g Ca

1 mol Ca= 5.7 g Ca

Mass-Mass ProblemsExample:

2 H2 + O22 H2O

How many grams of water can be formed from 48.0 g O2?48.0 g O2 x

1 mol O2

2 mol H2O=

32.00 g O2

1 mol O2x x

18.02 g H2O

1 mol H2O

54.1 g H2O

Mass-Mass PracticeCuSO4 Al Al2(SO4)3 Cu3 2 3+ +

1. a. 8.5 g Al 1 mol Al

26.98 g Al

=x

75 g CuSO4

b. 8.5 g Al 1 mol Al2(SO4)3

2 mol Al=x

54 g Al2(SO4)3

Mole ratio

x3 mol CuSO4

2 mol Al

1 mol Al

26.98 g Al

x

1 mol CuSO4

342.14 g Al2(SO4)3

1 mol Al2(SO4)3

x

x

159.61 g CuSO4

Mass-Mass Practicec. 8.5 g Al 3 mol Cu

2 mol Al=x

30. g Cu

1 mol Al

26.98 g Al

x x63.55 g Cu

1 mol Cu