Stoichiometry – Chemical Quantities Notes Stoichiometry Stoichiometry – Study of quantitative relationships that can be derived from chemical formulas.

Stoichiometry – Chemical Quantities Notes

StoichiometryStoichiometry – Study of quantitative relationships that can be derived

from chemical formulas and chemical equationsMole ─ Mole Relationship

need a balanced equationMole Ratio – the ratio of moles of one substance to moles of another

substance in a balanced chemical equationThe coefficients in a balanced equation give the relative numbers of

molecules, as well as, the relative number of moles.

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

Ex: How many moles of O2 are required to produce 10. moles of CO2?2 CO + O2 2 CO2

1 mol CO = 2 mol H2 = 1 mol CH3OH

10. mol CO2 x __________ mol CO2

mol O2

2

1 = 5.0 mol O2

What other relationships do we have for the mole?

• 1 mol = 6.02 x 1023 atoms / molecules / particles

• 1 mol = [molar mass] g

• 1 amu = 1.66x10-24 g

We can add these mole relationships on either end of the mole ratio:

# unit A x 1 mol A x mol B x __ unit B = # unit B

_ unit A _ mol A 1 mol B

mole relationship mole ratio mole relationship

(switch units) (switch substances) (switch units)

*A is the GIVEN substance & B is the WANTED substance

Mass A – Mole B

Ex: Calculate moles of O2 produced if 2.50 g KClO3 decomposes completely:

2 KClO3 2 KCl + 3 O2

x ______________mol O2

mol KClO3

3

2

x ________________mol KClO3

122.55

1

g KClO3

2.50 g KClO3 =

0.0306 mol O2

K 1 x 39.10 = 39.10

Cl 1 x 35.45 = 35.45

O 3 x 16.00 = 48.00

122.55 g/mol

Mass A – Mass B

Ex: Determine the mass of NaCl that decomposes to yield 355 g Cl2

2NaCl 2 Na + Cl2355 g Cl2

mol Cl2

g Cl270.90

1 x _____________ x ______________ mol Cl2

mol NaCl

1

2 x ______________ mol NaCl

g NaCl58.44

1=

585 g NaCl

Cl 2 x 35.45 = 70.90 g/mol

Na 1 x 22.99 = 22.99

Cl 1 x 35.45 = 35.45

58.44 g/mol

Mole A – Mass BEx: Calculate the number of grams of oxygen

required to react exactly with 4.30 mol of propane, C3H8, in the reaction by the following balanced equation:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

4.30 mol C3H8 mol O2

mol C3H81

5 x _____________ x __________ mol O2

g O2

1

32.00=

688 g O2O 2 x 16.00 = 32.00 g/mol

Learning Check

How many grams of water are needed to produce 9.23 moles of oxygen?____ Na2O2 + ____ H2O ____ NaOH + ____ O2

2. Limiting Reactant & Percent Yield

Background Knowledge Check

Label the reactant(s) and product(s) in the following reaction:

2 Mg + O2 2MgO

Reactant(s):

Product(s):

Mg and O2

MgO

Limiting ReactantManufacturers of cars and bicycles order parts in the same

proportion as they are used in their product. Car manufacturers order four times as many wheels as engines and bicycle manufacturers order twice as many pedals as seats. In the same manner, when chemicals are mixed together so they can undergo a reaction, they are often mixed in stoichiometric quantities – exactly the correct amounts so that all the reactants “run out” at the same time.

If the chemicals aren’t mixed to run out at the same time, one of the chemicals will limit or halt the reaction from taking place any further. The reactant that “runs out” or limits the reaction is called the limiting reactant. The reactant that still remains or is extra is called the excess reactant.

In any stoichiometric problem, where reactants are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting in order to calculate correctly the amounts of products that will be formed.

Analogy: Baking CookiesA recipe calls for 2 cups of flour for every egg. You have 5

cups of flour and 3 eggs.What is your limiting ingredient? What is your excess ingredient?

Steps for Solving Stoichiometry Problems Involving Limiting Reactants1. Write and balance the equation for the reaction, if necessary.2. For each reactant, convert grams reactant to grams product.3. Compare grams of product:

• The smaller grams of product is the theoretical yield and is the amount of product made

• The smaller grams of product came from the limiting reactant• The larger grams of product came from the excess reactant

flour

eggs

x ______________ mol Mg

mol MgO2

2

x _____________mol O2

mol MgO 2

1

7.24 mol Mg=

=3.86 mol O2

292 g MgO

Ex: 7.24 moles of Mg and 3.86 moles of O2 react to form MgO. 2 Mg + O2 2MgO How many grams of MgO are formed ?

What is the limiting reactant ? What is the excess reactant ? Mg O2

Mg 1 x 24.31 = 24.31

O 1 x 16.00 = 16.00

40.31 g/mol

x ____________ mol MgO

g MgO40.31

1

x ____________ mol MgO

g MgO40.31

1311 g MgO

x ___________ g N2

mol N21

28.02 x ___________

mol N2

mol NH3 2

1

x ___________ g H2

mol H21

2.02

x ______________ mol NH3

g NH317.04

1

x ___________ mol H2

mol NH32

3

2.50 x 104 g N2

3.04 x 104 g NH3=

=

5.00 x 103 g H2

2.81 x 104 g NH3

Ex: Suppose 2.50 x 104 g of N2 and 5.00 x 103 g of H2 are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when the reaction is run to completion. N2 + 3 H2 2 NH3

What is the limiting reactant? What is the excess reactant?

N2H2

N 2 x 14.01 = 28.02 g/mol

H 2 x 1.01 = 2.02 g/mol

N 1 x 14.01 = 14.01

H 3 x 1.01 = 3.03

17.04 g/mol

x ______________ mol NH3

g NH317.04

1

Percent Yield Theoretical yield – amount of product predicted

from the amounts of reactants used, calculated from the limiting reactant

Actual yield – amount of product actually obtained through experiment

Percent yield – comparison of actual and theoretical yield

Percent Yield = Actual yield X 100%

Theoretical yield

Example: Methanol, CH3OH, can be produced by the reaction between carbon monoxide and hydrogen. Suppose 6.85 x 104 g of CO is reacted with 8.60 x 103 g of hydrogen. CO + 2H2 CH3OH1. Calculate the theoretical yield of methanol.

2. If 3.57 x 104 g of CH3OH is actually produced, what is the percent yield of methanol?

x ___________ g CO

mol CO1

28.01 x ___________

mol CO

mol CH3OH 1

1

x ___________ g H2

mol H21

2.02 x ___________

mol H2

mol CH3OH1

2

6.85 x 104 g CO

7.83 x 104 g CH3OH=

=

8.60 x 103 g H2

6.82 x 104 g CH3OH

C 1 x 12.01 = 12.01

O 1 x 16.00 = 16.00

28.01 g/mol

H 2 x 1.01 = 2.02 g/mol C 1 x 12.01 = 12.01

H 4 x 1.01 = 4.04

O 1 x 16.00 = 16.00

32.05 g/mol

% Yield =3.57 x 104 g

6.82 x 104 gX 100% = 52.3 %

6.82 x 104 g CH3OH

x ______________ mol CH3OH

g CH3OH32.05

1

x ______________ mol CH3OH

g CH3OH32.05

1

Learning CheckLearning Check - If 1.30 grams of oxygen and 3.10 grams of iron are reacted, then what is the theoretical yield of iron (III) oxide. What is the limiting reactant? If 3.00 grams of iron (III) oxide is produced, then what is the percent yield?

Balanced equation: 4 Fe + 3 O2 2 Fe2O3

Theoretical Yield?

Limiting Reactant? Excess Reactant?

Percent Yield?