Stoichiometry

Applying equations to calculating

quantities

Learning objectives

Use chemical equations to predict quantity

of one substance from given quantity of

another

Determine percentage yield

Identify limiting reactant

The chemical equationaA + bB = cC + dD

The Law of Conservation of Matter states that matter

is neither created nor destroyed

All atoms on left must be same as those on right

Reactant

side

Product

side

coefficient ELEMENT or

COMPOUND

Working with equations:

STOICHIOMETRY

Predict how much product is obtained from

given amount of reactant

Predict how much reactant is needed to give

required amount of product

Predict how much of one reactant is

required to give optimum result with given

amount of another reactant

Stoichiometry with equations:

The roadmap

Equations are in moles, but we measure in grams

Three conversions required:

A is given substance (reactant or product); B is target

substance (reactant or product)

1. Must convert grams A to moles A using molar mass

2. Use coefficients in equation to get moles B from moles A

3. Convert moles B to grams B using molar mass

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

Remember the mole triangle

Mole:mole ratio: the central step

Tells us molar ratio of substances in

balanced chemical equation

xX + yY = cC + dD

Mole:mole ratio (target:given) B:A = b/a

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

Mole:mole ratio examples

Types of problems:

Moles A → moles B

(A is given; B is target)xX + yY = cC + dD

Single step

Require mole:mole ratio:

Always target/given (B/A)

In balanced equation a mol A ≡ b mol B

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

moles B b mol B=

moles A a mol A

Mole:mole ratio problems

CH4 + 2O2 = CO2 + 2H2O

How many moles of H2O are produced from 5 moles of CH4?

Identify given substance: A = CH4

Identify target substance: B = H2O

Mole:mole ratio is

Identify coefficients: CH4 a = 1; H2O b = 2

moles B b mol B=

moles A a mol A

4

2

4

2

CH mol 1

OH mol 2

CH moles

OH moles

Moles A → mass BxX + yY = cC + dD

Two steps on road map

1. Convert moles A → moles B:

Mole:mole ratio (target/given):

2. Convert moles B → mass B using molar mass B

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

moles B b mol B=

moles A a mol A

mass B (g) = moles B (mol) x molar mass B (g/mol)

Mole:mass/mass:mole problems

CH4 + 2O2 = CO2 + 2H2O

What mass of H2O is produced from burning 5 moles CH4?

How many moles of CO2 is produced when 64 g O2 is consumed?

Mass A → mass BxX + yY = cC + dD

All three steps

1. Mass A → moles A using molar mass A

2. Moles A → moles B using mole:mole ratio

3. Moles B → mass B using molar mass B

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

moles B b mol B=

moles A a mol A

mass B (g) = moles B (mol) x molar mass B (g/mol)

mass A (g)moles A (mol) =

molar mass A (g/mol)

2 NaOH + Cl2 = NaOCl + NaCl + H2O What mass NaOH required to react with 142 g Cl2?

Molar mass Cl2 = 35.5 x 2 = 71.0 g/mol

Molar mass NaOH = 23.00 + 16.00 + 1.01 = 40.01 g/mol

Summary of stoichiometry problems

Maximum of three conversions required1. Must convert grams A to moles A using molar mass

2. Use coefficients in equation to get moles B from moles A

3. Convert moles B to grams B using molar mass

Maximum of three pieces of information required

1. Molar mass of given substance (maybe)

2. Molar mass of target substance (maybe)

3. Balanced chemical equation (always)

Work this example

CH4 + 2O2 = CO2 + 2H2O

What mass of CO2 is produced by the

complete combustion of 16 g of CH4

Atomic weight H = 1, C = 12, O = 16

44 g

Do stoichiometry exercises

Reaction Yield

The actual yield from a chemical reaction is

normally less than predicted from the

stoichiometry. Incomplete reaction

Product lost in recovery

Competing side reactions

Percent yield is:

actual yield% yield=

theoretical yield

Worked example

Actual yield of product is 32.8 g after reaction of 26.3 g of C4H8 with excess CH3OH to give C5H12O.

What is theoretical yield? Use stoichiometry to get mass of product:

convert mass moles moles mass

Theoretical yield = 41.4 g

Percent yield = 32.8/41.4 x 100 %

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

CH4 + 2O2 = CO2 + 2H2O What is percent yield if 40 g of CO2 is

produced by the complete combustion of 16

g of CH4?

First calculate theoretical yield as previously

Theoretical yield CO2 = 44 g

Actual yield CO2 = 40 g

Percent yield = 40 g/44 g x 100% = 91%

actual yield% yield=

theoretical yield

Limiting reactant

Balanced equations involve precisely the

right amounts of each reactant. None is left

over

Real-life situations usually involve an excess

of one reactant

Burning CH4 in excess O2

Reacting Zn with excess HCl

Identifying the limiting reactant

In balanced equations

each reactant will give

the same amount of

products

When one reactant is

limiting (deficient) the

amount of product will

be determined by its

amount

Identifying the limiting reactant

Which reactant gives the least amount of product

CO(g) + 2 H2(g) = CH3OH(g) Given 3 mol CO and 5 mol H2

3 mol CO → 3 mol CH3OH (mol:mol ratio CH3OH:CO = 1)

5 mol H2 → 2.5 mol CH3OH (mol:mol ratio CH3OH:H2 = 0.5)

H2 is limiting

Maximum product = 2.5 mol CH3OH