Stoichiometry: Calculations with Chemical Equations

Stoichiometry:Stoichiometry:

Calculations with Chemical Calculations with Chemical Equations Equations

ObjectivesObjectives

Use chemical equations to predict amount of Use chemical equations to predict amount of product from given reactantsproduct from given reactants

Determine percentage yieldDetermine percentage yield Determine limiting reactantDetermine limiting reactant

Working with equations:Working with equations:STOICHIOMETRYSTOICHIOMETRY

Predict how much product is obtained from Predict how much product is obtained from given amount of reactantgiven amount of reactant

Predict how much reactant is needed to give Predict how much reactant is needed to give required amount of productrequired amount of product

Predict how much of one reactant is Predict how much of one reactant is required to give optimum result with given required to give optimum result with given amount of another reactantamount of another reactant

Relating moles, masses and Relating moles, masses and moleculesmolecules

Note: Note: Conservation of mass (4 g + 32 g = 36 g)Conservation of mass (4 g + 32 g = 36 g) But not necessarily conservation of moles (2 But not necessarily conservation of moles (2

moles + 1 mole = 2 moles)moles + 1 mole = 2 moles)

Stoichiometry with equations:Stoichiometry with equations:The roadmapThe roadmap

Equations are in moles, but we measure in Equations are in moles, but we measure in gramsgrams Three conversions required:Three conversions required: A is given substance; B is target substanceA is given substance; B is target substance

1.1. Must convert grams A to moles A using molar massMust convert grams A to moles A using molar mass

2.2. Use coefficients in equation to get moles B from moles AUse coefficients in equation to get moles B from moles A

3.3. Convert moles B to grams B using molar massConvert moles B to grams B using molar mass

Mass/molar mass Mole:mole ratioMoles x molar

massMass A Moles A Moles B Mass B

Types of problems:Types of problems:Moles A Moles A → moles B→ moles B

aaA + A + bbB = B = ccC + C + ddDD Single stepSingle step Mole:mole ratio:Mole:mole ratio: aa mol A ≡ mol A ≡ bb mol B mol B Target/givenTarget/given



/

moles B bmol B mol A

moles A a

Mole:mole ratio problemsMole:mole ratio problems

CHCH44 + 2O + 2O22 = CO = CO22 + 2H + 2H22OO How many moles of HHow many moles of H22O are produced from 5 moles of CHO are produced from 5 moles of CH44??

Moles A Moles A → mass B→ mass B

1.1. Convert moles A Convert moles A → moles B:→ moles B:Mole:mole ratio (target/given): Mole:mole ratio (target/given):

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



/

moles B bmol B mol A

moles A a

Mole:mass/mass:mole problemsMole:mass/mass:mole problems

CHCH44 + 2O + 2O22 = CO = CO22 + 2H + 2H22OO What mass of HWhat mass of H22O is produced from burning 5 moles CHO is produced from burning 5 moles CH44??

Mass A Mass A → mass B→ mass B

1.1. Mass A Mass A → moles A using molar mass A→ moles A using molar mass A

2.2. Moles A → moles B using mole:mole ratioMoles A → moles B using mole:mole ratio

3.3. Moles B → mass B using molar mass BMoles B → mass B using molar mass B



Summary of stoichiometry problemsSummary of stoichiometry problems

Maximum of three conversions requiredMaximum of three conversions required1.1. Must convert grams A to moles A using molar massMust convert grams A to moles A using molar mass2.2. Use coefficients in equation to get moles B from Use coefficients in equation to get moles B from

moles Amoles A3.3. Convert moles B to grams B using molar massConvert moles B to grams B using molar mass

Maximum of three pieces of information Maximum of three pieces of information requiredrequired

1.1. Molar mass of given substance (maybe)Molar mass of given substance (maybe)2.2. Molar mass of target substance (maybe)Molar mass of target substance (maybe)3.3. Balanced chemical equation (always)Balanced chemical equation (always)

Work this exampleWork this example

CHCH44 + 2O + 2O22 = CO = CO22 + 2H + 2H22OO

What mass of COWhat mass of CO22 is produced by the is produced by the

complete combustion of 16 g of CHcomplete combustion of 16 g of CH44

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

44 g44 g

Do stoichiometry exercisesDo stoichiometry exercises

Reaction YieldReaction Yield

Actual yield from chemical reaction is Actual yield from chemical reaction is normally less than predicted by normally less than predicted by stoichiometry.stoichiometry. Incomplete reactionIncomplete reaction Product lost in recoveryProduct lost in recovery Competing side reactionsCompeting side reactions

Percent yield is:Percent yield is:

Actual yield/Theoretical yield x 100 %Actual yield/Theoretical yield x 100 %



Worked exampleWorked example ActualActual yield of product is 32.8 g after reaction of 26.3 g of yield of product is 32.8 g after reaction of 26.3 g of

CC44HH88 with excess CH with excess CH33OH to give COH to give C55HH1212O.O.

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

convert mass (26.3 g) convert mass (26.3 g) moles moles moles moles mass mass TheoreticalTheoretical yield = 41.4 g yield = 41.4 g Percent yield = 32.8/41.4 x 100 %Percent yield = 32.8/41.4 x 100 %

Do percent yield exercisesDo percent yield exercises

Percent yield practicePercent yield practice

Limiting ReactantLimiting Reactant Exact quantities of reactants dictated by the Exact quantities of reactants dictated by the

reaction stoichiometry are not the normreaction stoichiometry are not the norm Usually one reactant is reacted with an Usually one reactant is reacted with an

excess of the other(s)excess of the other(s) Burning natural gas in furnaceBurning natural gas in furnace

This reactant is the This reactant is the limiting reactantlimiting reactant – – amount of products limited by this reactantamount of products limited by this reactant

Limiting reactant at molecular levelLimiting reactant at molecular level

In the reaction to produce ethylene glycol:In the reaction to produce ethylene glycol: 1 mole of ethylene oxide + 1 mole of water 1 mole of ethylene oxide + 1 mole of water → 1 → 1

mole glycolmole glycol Here, 2 moles of water remain after Here, 2 moles of water remain after

conversion of all ethylene oxide into glycolconversion of all ethylene oxide into glycol Ethylene oxide is limitingEthylene oxide is limiting

More product would be obtained by More product would be obtained by increasing the ethylene oxide until the water increasing the ethylene oxide until the water became limitingbecame limiting

Determination of limiting reactantDetermination of limiting reactant

Compare reaction stoichiometry with actual Compare reaction stoichiometry with actual reactant mole ratiosreactant mole ratios

Reaction aA + bB = productsReaction aA + bB = products Stoichiometric mole ratio reactants = a/bStoichiometric mole ratio reactants = a/b

Actual ratio of reactants = a’/b’Actual ratio of reactants = a’/b’ If a’/b’ > a/b, B is limitingIf a’/b’ > a/b, B is limiting If a’/b’ < a/b, A is limitingIf a’/b’ < a/b, A is limiting

Do limiting reagent exercisesDo limiting reagent exercises

Practice limiting reactant problemPractice limiting reactant problem 30 g CH30 g CH44 and 30 g O and 30 g O22 are reacted. Which is the limiting are reacted. Which is the limiting

reactant? How much COreactant? How much CO22 is produced? is produced?

Case of three (or more) reactantsCase of three (or more) reactants

Determine theoretical molar ratios for each Determine theoretical molar ratios for each pair based on reaction stoichiometrypair based on reaction stoichiometry

a/b a/c b/c…a/b a/c b/c…

Calculate actual molar ratios for all reactant Calculate actual molar ratios for all reactant pairs:pairs:

a’/b’ a’/c’ b’/c’…a’/b’ a’/c’ b’/c’…

The smallest ratio compared to the The smallest ratio compared to the theoretical will identify limiting reactanttheoretical will identify limiting reactant

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Stoichiometry: Calculations with Chemical Equations