STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE.

STOICHIOMETRY

USING THE REACTION EQUATION LIKE A

RECIPE

Cool word, but it’s all Greek to me!

• Stoicheion = element• Metron = measure

• STOICHIOMETRY…measuring and calculating the chemical elements

• YOU WILL NEED YOUR CALCULATOR EVERY DAY!

• Nearly everything we use is manufactured from chemicals.– Soaps, shampoos, conditioners,

cd’s, cosmetics, medications, and clothes.

• For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them.

• Chemical processes carried out in industry must be economical, this is where balanced equations help.

USING EQUATIONS

• Equations are a chemist’s recipe.– Eqs tell chemists what amounts of

reactants to mix and what amounts of products to expect.

• When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn.– Quantity meaning the amount of a

substance in grams, liters, molecules, formula units, or moles.

USING EQUATIONS

• The calculation of quantities in chem-ical reactions is called stoichiometry.

• Imagine you are in charge of manu-facturing for Rugged Rider Bicycle Company.

• The business plan for Rugged Rider requires the production of 128 custom-made bikes each day.

• You are responsible for insuring that there are enough parts at the start of each day.

USING EQUATIONS

• Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P).

• The finished bike has a “formula” of FSW2HP2.

• The balanced equation for the production of 1 bike is.

USING EQUATIONS

F +S+2W+H+2P FSW2HP2

• Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes?

• What do we know?– Number of bikes = 640 bikes– 1 FSW2HP2=2W (balanced eqn)

• What is unknown?–# of wheels = ? wheels

USING EQUATIONS

• The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation.

640 FSW2HP2

1 FSW2HP2

2 W= 1280

wheels

.

• We can make the same kinds of connections from a chemical rxn eqn.

• N2(g) + 3H2(g) 2NH3(g)

• The key is the “coefficient ratio”.

COEFFICIENT RATIOS or MOLE RATIOS

• N2(g) + 3H2(g) 2NH3(g)

• 1 mol N2 or 3 mol H2 3 mol H2 1 mol N2• 1 mol N2 or 2 mol NH3 2 mol NH3 1 mol N2• 3 mol H2 or 2 mol NH3 2 mol NH3 3 mol H2

– The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical rxn.

• 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.– N2 and H2 will always react to form

ammonia in this 1:3:2 ratio of moles.• So if you started with 10 moles of

N2 it would take 30 moles of H2 and would produce 20 moles of NH3

• Using the coefficients, from the balan-ced rxn equation to make connections between reactants and products, is the most important information that a rxn equation provides.– Using this information, you can

calculate the amounts of the reactants involved and the amount of product you might expect.

– Any calculation done with the next process is a theoretical number, the real world isn’t always perfect.

• Using the coefficients of balanced rxn equations and our knowledge of mole conversions we can perform powerful calculations. A.K.A. stoichiometry.

• A balanced rxn equation is essential for all calculations involving amounts of reactants and products.– If you know the number of moles of

1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn equation.

MOLE – MOLE EXAMPLE• The following rxn shows the

synthesis of aluminum oxide.

3O2(g) + 4Al(s) 2Al2O3(s)

• If you only had 1.8 mols of Al how much product could you make?

Given: 1.8 moles of Al

Uknown: ____ moles of Al2O3

3O2(g) + 4Al(s) 2Al2O3(s)

• Solve for the unknown:

1.8 mol Al4 mol Al

2 mol Al2O3 = 0.90mol Al2O3

MOLE – MOLE EXAMPLE

3O2(g) + 4Al(s) 2Al2O3(s)

Mole Ratio

MOLE – MOLE EXAMPLE 2• The following rxn shows the

synthesis of aluminum oxide.

3O2(g) + 4Al(s) 2Al2O3(s)

• If you wanted to produce 24 mols of product how many mols of each reactant would you need?Given: 24 moles of Al2O3

Uknown: ____ moles of Al ____ moles of O2

• Solve for the unknowns:

24 mol Al2O32 mol Al2O3

4 mol Al= 48 mol Al

MOLE – MOLE EXAMPLE 2

3O2(g) + 4Al(s) 2Al2O3(s)

24 mol Al2O32 mol Al2O3

3 mol O2 = 36 mol O2

Cross the Mole-Mole Bridge

With thanks to Rossini, Willam Tell, and the Lone RangerWilliam Tell Overture GuitarHiho, Stoichiometry...Away!

Backwards

MASS – MASS CALCULAT’NS• No lab balance measures moles

directly, generally mass is the unit of choice.

• From the mass of 1 reactant or prod-uct, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced rxn equation.

• As in mole-mole calcs, the unknown can be either a reactant or a product.

Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2).

How many grams of C2H2 are produced by adding water to 5.00

g CaC2?

CaC2 + 2H2O C2H2 + Ca(OH)2

MASS – MASS CALCULAT’NS 1

• What do we know?– Given mass = 5.0 g CaC2

– Mole ratio: 1 mol CaC2 = 1 mol C2H2

– MM of CaC2 = 64 g CaC2

– MM of C2H2 = 26 g C2H2

• What are we asked for?– grams of C2H2 produced


mass A moles A moles B mass B

5.0 g CaC2 64 g CaC2

1 mol CaC2


1mol CaC2

1 mol C2H2

1mol C2H2

26 g C2H2

= 2.0 g C2H2

You’ve recently learned that Copper will replace silver ions out of solution.

You’re eyes light up with this money making opportunity. However, you

decide it might be best if you did some preliminary calculations to determine to the feasibility of this get rich scheme.

Copper is not very hard to find, however the largest size of Silver

nitrate found in the Flinn Catalog is the 500 g size and it costs $305.91.

Currently Silver sells for $9.00/ounce on the stock market. How much money

could you sell your manufactured Silver for?


• What do we know?– Given mass = 500. g of AgNO3

– Mole ratio: 2 mol AgNO3 = 2 mol Ag

– MM of AgNO3: 169.84g = 1mol– MM of Ag: 107.87 g = 1mol– Price of Silver: $9.00 = 1 ounce–

Conversion g to oz: 28.23g = 1 oz


Cu + 2AgNO3 2Ag + Cu(NO3)2Cu + 2AgNO3 2Ag + Cu(NO3)2

500. g AgNO3 169.84gAgNO3

1mol AgNO3


2 mol AgNO3

2 mol Ag

1mol Ag

107.87g Ag

= $101.26

=$101

28.23 g

1 oz

1 oz

$9.00

• A balanced reaction equation indicates the relative numbers of moles of reactants and products.

• We can expand our stoichiometric calculations to include any unit of measure that is related to the mole.

• The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP.

• The problems can include mass-volume, volume-volume, and particle-mass calculations.

• In any of these problems, the given quantity is first converted to moles.

• Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown

• Then the moles of the unknown are converted to the units that the problem requests.

• The next slide summarizes these steps for all typical stoichiometric problems

MORE MOLE EXAMPLESHow many molecules of O2

are produced when a sample of 29.2 g of H2O is

decomposed by electrolysis according to this balanced

equation:2H2O 2H2 + O2

• What do we know?– Mass of H2O = 29.2 g H2O – 2 mol H2O = 1 mol O2 (from

balanced equation)– MM of H2O = 18.0 g H2O – 1 mol O2 = 6.02x1023 molecules of

O2

•What are we asked for?– molecules of O2

MORE MOLE EXAMPLES

mass A mols A mols B molecules B

29.2 g H2O 18.0 g H2O

1 mol H2O

2 mol H2O

1 mol O2

1 mol O2

6.02x1023 molecules O2

= 4.88 x 1023 molecules O2

The last step in the production of nitric acid is the reaction of

NO2 with H2O.3NO2+H2O2HNO3+NO

How many liters of NO2 must react with water to produce 5.0x1022 molecules of NO?

MORE MOLE EXAMPLES

• What do we know?– Molecules NO = 5.0x1022 molecules

NO – 1 mol NO = 3 mol NO2 (from

balanced equation)– 1 mol NO = 6.02x1023 molecules NO – 1 mol NO2 = 22.4 L NO2

• What are we asked for?– Liters of NO2

MORE MOLE EXAMPLES

molecules A mols mols B volume B

5.0x1022 mol-ecules NO

6.02x1023 mol-ecules NO

1 mol NO

1 mol NO

3 mol NO2

= 5.6 L NO2

1 mol NO2

22.4 L NO2

Aspirin can be made from a chemical rxn between the reactants salicylic

acid and acetic anhydride. The products of the rxn are acetyl-salicylic

acid (aspirin) and acetic acid (vinegar). Our factory makes 125,000

100-count bottles of Bayer Aspirin/day. Each bottle contains 100

tablets, and each tablet contains 325mg of aspirin. How much in kgs + 10% for production problems, of each

reactant must we have in order to meet production?

C7H6O3 + C4H6O3 C9H8O4 + HC2H3O2

Salicylic acid

Acetic anhydride aspirin vinegar

• What do we know?– Make 125,000 aspirin bottles/day– 100 aspirin/bottle– 325 mg aspirin/tablet– Mole ratio of aspirin to salicylic acid

(1:1) and acetic anhydride (1:1)– MM aspirin C9H8O4 = 180.11g– MM SA C7H6O3 = 138.10g– MM AA C4H6O3 = 102.06g

• What are we asked for?– Mass of salicylic acid in kgs + 10%– Mass of acetic anhydride in kgs +

10%

125,000 bottles

1 bottle

100 tablets

1 tablet

325mg asp.

1000 mg asp

1 g asp

= 22,549.4 mols aspirin

180.16g asp

1mol asp.

22,549.4 mols

aspirin 1 mol asp

1 mol SA C7H6O3

1 mol SA C7H6O3

138.10g SA C7H6O3

1000 g SA C7H6O3

1 kg SA C7H6O3

= 3068.97 kg salicylic acid + (306.897 g)

Salicylic Acid:

= 3380 kg of salicylic acid

22,549.4 mols

aspirin 1 mol asp

1 mol AA C4H6O3

1 mol AA C4H6O3

102.06g AA C4H6O3

1000 g AA C4H6O3

1 kg AA C4H6O3 = 2301.39 kg Acetic anhydride + 230.139 kg

Acetic Anhydride:

= 2530 kg Acetic anhydride

STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE.

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