D) Chemical Stoichiometry Chemicalreactions · D) Chemical Stoichiometry ... Excess Excess Lim. Item 40 Theoretical Yield n ... Recipes vs Chemical Reactions

2013-‐08-‐04

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D) Chemical Stoichiometry

The study of the amounts of substances that are consumed and produced during chemical reactions.* * A condensed definition from different sources

Recipes

• Quantities of ingredients (grams, millilitres, pinches, etc.)

• Proportions of ingredients

• Limiting ingredient • Recipe yields

Chemical reactions

• Amounts of matter (grams, millilitres, number of particles, moles)

• Molar ratios or reactants/

products • Limiting reactant • Reaction yields

(theoretical, experimental, and percent)

"One day in McDonald’s Kitchen!"

Big Mac = BM Pieces of Meat = PM Pieces of Bread = PB

The « Equation » of a Big Mac:

3 PB + 2 PM → 1 BM

The Stoichiometric Table (or: How to structure a proportion problem)

The « Equation » of a Big Mac

nd: during process

Amount:pieces 3 PB + 2 PM → 1 BM

nd

The « Equation » of a Big Mac

Amount:pieces 3 PB + 2 PM → 1 BM

nd 45 30 15

PM 30PB 3PM 2PB 45

PB 3PM 2 PB ofNumber PM ofNumber =⎟

⎠

⎞⎜⎝

⎛×=⎟⎠

⎞⎜⎝

⎛×=

nd: during process

BM 15PB 3BM 1PB 54

PB 3BM 1 PB ofNumber BM ofNumber =⎟

⎠

⎞⎜⎝

⎛×=⎟⎠

⎞⎜⎝

⎛×=

"One Evening at the Bar!" Preparing a Harvey Wallbanger

Galliano = GA (in cL) Vodka = VO (in cL) Orange Juice = OJ (in cL) Harvey Wallbanger = HWB (in "Drink")

The « Equation » of a Harvey Wallbanger:

2 GA + 4 VO + 10 OJ → 1 HWB

The Stoichiometric Table (or: How to structure a stoichiometric problem)

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The « Equation » of a DRINK Amount:

cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb

nd

na nb: before process; nd: during process; na: after process


cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd ? ? ? ?



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 150 300 750 75



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 150 300 750 75

na impossible

nb: before process; nd: during process; na: after process


cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 400 ? ? ?



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 400 800 2000 200


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cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 400 800 2000 200

na impossible impossible



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd ? ? 400 ?



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 80 160 400 40



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 80 160 400 40

na 320 140 0 40



cL 2 GA + 4 VO + 10 OJ → 1 HWB

nb 400 300 400 0

nd 80 160 400 40

na 320 140 0 Excess Excess Lim. Item

40 Theoretical

Yield nb: before process; nd: during process; na: after process

Recipes vs Chemical Reactions Cups of flour or sugar, Pieces of bread or meat, mL of milk, number of eggs, cL of liquids; these are “Amounts of Ingredients” in recipes.

In chemistry, how do we express “amounts” of elements or compounds?

in Moles (or “mol”)!

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Relationships between Amounts of Matter

Mass of reactants ⇔ mass of products Moles of reactants ⇔ mass of products Mass of reactants ⇔ moles of products Vol. of reactant ⇔ vol. of products (pure liquids) Vol. of reactant ⇔ vol. of products (solutions) Volume of reactant solution ⇔ Concentrations And all the other possible combinations… The ultimate “amount of matter” unit to use in any of the above cases should always be the “mole”.

What if amounts of matter are available or required in other units?

1.  We need to work with number of moles (n);

2.  We can get the number of moles from mass (n = m / M);

3.  For pure liquids, we can get masses from densities (m = d V) and then calculate the number of moles (n);

4.  For solutions, we can get the number of moles (n) from the molar concentration (C) and the volume (n = C V);

5.  For gases, we can get the number of moles (n) from the ideal gas law (n = PV/RT)

6.  How to integrate all these parameters in a single structure?

Example 1: Given the mass of a reactant in a reaction, find the mass of products formed.

Methyl alcohol, CH3OH, is commonly used as fondue fuel (back in the kitchen, we are!). Its combustion in air leads to carbon dioxide and water as shown in the following balanced equation:

2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)

What masses of CO2 and H2O are produced when 75.3 g of CH3OH are burned? Mass of CH3OH: m(CH3OH) = 75.3 g

moles 2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)

M (g/mol)

32.04 32.00 44.01 18.02

m (g) 75.3 g ? ? ?

nd



M (g/mol)

32.04 32.00 44.01 18.02

m (g) 75.3 g ? ? ?

nd 2.35 mol 3.53 mol 2.35 mol 4.70 mol



M (g/mol)

32.04 32.00 44.01 18.02

m (g) 75.3 g 113 g 103 g 84.7 g

nd 2.35 mol 3.53 mol 2.35 mol 4.70 mol


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Example 2: Given the volume of a liquid reactant in a reaction, find the volume of a liquid product formed.

Methyl alcohol, CH3OH, is a liquid and has a density of 0.7866 g/mL at 25°C and atmospheric pressure. The density of water under the same conditions is 0.997 g/mL.

2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)

What volume of water (at SATP) is produced when a 500. mL bottle of methyl alcohol is burned?


M (g/mol) [d (g/mL)]

32.04 32.00 44.01 18.02 [0.7866] --- --- [0.997]

V (mL) m (g)

500. mL ? ? ? ?

nd



32.04 32.00 44.01 18.02 [0.7866] --- --- [0.997]

V (mL) m (g)

500. mL 393 g ? ? ?

nd



32.04 32.00 44.01 18.02 [0.7866] --- --- [0.997]

V (mL) m (g)

500. mL 393 g ? ? ?

nd 12.3 mol 18.4 mol 12.3 mol 24.6 mol



32.04 32.00 44.01 18.02 [0.7866] --- --- [0.997]

V (mL) m (g)

500. mL 393 g 589 g 541 g 443 g

nd 12.3 mol 18.4 mol 12.3 mol 24.6 mol



32.04 32.00 44.01 18.02 [0.7866] --- --- [0.997]

V (mL) m (g)

500. mL 445 mL 393 g 589 g 541 g 443 g

nd 12.3 mol 18.4 mol 12.3 mol 24.6 mol

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Example 3: Given masses of reactants for a reaction, find the limiting reactant and the maximum possible mass of products that can be obtained.

N2 (g) + 3 H2 (g) → 2 NH3 (g)

What mass of NH3 can be produced by the reaction of 54.7 g N2 and 29.5 g H2?

nb or mb: before process; nd: during process; na or ma: after process

Amount: moles

N2 (g) + 3 H2 (g) → 2 NH3 (g)

M (g/mol)

mb (g)

nb

nd

na

ma (g)


Amount: moles

N2 (g) + 3 H2 (g) → 2 NH3 (g)

M (g/mol) 28.02 2.016 17.03

mb (g) 54.7 g 29.5 g 0

nb

nd ? ? ?

na

ma (g) nb or mb: before process; nd: during process; na or ma: after process

Amount: moles

N2 (g) + 3 H2 (g) → 2 NH3 (g)

M (g/mol) 28.02 2.016 17.03

mb (g) 54.7 g 29.5 g 0

nb 1.95 mol 14.6 mol 0

nd ? ? ?

na

ma (g)


Amount: moles

N2 (g) + 3 H2 (g) → 2 NH3 (g)

M (g/mol) 28.02 2.016 17.03

mb (g) 54.7 g 29.5 g 0

nb 1.95 mol 14.6 mol 0

nd 1.95 mol 5.86 mol 3.90 mol

na

ma (g) nb or mb: before process; nd: during process; na or ma: after process

Amount: moles

N2 (g) + 3 H2 (g) → 2 NH3 (g)

M (g/mol) 28.02 2.016 17.03

mb (g) 54.7 g 29.5 g 0

nb 1.95 mol 14.6 mol 0

nd 1.95 mol 5.86 mol 3.90 mol

na 0 mol 8.74 mol 3.90 mol

ma (g) Lim. Reactant 17.6 g 66.5 g

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Refer to the notes provided for a series of stoichiometric questions and applications we can do as a group.

STOICHIOMETRY EXERCICES CONCLUSION 1.  We went through several « nuts and bolts » chemistry tips

together:

ü Lewis structures drawing ü Oxidation state determination ü A nomenclature proposal ü Stoichiometric calculations.

2.  We all have our own ways to work on these topics in our classes.

3.  It was a pleasure to share mine with you and I hope you will find them useful.

4.  Thank you very much for your participation at this workshop.

5.  If you pass by Winnipeg, feel free to drop by at the Université de Saint-Boniface for a chat!

François Gauvin ([email protected])

Thank you! Questions/Comments

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D) Chemical Stoichiometry Chemicalreactions · D) Chemical Stoichiometry ... Excess Excess Lim. Item 40 Theoretical Yield n ... Recipes vs Chemical Reactions