20130804 1 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 n d : during process Amount : pieces 3 PB + 2 PM → 1 BM n d The « Equation » of a Big Mac Amount : pieces 3 PB + 2 PM → 1 BM n d 45 30 15 PM 30 PB 3 PM 2 PB 45 PB 3 PM 2 PB of Number PM of Number = × = × = n d : during process BM 15 PB 3 BM 1 PB 5 4 PB 3 BM 1 PB of Number BM of Number = × = × = "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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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
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd ? ? ? ?
na nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd 150 300 750 75
na nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
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
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd 400 ? ? ?
na nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd 400 800 2000 200
na nb: before process; nd: during process; na: after process
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The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd 400 800 2000 200
na impossible impossible
nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd ? ? 400 ?
na nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd 80 160 400 40
na nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
cL 2 GA + 4 VO + 10 OJ → 1 HWB
nb 400 300 400 0
nd 80 160 400 40
na 320 140 0 40
nb: before process; nd: during process; na: after process
The « Equation » of a DRINK Amount:
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
Example 1: Given the mass of a reactant in a reaction, find the mass of products formed.
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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?