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Chemical Reactions
Chapter 4 Part 1
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Reactants: Zn + I 2 Product: ZnI 2 Chemistry 221 Professor Michael Russell Last update: 8/9/
Chemistry as Cooking! - the Chemical Reaction "Recipe" and technique leads to successful creations Must know amounts to add, how much will be produced Haphazard additions can be disastrous!
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Chemical Equations
Depict the kind of reactants and products
and their relative amounts in a reaction.
4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s)
The numbers in the front are called
stoichiometric coefficients
The letters (s), (g), (aq) and (l) are the
physical states of compounds.
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Reaction of Phosphorus with Cl 2
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Notice the stoichiometric coefficients
and the physical states of the reactants
and products.
Reaction of Iron with Cl 2
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Evidence of a chemical reaction: heat change, precipitate formation, gas evolution, color change
Chemical Equations
4 Al(s) + 3 O 2 (g) → 2 Al 2 O 3 (s) This equation means:
4 Al atoms + 3 O 2 molecules ---give---> 2 molecules of Al 2 O 3 or 4 moles of Al + 3 moles of O 2 ---give--->
MAR^ 2 moles of Al^2 O^3
Because the same atoms
are present in a
reaction at the
beginning and at the
end, the amount of
matter in a system
does not change.
The Law of the
Conservation of Matter
Also known as the Law
of Mass Action
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Chemical Equations
Because of the principle of the conservation of matter,
an equation must be
balanced.
It must have the same number of atoms of the same kind on both sides.
Lavoisier, 1788
Chemical Equations / Lavoisier
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Balancing Equations
_ Al(s) + _ Br 2 (liq) ---> _ Al 2 Br 6 (s)
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2 Al(s) + 3 Br 2 (liq) ---> Al 2 Br 6 (s)
Balancing Equations
____C 3 H 8 (g) + _____ O 2 (g) ----> _____CO 2 (g) + _____ H 2 O(g)
____B 4 H 10 (g) + _____ O 2 (g) ----> ___ B 2 O 3 (g) + _____ H 2 O(g)
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C 3 H 8 (g) + 5 O 2 (g) ----> 3 CO 2 (g) + 4 H 2 O(g)
2 B 4 H 10 (g) + 11 O 2 (g) ----> 4 B 2 O 3 (g) + 10 H 2 O(g)
Balancing Equations - Hints
Balance those atoms which occur in only one
compound on each side last (i.e. O 2 in previous
examples)
Balance the remaining atoms first
Reduce coefficients to smallest whole integers
Check your answer if uncertain
Helpful but optional: Check that charges are
balanced
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STOICHIOMETRY
Stoichiometry is the study of the quantitative aspects of chemical reactions. Stoichiometry rests on the principle of the conservation of matter. MAR
STEP 5 How much N 2 O is formed?
Total mass of reactants = total mass of products 454 g NH 4 NO 3 = ___ g N 2 O + 204 g H 2 O mass of N 2 O = 250. g law of mass action!
could also turn mol NH 4 NO 3 into mol N 2 O, then grams of N 2 O:
MAR^ 5.68 mol N^2 O * 44.01 g/mol = 250. g
454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O
Mass is conserved!
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Compound NH 4 NO 3 N 2 O H 2 O
Initial (g) 454 g 0 0
Initial (mol) 5.68mol 0 0 Change (mol) -5.68 +5.68 +2(5.68)
Final (mol) 0 5.68 11. Final (g) 0 250. 204
454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O
STEP 6 Calculate the percent yield
We predicted a yield of 250. g of N 2 O. If you isolated only 131 g of N 2 O, what is the percent yield of N 2 O? This compares the theoretical yield (250. g) and actual yield (131 g) of N 2 O.
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454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O
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% yield =
actual yield theoretical yield
% yield =
131 g
- g
- 100% = 52.4%
STEP 6 Calculate the percent yield
454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O
GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
MAR Molarity in next chapter - See^ Stoichiometry Guide
We use the "grams - moles - moles - grams" conversion often in chemistry (mass is grams) Sometimes I call this the "grams - moles - moles - grams" dance lol
PROBLEM: Using 5.00 g of
H 2 O 2 , what mass of O 2 and of
H 2 O can be obtained?
2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2
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PROBLEM: Using 5.00 g of
H 2 O 2 , what mass of O 2 and of
H 2 O can be obtained?
2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2 (2.35 g) Step 4: mass of H 2 O (2.65 g) Try this problem yourself! MAR
Reactions Involving a LIMITING REACTANT
In a given reaction, there is not
enough of one reagent to use up the
other reagent completely.
The reagent in short supply LIMITS
the quantity of product that can be
formed.
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LIMITING REACTANTS
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LIMITING REACTANTS
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R eactants Products
2 NO(g) + O 2 (g) 2 NO 2 (g)
Limiting reactant = ___________
Excess reactant = ____________
LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq)
Zn(s) + 2 HCl(aq) ---> ZnCl (^) 2(aq) + H2(g)
Left: Balloon inflates fully, some Zn left
- More than enough Zn to use up the 0.100 mol HCl Center: Balloon inflates fully, no Zn left
- Right amount of each (HCl and Zn) Right: Balloon does not inflate fully, no Zn left. MAR^ * Not enough Zn to use up 0.100 mol HCl
Left Center Right mass Zn (g) 7.00 3.27 1. mol Zn 0.107 0.050 0. mol HCl 0.100 0.100 0. mol HCl/mol Zn 0.93 2.00 5. Lim Reactant LR = HCl no LR LR = Zn
LIMITING REACTANTS React solid Zn with 0. mol HCl (aq) Zn(s) + 2 HCl(aq) ---> ZnCl (^) 2(aq) + H2(g) 0.100 mol HCl [1 mol Zn/2 mol HCl] = 0.0500 mol Zn
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Limiting reactant = Cl 2
Base all calculations on Cl 2
moles Cl 2
moles Al 2 Cl 6
grams Cl 2
grams Al 2 Cl 6
1 mol Al 2 Cl 6 3 mol Cl 2
Mix 5.40 g of Al with 8.10 g of Cl 2.
What mass of Al 2 Cl 6 can form?
MAR 2 Al + 3 Cl 2 --->^ Al 2 Cl 6
Step 1: Calculate moles of Al 2 Cl 6 expected using chlorine:
CALCULATIONS: calculate mass of
Al 2 Cl 6 expected using limiting reactant.
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Step 2: Calculate mass of Al 2 Cl 6 expected based on chlorine:
2 Al + 3 Cl 2 ---> Al 2 Cl 6
0.114 mol Cl 2 • 1 mol Al^2 Cl^6
3 mol Cl 2
= 0.0380 mol Al 2 Cl 6
0.0380 mol Al 2 Cl 6 • 266.4 g Al^2 Cl^6 mol
= 10.1 g Al 2 Cl (^6)
Alternate Limiting Reactant Method
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Calculate theoretical yield of product based on both reactants. Smaller theoretical yield comes from limiting reactant, greater yield from excess reactant.
10.1 g < 26.6 g , so: limiting reactant = Cl 2 , theoretical yield = 10.1 g, excess reactant = Al
8.10 g Cl 2 • 1 mol 70.9 g
- 1 mol Al^2 Cl^6 3 mol Cl (^2) - 266.4 g 1 mol
= 10.1 g Al 2 Cl (^6)
5.40 g Al • 1 mol 27.0 g
- 1 mol Al^2 Cl^6 2 mol Al - 266.4 g 1 mol
= 26.6 g Al 2 Cl (^6)
2 Al + 3 Cl 2 ---> Al 2 Cl 6
Cl 2 was the limiting reactant.
Therefore, Al was present in
excess. But by how much?
First find how much Al was required
based on limiting reactant (Cl 2 ).
Then find how much Al is in excess.
How much of which reactant will
remain when reaction is complete?
MAR^ 2 Al + 3 Cl 2 --->^ Al 2 Cl 6
Excess Al = Al available - Al required = 5.40 g - 2.05 g = 3.35 g Al unused in reaction
2 Al + 3 Cl 2 products
0.200 mol 0.114 mol = LR
Calculating Excess Al
MAR 2 Al + 3 Cl 2 --->^ Al 2 Cl 6
8.10 g Cl 2 • 1 mol 70.9 g
- 2 mol Al 3 mol Cl (^2) - 26.98 g 1 mol
= 2.05 g Al
Using Stoichiometry to
Determine a Formula
Hydrocarbons, Cx H (^) y , can be burned in oxygen to give CO 2 and H 2 O (combustion reaction). The CO 2 and H 2 O can be collected to determine the empirical formula of the hydrocarbon. Cx H (^) y + O 2 ---> CO 2 + H 2 O
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Using Stoichiometry to
Determine a Formula
What is the empirical formula of a hydrocarbon, Cx H (^) y , if burning 0.115 g produces 0.379 g CO 2 and 0.1035 g H 2 O? Cx H (^) y + some O 2 ---> 0.379 g CO 2 + 0.1035 g H 2 O
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Puddle of Cx H (^) y 0.115 g
+O 2 0.379 g CO 2
+O (^2) 0.1035 g H 2 O 1 H 2 O molecule forms for each 2 H atoms in Cx H (^) y
1 CO 2 molecule forms for each C atom in C (^) x H (^) y
Using Stoichiometry to
Determine a Formula
First , recognize that all C in CO 2 and all H in H 2 O comes from C (^) x H (^) y.
- Calculate amount of C in CO 2 8.61 x 10-3^ mol CO 2 --> 8.61 x 10 -3^ mol C 1 mol C per 1 mol CO 2
- Calculate amount of H in H 2 O 5.744 x 10-3^ mol H 2 O -- >1.149 x 10 -2^ mol H 2 mol H per 1 mol water! MAR
C (^) x H (^) y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O
Using Stoichiometry to
Determine a Formula
Now find ratio of mol H/mol C to find values
of x and y in Cx H y.
1.149 x 10 -2^ mol H/ 8.61 x 10-3^ mol C
= 1.33 mol H / 1.00 mol C
= 4 mol H / 3 mol C
Empirical formula = C 3 H 4
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C (^) x H (^) y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O
Caproic acid, the substance responsible for "dirty gym socks" smell, contains C, H and O. Combustion analysis of 0.450 g caproic acid gives 0.418 g H 2 O and 1.023 g CO 2 , and the molar mass was found to be 116.2 g mol-^. What is the molecular formula of caproic acid? C (^) x H (^) y Oz + some oxygen ---> 1.023 g CO 2 + 0.418 g H 2 O Careful: oxygen comes from caproic acid and O 2 , need special technique
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Formulas with C, H and O
Combustion analysis of 0.450 g caproic acid gives 0.418 g H 2 O and 1. g CO 2 , and the molar mass is 116.2 g mol-1^. What is the molecular formula? Start with "regular" approach for mol H & mol C: 0.418 g H 2 O * (mol/18.02 g) * (2 mol H/mol H 2 O) = 0.0464 mol H 0.0464 mol H * (1.01 g/mol H) = 0.0469 g H 1.023 g CO 2 * (mol/44.01 g) * (1 mol C/mol CO 2 ) = 0.02324 mol C 0.02324 mol C * (12.01 g/mol C) = 0.2791 g C Why did we convert to grams? Law of Mass Action! MAR
Formulas with C, H and O
0.450 g caproic acid: 0.418 g H 2 O (0.0464 mol H, 0.0469 g H) and 1.023 g CO 2 (0.02324 mol C, 0.2791 g C), molar mass = 116.2 g/mol. What is the molecular formula? Realize that 0.450 g of caproic acid equals all the g C, g H and g O in the complex. Converting mol H and mol C to grams, then subtracting from 0.450 g, gives g O in caproic acid: 0.450 g - 0.0469 g - 0.2791 g = 0.124 g O
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caproic acid g of H in acid g of C in acid g of O in acid
0.124 g O * (mol O / 16.00 g) = 0.00775 mol O
Formulas with C, H and O
