Download Chapt. 3

Chem 31
Fall 2002
Chapter 3
Stoichiometry: Calculations
with Chemical Formulas and
Equations
Writing and Balancing
Chemical Equations
1. Write Equation in Words
-you cannot write an equation unless you know
what happens in the reaction!
2. Substitute Chemical Formulas for Words
3. Balance the Equation
-Conservation of Mass
4. Indicate Physical States
-solid (s), liquid (l), gas (g), aqueous (aq)
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Example
„ One type of rocket fuel reacts hydrazine and
dinitrogen tetroxide and produces nitrogen gas
and water
1.
hydrazine + dinitrogen tetroxide → nitrogen + water
2.
N2H4 + N2O4 → N2 + H2O
3.
2N2H4 + N2O4 → 3N2 + 4H2O
4.
2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (l)
Formulas
Balanced
Done!
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Another Example
„ A solution of sodium chloride was added to a solution of
silver nitrate, forming a precipitate of silver chloride
1. sodium choride + silver nitrate → silver chloride + sodium
nitrate
2. NaCl + AgNO3 → AgCl + NaNO3
Formulas
3. NaCl + AgNO3 → AgCl + NaNO3
Balanced
4. NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) →
AgCl (s) + Na+ (aq) + NO3- (aq)
Ag+ (aq) + Cl- (aq) → AgCl (s)
Net Ionic Equation
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Let’s Get Quantitative
„Recall:
1 mole = 6.02 x 1023 particles
Atomic Masses (Weights): mass/atom (amu)
Molecular/Formula Weights: mass/compound (amu)
Molar Masses: mass of a mole of atoms or
compound (g)
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Why Fractional Molar Masses?
„Need to consider the natural
abundances of isotopes
Example: Chlorine
75.5%
35Cl
+ 24.5%
37Cl
(0.755)(34.97) + (0.245)(36.97) = 35.45 g/mol
-This is a weighted average; 1 mol of Cl will
have a mass of 35.45 grams
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3
General Strategy
Be Careful!
moles of: atoms? molecules?
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Mole-Based Calculations
„How many grams of Phosphorous are
there in 0.010 mol P2O5?
Strategy: mol P2O5 → mol P → g P
0.010 mol P2O5 x 2 mol P x 30.974 g P = 0.61948 g P
1 mol P2O5
1 mol P
Round to:
0.62 g Phosphorous
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How Many Atoms?
„How many Phosphorous atoms are there
in 0.010 mol P2O5?
Strategy: mol P2O5 → mol P → #P atoms
0.010 mol P2O5 x 2 mol P x 6.022 x 1023 P atoms
1 mol P2O5
1 mol P
=
= 1.20440 x 1022 P atoms
= 1.2 x 1022 P atoms
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Strategy: Empirical Formula
from Percent Composition
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Empirical Formula from
Percent Composition
„ What is the empirical formula for a binary
compound which is found to be:
56.4% Oxygen (by mass)
43.6% Phosphorous (by mass)?
Strategy: % → grams → mol
(% is a relative measure, so DEFINE a sample size (100 g))
In a 100-g sample:
56.4 g O x 1 mol O
43.6 g P
15.999 g O
x
1 mol P
30.974 g P
=
3.525 mol O
=
1.4076 mol P
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Empirical Formula continued
This gives:
P1.4076O3.525
Dividing: PO2.50
→ P2O5
•What about a MOLECULAR formula?
-need the molar mass (MW) of the compound
Example:
MW of P2O5 cmpd is 284 g/mol
Empirical Formula Mass ≈ 2x31 + 5x16 = 142 g
MW/Emp Form Mass = 284/142 = 2
So:
2 x P2O5 =
P4O10
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Quantifying Reaction
Chemistry
„With a balanced equation, we can
relate amounts of reactants and
products via molar relationships
Let’s look at this combustion reaction:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
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Reactant Quantities
„How many moles of CH4 will react
with 3.62 mol O2?
convert:
mol O2 → mol CH4
from rxn we know: 2 mol O2 react with 1 mol CH4
3.62 mol O2 x 1 mol CH4 = 1.81 mol CH4
2 mol O2
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Product from
Reactant Moles
„How many moles of CO2 are produced
when 1.24 mol O2 react?
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
↑
↑
1.24 mol O2 x 1 mol CO2 = 0.62 mol CO2
2 mol O2
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Product Mass from
Reactant Moles
„ How many grams of CO2 are produced when
1.24 mol O2 are reacted?
mol O2 → mol CO2 → g CO2
1.24 mol O2 x 1 mol CO2 x 44.01 g CO2 =
2 mol O2
mol CO2
= 2.72862 x 101 g CO2
= 2.73x 101 g CO2
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Product Mass from Reactant
Mass
„ How many grams of CO2 are produced when
3.47 grams of O2 react?
g O2 → mol O2 → mol CO2 → g CO2
3.47 g O2 x 1 mol O2 x 1 mol CO2 x 44.01 g CO2 =
32.00 g O2
2 mol O2
mol CO2
= 2.38616719 g CO2
= 2.39 g CO2
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The Big Picture Strategy
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What about
Non-stoichiometric Amounts?
¾ Huh?
¾ EXAMPLE:
3Fe + 4H2O → Fe3O4 + 4H2
-How many moles of H2 can be prepared from
4.00 mol Fe and 5.00 mol H2O?
-BUT: Fe and H2O react in a 3:4 ratio but they are
available in a 4:5 ratio
We will run out of ONE of the reactants before the
other is completely reacted.
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Limiting Reagent!
¾Define: reagent that is completely
consumed before any other
In our example:
4.00 mol Fe = 1.33 > 1.25 = 5.00 mol H2O
3 mol Fe
4 mol H2O
So, H2O is our limiting reagent:
5.00 mol H2O x 4 mol H2 = 5.00 mol H2
4 mol H2O
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Ok, Let’s Try Another One
¾ How many grams of N2F4 can be prepared from
4.00 g NH3 and 14.0 g F2?
2NH3 + 5F2 → N2F4 + 6HF
Which one is limiting reagent?
4.00 g NH3 x 1 mol NH3 = 0.23488 mol NH3
17.03 g NH3
14.0 g F2 x
1 mol F2 = 0.36845 mol F2
37.997 g F2
So: 0.23488 mol NH3 = 0.117 > 0.0737 = 0.36845 mol F2
2 mol NH3
5 mol F2
*F2 is the Limiting Reagent*
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On to the answer:
Strategy:
g F2 → mol F2 → mol N2F4 → g N2F4
0.36845 mol F2 x 1 mol N2F4 x 104.01 g N2F4 =
5 mol F2
1 mol N2F4
= 7.6644998 g N2F4
= 7.66
g N2F4
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Percent Yield
„ Suppose the previous reaction was
performed, but only 4.80 g of N2F4 were
produced? Calculate the percent yield of
the reaction.
%-yield = Actual (exptl) Yield x 100
Theoretical Yield
= 4.80 g N2F4 x 100 = 62.7%
7.6645 g N2F4
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