Transcript CHEM1310 Lecture

Chapter 3
Stoichiometry
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Atomic Masses
The Mole
Molar Mass
Percent Composition of Compounds
Determining the Formula of a Compound
Chemical Equations
Balancing Chemical Equations
Stoichiometric Calculations: Amounts of
Reactants and Products
3.9 Calculations Involving a Limiting Reactant
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Atomic Mass
• We measure mass, volume, pressure, etc.
• We want number of atoms (or number of
moles)
• grams x (grams/mole)-1 = moles
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Atoms
• Avogadro’s Number is the number of 12C atoms in
exactly 12 grams of carbon
N0 = 6.02 X 1023
• The mass, in grams, of Avogadro's number of atoms of
an element is numerically equal to the relative atomic
mass of that element (relative to carbon)
•
•
•
•
•
The mass of 6.02 X 1023 atoms of 12C = 12.00 g (defn)
The mass of 1 atom of 12C = 12.00 amu (12.00 Daltons)
The mass of 6.02 X 1023 atoms of C = 12.01 g
The mass of 6.02 X 1023 atoms of 35Cl = 35.0 g
The mass of 6.02 X 1023 atoms of Cl = 35.4 g
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Molar Mass
• Molecular Mass (aka the molecular weight) of a
molecule equals the sum of the atomic masses of all of
the atoms making up the molecule.
• Examples
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Fructose (C6H12O6)
Benzene (C6H6)
Water (H20)
Sodium Chloride (NaCl)
Hemoglobin (best to look that one up)
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Moles
• The # of moles of chemical is its amount.
• One mole of a substance equals the amount that
contains Avogadro's number of atoms or molecules.
• One mole of an element or molecule has a Molecular
Weight (MWt) of that element or molecule, expressed in
grams
• For example, the Molecular Weight of Glucose
(Glucose C6H12O6) is
MWt =
6(12.0 g/mol)
+ 12(1.00 g/mol)
+ 6(16.0 g/mol)
= 180 g/mol
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Isoamyl acetate have the formula C7H14O2.
Calculate (a) how many moles and (b) how
many molecules are contained in 0.250
grams of isoamyl acetate.
Strategy: Use the units. You are given grams. You need
moles. The molecular weight is given in g / mol). Then you
need molecules. Avo’s # is molecules / mole.
1. Calculate MWt of C7H14O2
2. Calculate the number of moles in 0.250 grams
3. Using Avogadro’s number to calculate the
number of molecules in the number of moles of
C7H14O2
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Percentage Composition (g/g) from Empirical or
Molecular Formula
Tetrodotoxin, a potent poison found in the
ovaries and liver of the globefish, has the
empirical formula C11H17N3O8. Calculate the
mass percentages of the four element in this
compound.
Strategy:
1. Calculate molar mass of C11H17N3O8, by
finding the mass contributed by each
element.
2. Assume you have one mole of the
compound.
3. Divide the mass of each element by the
total mass of the
compound.
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Tetrodotoxin has the empirical formula C11H17N3O8.
Calculate the mass percentages (g/g) of the four element in
this compound.
Solution:
1.
2.
3.
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Calculate molar mass of C11H17N3O8, by finding the
mass contributed by each element
C : 11x12 =
132 g / mol
H : 1x17 =
17
N : 3x14 =
42
O : 8x16 =
126
tot =
317 g / mol
Calculate the mass of one mole of tetrodotoxin
317 g / mol x 1 mol = 317 g
Divide the mass of each element by the mass of the
compound.
Example C: 132 g / 317 g = 0.42 g/g (42 %)
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Milli, Micro, etc
1 mmol = 1 millimole =1x10-3 mol
1mg = 1 milligram =1x10-3 g
1msec = 1 millisec =1x10-3 sec
1 u mol = 1 micromole =1x10-6 mol
1u g = 1 microgram =1x10-6 g
Also nano (10-9)
pico (10-12)
femto (10-15)
An Angstrom (Å)= 10-15 meters = 0.1 nm
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The Law of Conservation of Mass
• The mass/matter of a closed system is constant.
• Mass can be rearranged but not created or
/destroyed.
• In a chemical reaction the mass of the reactants
equals the mass of the products.
• Mass is conserved during a change of state (solid
to liquid to gas)
(This law holds in this class, but maybe not in your
physics class)
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Using Conservation of Mass to
determine an Empirical Formula
Moderate heating of 97.4 mg of a
compound containing nickel, carbon
and oxygen and no other elements
drives off all of the carbon and
oxygen in the form of carbon
monoxide (CO) and leaves 33.5 mg
of metallic nickel (Ni) behind.
Determine the empirical formula of
the compound.
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Using Conservation of Mass to
determine an Empirical Formula
1. Write the reaction. NixCyOy -> X Ni + Y CO
2. Use conservation of mass to find the mass of CO.
97.4 mg (mass tot) – 33.5 mg (mass Ni) = 63.9 g (mass CO)
3. Find the number of moles of CO and of Ni.
CO : 63.9 mg / (12.0+16.0 g/mol) = 2.28 mmol
Ni : 33.5 mg / 58.7 g / mol) = 0.57 mmol
4. Find the ratios of the moles by dividing each by the smallest
one, i.e., normalize to the smallest.
2.28 mmol CO /.57 mmol Ni = 4
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5. Y/X = 4: answer is NiC4O4
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Chemical Equations
•
Chemical Reactions tell us three things
• What atoms or molecules react together to form
what products.
• How much reactant how much product.
• The state of each species
aA (l) + bB (s)  cC (s) + dD (g)
Reactants
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Products
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Chemical Equations
The Law of Conservation of Mass says that a
chemical equation must have the same
number of atoms of a given kind on each
side (a chemical reaction cannot create or
destroy carbon, or oxygen, or hydrogen, or
etc.)
Chemical equations must be balanced!
H2 + O2  H2O (not balanced)
2H2 + O2  2H2O (balanced)
Chemical Equations
4 Al(s) + 3 O2(g)
→ 2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules
---produces--->
2 molecules of Al2O3
or
4 moles of Al + 3 moles of O2
---produces--->
2 moles of Al2O3
Br2 (l) + ___ Al (s) →
__Al2Br6 (s)
Balancing Chemical Equations
• The same atoms are present in a
reaction at the beginning and at the
end.
KClO3 (s)

KCl (s) +
O2 (g)
KClO3 (s)

KCl (s) +
3/2O2 (g) better
2KCl (s) +
3O2 (g) best
2KClO3 (s) 
no
To Balance an Equation
Step 1: Set the stoichimetric coefficient of the most complicated
molecule (with the largest number of different elements ) to 1.
Step 2: Balance as many atoms as possible in the second most
complicated molecule. Ignore atoms that show up elsewhere in
homonuclear species like O2 and H2.
Step 3: Balance the atoms in the in the homonuclear species (O2
and H2).
Step 4: Eliminate fractional coefficients.
Step 5: Count the each atom type on each side of the equation.
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Balancing
Equations
Combustion of Propane
C3H8(g) + O2(g) 
CO2(g) +
H2O (g)
Balancing
Equations
Combustion of Propane
1C3H8(g) + 5O2(g) 
3CO2(g) + 4H2O (g)
Writing Balanced Chemical Equations
PbO2 + Pb + H2SO4 →
PbSO4 + H2O

PbO2 + Pb + 2 H2SO4 → 2 PbSO4 + 2 H2O
2 Pb Check for balance
10 O
4H
Balanced
2 Pb
10 O
4H
Problem: Suppose we have 1.45 grams of Pb in the
presence of excess lead oxide and sulfuric acid. How many
grams of Lead Sulfate are produced?
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Limiting Reactants
3 Br2 (l) + 2 Al (s) → 1 Al2Br6 (s)
excess
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limiting
Given the amounts below
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Limiting Reagent
1. Balance the reaction.
2. Convert reactant masses to moles [g(g/mol)1=mol]
3. Normalize the moles of each reactant by its
stoichiometric coefficient.
4. Find the smallest normalized number of moles
5. Use the ratio of stoichiometric coefficients to
find the moles of product.
6. Convert to mass (if necessary).
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Example Calculation Involving a Limiting Reactant
Suppose that 1.00 g of sodium and 1.00 g of chlorine react to form
sodium chloride (NaCl). Which of these is limiting, and what is the mass
of product.
2 Na + Cl2 → 2 NaCl
nNa = 1.00 g x (1 mol Na / 23.0 g Na) = 0.0435 mol Na
nCl2 = 1.00 g × (1 mol / 70.9 g Cl2) = 0.0141 mol Cl2
nNa/2 = 0.022 (normalized) nCl2/1 = 0.0141 (normalized)
Cl2 is the limiting reagent
0.0141 moles of Cl2 gives 0.0282 moles of NaCl. Use the molecular
weight of NaCl to find the mass of NaCl produced.
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What mass (in grams) of xenon tetrafluoride would
be required to react completely with 1.000 g of
water?
XeF4 + 2 H2O → Xe + 4 HF + O2
=
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At one point in the purification of silicon, gaseous SiHCl3
reacts with gaseous H2 to give gaseous HCl and solid Si.
(a) Determine the chemical amount (in moles) of H2 required
to react with 160.4 mol of SiHCl3.
(b) Determine the chemical amount of HCl that is produced.
(c) Determine the mass (in grams) of Si that is produced.
SiHCl3 (g) + H2 (g) → 3 HCl (g) + Si (s)
(a)
(b)
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At one point in the purification of silicon, gaseous SiHCl3 reacts
with gaseous H2 to give gaseous HCl and solid Si.
(a) Determine the chemical amount (in moles) of H2 required to
react with 160.4 mol of SiHCl3.
(b) Determine the chemical amount of HCl that is produced.
(c) Determine the mass (in grams) of Si that is produced.
1 mol SiHCl3
1 mol H2
3 mol HCl
1 mol Si
SiHCl3 (g) + H2 (g) → 3 HCl (g) + Si (s)
(c)
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Isotopes of Cl:
031
150 msec
EC/ECp,11.98
0
S-31/P-30
30.9924
032
298 msec
EC/ECa/ECp,12.6
85
S-32/Si-28/P-31
31.985688
033
2.511 sec
EC,5.583
MeV
S-33
32.97745
034
1.5264 sec
EC,5.492
MeV
S-34
33.97376
035
75.77%
Stable
036
3.01E+5 yr
B/EC,10.413
Ar-36/S-36
35.9683
037
24.23%
043
3.3 sec
B-,7.950
MeV
Ar-43
42.974202
044
0.43 sec
B-/Bn,3.920
Ar-44/Ar-43
43.978539
34.9688
038
37.24 min
B-,4.917 MeV
Ar-38
37.968010
039
55.6 min
B-,3.442 MeV
Ar-39
38.968008
040
1.35 min
B-,7.480
MeV
Ar-40
39.970413
041
38.4 sec
B-,5.730
MeV
Ar-41
40.970649
042
6.8 sec
B-,9.430
MeV
Ar-42
41.973172
045
400 msec
B-/B-n,10.800
Ar-45/Ar-44
44.979710
046
0.22 sec
B-/B-n,6.900
Ar-46/Ar-45
45.984111
047
200 nsec
B-/Bn,14.700
Ar-47/Ar-46
46.987976
MASS
abund.
Halflife
Particle, Energy
Decay Product(s)
Isotopic Mass
35Cl
contains 17 protons and 18 neutrons
37Cl
contains 17 protons and 20 neutrons
Stable
36.9659
Average Relative Atomic Mass = A1P1 + A2P2 + ... + AnPn
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Chlorine = 35.45
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