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Transcript of Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations Chapter Three:...
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Chapter Three:
Stoichiometry: Calculations with Chemical Formulas and Equations
Chapter Three:
Stoichiometry: Calculations with Chemical Formulas and Equations
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Overview
Chemical Equations Patterns/Reactions Atomic/Molecular Weights Moles/Molar Mass Empirical/Molecular Formulas Quantitative Relationships Limiting Reactants/Theoretical Yields
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Chemical Equations
chemical ‘sentences’– reactants and products described by formulas or
symbols combined with “punctuation”
2 H2(g) + O2(g) 2 H2O(l)
reactant formulas
product formula
coefficients physical state
‘react to form’
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“atoms can be neither created nor destroyed”– all equations must be ‘balanced’ with the same number of
atoms on both sides of the reaction arrow
H2O + O2 H2O2
unbalanced
2 H & 3 O 2 H & 2 O
2H2O + O2 2H2O2
balanced
4 H & 4 O = 4 H & 4 O
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unbalanced
balanced
two formula units
one formula unit
H2O O2 H2O2
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Examples CH3OH(l) + O2(g) CO2(g) + H2O(l)
Na(s) + H2O(l) NaOH(aq) + H2(g)
HBr(aq)+ Ba(OH)2(aq) H2O(l) + BaBr2(aq)
2 2 43
22 2
2 2
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Patterns of Chemical Reactivity
Because elements are grouped by chemical properties, their reactions can also be grouped: – alkali metals and water
2K(s) + 2H2O(l) 2KOH(aq) + H2(g) specific
2M(s) + 2H2O(l) 2MOH(aq) + H2(g) general
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– Combustion in air
C3H8(g) + O2(g) CO2(g) + H2O(l) 3 45
CxHy + O2(g) CO2(g) + H2O(l)
specific
general
hydrocarbon
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– Combination Reactions
2Mg(s) + O2(g) 2MgO(s)
X + Y XY
specific
general
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– Decomposition Reactions
CaCO3(s) CaO(s) + CO2(g)
XY X + Y
specific
general
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Name the Reaction
PbCO3(s) PbO(s) + CO2(g) decomposition
C(s) + O2(g) CO2(g) combination
2NaN3(s) 2Na(s) + 3N2(g)
decomposition
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) combustion
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Atomic and Molecular Masses
Amu scale – defined by assigning the mass of 12C as 12 amu exactly
– 1 amu = 1.66054 x 10-24 g– 1 g = 6.02214 x 1023 amu
Average Atomic Masses– 12C 98.892% abundant 13C 1.1108% abundant
(0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011
amu
atomic mass
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Formula and Molecular Masses– sum of all atomic masses in the formula of an
ionic or molecular compound
vitamin C C6H8O6
6 x 12.0 = 72.0 amu
8 x 1.0 = 8.0 amu
6 x 16.0 = 96.0 amu
176.0 amu
formula mass of vitamin C (often called molecular mass)
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Percentage Composition Calculate the percent mass that each type
of atom contributes to a molecule– % X = (no. X atoms)(X amu) x 100
formula mass cmpd
– C6H8O6
% C = (6)(12.01amu) x 100 = 40.94% C
176.0 amu% H = (8)(1.01amu) x 100 = 4.59% H
176.0 amu
% O = (6)(16.00 amu) x 100 = 54.55% O
176.0 amu
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The Mole We can measure masses in amu but how do we
relate that to mass in grams? We define a quantity of atoms – a mole – which has the same mass in grams as the mass of the element in amu.
So how many atoms does it take to make, say, 1.00 g of H?
1.0 g H x 1 atom H 6.0 x 1023 atoms of H
1.7 x 10-24 g H
12.0 g C x 1 atom C 6.0 x 1023 atoms of C
2.0 x 10-23 g C
a mole
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Avogadro’s Number
6.02214 x 1023 units/mole
– No. of atoms per mole of an element– No. of molecules per mole of molecular
cmpd.– No. of formula units per mole of ionic cmpd.– No. of cows per mole of cows
Memorize this number & what it means!Memorize this number & what it means!
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1 C atom = 12 amu 1 mole C atoms = 12 g
1 Mg atom = 24 amu 1 mole Mg atoms = 24 g
1 CO molecule = 28 amu 1 mole CO molecules = 28
g
1 NaCl fm. unit = 58 amu 1 mole NaCl fm.units = 58g
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Molar Mass From this information we can define something
called the molar mass (MM) of an atom (or molecule or formula):
from the equality: 1 mole C = 12.0 g C
we define the molar mass of a substance
12.0 g C = MM or Molar Mass1 mole C (Atomic Mass)
(Molecular Mass)(Formula Mass)conversion
factor
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Problems
Practice Ex. 3.9:– How many mole in 508 g of NaHCO3?
Given: MM = 84.02 g/mol NaHCO3 508 g NaHCO3 508 g NaHCO3 x 1 mole = 6.05 mole NaHCO3
84.02 g NaHCO3
– How many formula units of NaHCO3?
Given: 6.02 x 1023 form. units/mole NaHCO3
6.05 mole NaHCO3 x 6.02 x 1023 fm. units = 3.6 x 1024 fm. 1 mole unitsNaHCO3
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Molar Mass converts between moles and grams of a substance
Avogadro’s number converts between moles of a substance and atoms (or molecules or formula units) of that substance
These are very important conversion factors, know & understand them!
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Problems How many moles of vitamin C are contained in 5.00 g of
vitamin C? C6H8O6 176.0 g/mol
17.5 mg of cocaine (C17H21NO4) per kg of body weight is a lethal dose. How many moles is that? How many molecules?
In 25 g of C12H30O2 THC (tetrahydrocannibinol) how many moles are there? How many molecules are there? How many C atoms are there?
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How many moles of O are contained in 1.50 moles of C6H5NO3?
How many grams of nitrogen are contained in 70.0 g of C6H5NO3? How many atoms?
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Calculate the number of H atoms in 50.0 mg of acetominophen, C8H9O2N.
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Determination Empirical Formulas
simplest ratio of atoms
– change g of each element to moles or
– assume 100 grams of substance & change the % of each element to moles
– change the mole ratio of atoms to the simplest ratio by dividing by the smallest number of moles
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Practice Ex. 3.12:– 5.325 g methyl benzoate contains 3.758 g C, 0.316 g H,
1.251 g O. Determine empirical formula.
3.758 g C x 1 mole = 0.313 mol C
12.01 g
0.316 g H x 1 mole = 0.313 mol H
1.01 g
1.251 g O x 1 mole = 0.0782 mol O
16.00 g
C0.313H0.313O0.0782
C4H4O
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Determination of Molecular Formulas
actual ratio of atoms
– determine the empirical formula
– divide the actual molar mass by the empirical formula mass to get ‘n’
– multiply the mole ratio in the empirical formula by ‘n’
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Practice Ex. 3.13:– Ethylene glycol is composed of 38.7% C, 9.7% H &
51.6% O by mass. Its true molar mass is 62.1 g/mol. What are the empirical and molecular formulas?
38.7 g C x 1 mole = 3.23 mole C
12.0 g
9.7 g H x 1 mole = 9.60 mole H 1.01 g
51.6 g O x 1 mole = 3.22 mole O 16.0 g
C3.23H9.60O3.22
CH3Oempirical formula
n = 2
molecular formula
C2H6O2
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Formulas from Combustion Data
Formulas determined from products of combustion products– Menthol is composed of C, H, and O. A 0.1005 g
sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g H2O. What is the empirical formula?
CxHyOz + O2 CO2 + H2O0.1005 g 0.2829 g 0.1159 g
– Calculate moles CO2 & C; moles H2O & H
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0.2829 g CO2 x 1 mol x 1 mol C = 0.00643 mol C 44.0 g 1 mol CO2
0.1159 g H2O x 1 mol x 2 mol H = 0.0129 mol H
18.0 g 1 mol H2O
total mass of C + H = 0.0902 g
mass of O = 0.1005 g - 0.0902 g = 0.0103 g O x 1 mol = 16.0 g
6.44 x 10-4 mol O
total mass of all C, H & O
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C0.00643H0.0129O0.000644
C10H20O
0.00643 mol C 0.0129 mol H 6.44 x 10-4 mol O
If the MM is 156 g/mol, what is the molecular formula?
n=1 therefore molecular formula is C10H20O
(empirical formula mass 156 g/mol)
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Quantitative Stoichiometry
Determination of quantities from balanced chemical reaction equations– mole ratios from balanced chemical equation
convert between species
– if quantities are given for more than one reactant, the limiting reactant must be determined
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– Given the following balanced equation:
1Mg(OH)2 + 2HCl 1MgCl2 + 2H2O
– Calculate the number of moles of HCl required to react completely with 0.42 mol of Mg(OH)2
0.42 mol Mg(OH)2 x 2 mol HCl = 0.84 mol HCl 1 mol Mg(OH)2
The mole ratio comes from the balanced chemical equation
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– How many grams of MgCl2 can be produced?
0.42 mol Mg(OH)2 x 1 mol MgCl2 x 95.3 g MgCl2 1 mol Mg(OH)2 1 mol
= 40.0 g MgCl2
Theoretical Yield -- maximum amount that can be produced
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General Sequence of Conversion:
grams of reactant
moles of reactant
MM reactant
moles of product
mole ratio
grams of product
MM product
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Practice Ex. 3.14:
How many grams of O2 can be prepared from 4.50 g of KClO3?
2KClO3 2KCl + 3O2
4.50 g KClO3 x 1 mol x 3 mol O2 x 32.0 g O2 = 1.76 g O2
122.6 g 2 mol KClO3 1 mol
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Limiting Reactant
given a non-stoichiometric amount of both reactants, you will have to determine which is the limiting reagent or reactant
example: you have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is the limiting “reagent”, and how much excess “reagent” do you have left over?
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—Balanced ‘Equation’
1 (mole) frame + 2 (moles) wheels 1 (mole) bicycles [10 (moles) frames] [16 (moles) wheels] [8(moles) bicycles]
– Limiting Reactant -- will produce the least amount of product
10 mol frames x 1 mol bicycles = 10 bicycles 1 mol frames
16 mol wheels x 1 mol bicycles = 8 bicycles 2 mol wheels
limiting reactant
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Practice Ex. 3.16:
A mixture of 1.5 mol of Al and 3.0 mol of Cl2 react. What is limiting & how many moles of AlCl3 are formed?
2Al(s) + 3Cl2(g) 2AlCl3(s) 1.5 mol 3.0 mol
1.5 mol Al x 2 mol AlCl3 = 1.5 mol AlCl3 2 mol Al
3.0 mol Cl2 x 2 mol AlCl3 = 2.0 mol AlCl3 3 mol Cl2
1.5 mol