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7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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We measure ordinary objects either
by counting or weighing them,
depending on which method is more convenient
Silberberg, M. 2010. Principles of General Chemistry. 2
nd
ed. New York: McGraw-Hill.http://weyume.com/wp-content/uploads/2011/04/rice.jpg
http://farm1.static.flickr.com/21/90994367_5613e69fd9.jpg
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Dozen = 12
Pair = 2
Certain nouns can be used to define
a collection of objects
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The mole
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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The mole (n or mol) is the amount of matter that
contains as many entities (atoms, molecules,
ions, or other particles) as there are atoms inexactly 12 g of the carbon-12 isotope (12C)
• The actual number of atoms in 12 g of
carbon-12 was determined experimentally
• Avogadro’s number (N A)
N A
= 6.02 x 1023
Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd.
Chan R. 2002. Chemistr 7th ed. Sin a ore: McGraw-Hill.
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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Just as 1 dozen of oranges contains 12 oranges,
1 mole of matter contains 6.02 x 1023 entities
Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd.
Chan R. 2002. Chemistr 7th ed. Sin a ore: McGraw-Hill.
1 mole 12C atoms = 6.02 x 1023 12C atoms
1 mole H2O molecules = 6.02 x 1023 H2O molecules
1 mole NO3
-
ions = 6.02 x 10
23
NO3
-
ions
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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Each of these contains one mole of the
substance
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
copper iron
carbon sulfur
mercury
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One mole (or an Avogadro’s number)
is an extremely big number
• One mole of softdrink cans would cover
the surface of the earth to a depth of over
300 kilometers
• If we were able to count the number of
atoms at a rate of 10 million per second, it
would take about 2 billion years to count a
mole of atoms
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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Molar mass
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The molar mass (M ) of a substance is the mass
of one mole of its entities (atoms, molecules,
ions, or other particles) in units of g/mol
M H2O = 18.0 g/mol
(one mole of H2O molecule weighs 18.0 g)
M NO3- = 62.0 g/mol
(one mole of NO3- ion weighs 62.0 g)
M C = 12.01 g/mol
(one mole of C atom weighs 12.01 g)
Brown, T., E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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The periodic table is indispensable for
calculating the molar mass of a substance
• Elements
–M is the numerical value from the
periodic table
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
M H = 1.008 g/mol
M O = 16.00 g/mol
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The periodic table is indispensable for
calculating the molar mass of a substance
• Compounds
–M is the sum of the molar masses of the
atoms of the elements in the formula
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
M SO2 = M S + (2 x M O)
= 32.07 g/mol + (2 x 16.00 g/mol)
= 64.07 g/mol
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The periodic table is indispensable for
calculating the molar mass of a substance
• Compounds
–M is the sum of the molar masses of the
atoms of the elements in the formula
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
M K2S = (2 x M K) + M S
= (2 x 39.10 g/mol) + 32.07 g/mol
= 110.27 g/mol
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Interconverting moles, mass,
and chemical entities
(atoms, molecules, ions, or other particles)
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The factor-label method is used to convert
from one unit to another
1 peso = 4 25-centavos
1 peso
4 25-centavos = 1
1 peso
4 25-centavos = 1
unit factor
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
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Alexa bought 18 fresh chicken’s eggs. How
many dozens of egg did she buy?
dozens of egg = x18 eggs 1 dozen egg
12 eggs
= 1.5 dozens of egg
unit factor
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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In order to convert between moles, mass, and
chemical entities, the factor label method is used
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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Methane (CH4) is the principal component of
natural gas. How many moles of methane
are present in 6.07 g of CH4?M CH4 = M C + (4 x M H)
= 12.01 g/mol + (4 x 1.01 g/mol)
= 16.05 g/mol
Report final answer with the correct number of significantfi ures!
nCH4 = x
6.07 g CH4 1 mol CH4
16.05 g CH4
= 0.378 mol CH4
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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How many molecules of methane are
present 6.07 g of CH4?
molecules CH4 = x
6.07 g CH4 1 mol CH4
16.05 g CH4
= 2.28 x 1023 molecules CH4
x6.02 x 1023 molecules CH4
mol CH4
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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Glucose (C6H12O6), also known as blood sugar,
is used by the body as energy source.
How many moles of glucose are present in1.75 x 1022 molecules of glucose?
nC6H12O6 = x1.75 x 1022 molecules C6H12O6 1 mol C6H12O6
6.02 x 1023
moleculesC6H12O6
= 0.0291 mol C6H12O6
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How many grams of glucose are present in
1.75 x 1022 molecules of glucose?
M C6H12O6 = (6 x M C) + (12 x M H) + (6 x M O)
= (6 x 12.01 g/mol) + (12 x 1.01 g/mol)
+ (6 x 16.00 g/mol)
= 180.18 g/mol
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How many grams of glucose are present in
1.75 x 1022 molecules of glucose?
nC6H12O6 = x1.75 x 1022 molecules C6H12O6 1 mol C6H12O6
6.02 x 1023
molecules
C6H12O6
180.18 g C6H12O6
x 1 mol C6H12O6
= 5.24 g C6H12O6
7/29/2019 Chemistry molecular Geometry and other significant studies plus electronic configuration
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1. Urea [(NH2)2CO] is used in animalfeeds, as fertilizer and in the
manufacture of polymers.a. Draw the Lewis structure of area. C is
surrounded by O and the N’s. (Where are the
H’s connected?) b. Calculate its molar mass.
c. Consider 25.6 g of urea. How many moles of
urea present?
d. How many moles of N are present?
e. How many moles of C are present?
f. How many molecules of urea are present?
g. How many atoms of N are present?
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2. Vitamin C, ascorbic acid, is oftensold as sodium ascorbate.
a. Calculate its molar mass.b. Consider a 500.-mg tablet. How many moles
of sodium ascorbate are present?
c. How many moles of C are present?
d. How many moles of Na are present?
e. How many formula units of sodium
ascorbate are present?
f. How many atoms of Na are present?
O
O
OH
-O
HO
OH
Na+
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Chemical reactions and
chemical equations
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A chemical reaction shows the process in
which a substance (or substances) is
changed into one or more new substances
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
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A chemical equation uses chemical symbols to
show what happens during a chemical reaction
reactants product
(g) (g) (l)
“Two molecules of hydrogen react with one molecule of
oxygen to yield two moles of water”
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
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The Law of Conservation of Mass states that
matter is neither created nor destroyed
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
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To conform with the Law of Conservation of Mass, there
must be the same number of each type of atom on both
sides of the arrow. Hence, we balance the equation by
adding coefficients before each chemical symbol
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
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Calculating the amounts of
reactant and product
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Stoichiometry a double
cheeseburger
2 bun slices + 2 cheese slices+ 2 burger patties =
I b l d ti th b f l f
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In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g) 2CO2(g)
2 mol CO = 1 mol O2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
2 mol CO
1 mol O2
= 11 mol O2
2 mol CO
= 1
I b l d ti th b f l f
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In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g) 2CO2(g)
2 mol CO = 2 mol CO2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
2 mol CO
2 mol CO2
= 12 mol CO2
2 mol CO
= 1
I b l d ti th b f l f
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In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g) 2CO2(g)
1 mol O2 = 2 mol CO2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
1 mol O2
2 mol CO2
= 12 mol CO2
1 mol O2
= 1
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The amount of one substance in a reaction
is related to that of any other
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
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All alkali metals react with water to produce
hydrogen gas and the corresponding
alkali metal hydroxide
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
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How many moles of H2 will be formed by the
complete reaction of 6.23 moles of Li with water?
nH2 = x6.23 mol Li 1 mol H2
2 mol Li
= 3.12 mol H2
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
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How many grams of H2 will be formed by the
complete reaction of 80.57 g of Li with water?
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
mH2 = x x80.57 g Li 1 mol Li 1 mol H
2
6.941 g Li 2 mol Li
2.016 g H2 x1 mol H2
= 11.70 g H2
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In a lifetime, the average American uses about
794 kg of copper in coins, plumbing, and wiring.
Copper is obtained from sulfide ores (such asCu2S) by a multistep process. After an initial
grinding, the first step is to “roast” the ore (heat it
strongly with O2) to form Cu2O and SO2
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
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How many moles of oxygen are required
to roast 10.0 mol of Cu2S?
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
nO2 = x10.0 mol Cu2S 3 mol O2
2 mol Cu2S
= 15.0 mol O2
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How many grams of SO2 are formed when
10.0 mol of Cu2S is roasted?
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
mSO2 = x x10.0 mol Cu2S 2 mol SO2 64.07 g SO2
2 mol Cu2S 1 mol SO2
= 641 g SO2
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Ch 2 F
• No meeting this Friday
• Lab discussion moved to March 2
• 1:30-3:30 pm• SOM 201
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How many grams of O2 are required to form
2.86 kg of Cu2O?
2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)
mO2 = x x2.86 kg Cu
2
O 1000 g Cu2
O 1 mol Cu2
O
1 kg Cu2O 143.10 g Cu2O
3 mol O2 32.00 g O2 x x2 mol Cu2O 1 mol O2
= 960 g O2
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Limiting Reactants
The reactant used up first in a chemical
reaction is called the limiting reactant. Excess
reactants are present in quantities greater than
necessary to react with the quantity of thelimiting reactant.
A + B --- C + D
Given the amounts of A and B, which is the
limiting reactant?
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Urea is prepared by reacting ammonia with
carbon dioxide:
2NH3(g) + CO2(g) -- (NH2)2CO(aq) + H2O(l)
In one process, 637.2 g of NH3 are allowed toreact with 1142 g of CO2.
(a) Which is the limiting reactant?
(b) How much urea (in grams) is produced?
(c) How much of the excess reactant (in grams)
is left at the end of the reaction?
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Strategy
• Convert mass of each reactant to moles
• Calculate the amount of product formed
from the each of the reactants.
• The reactant the produces the less
amount is the limiting reactant.
1. The reaction between aluminum and iron (III) oxide can generate
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( ) gtemperatures around 3000⁰C and is used in welding metals:
2Al + Fe2O3 -- Al2O3 + 2Fe
In one process, 124 g of Al are reacted with 601 g of ferric oxide.
(a)Which is the limiting reactant?
(b)How much Al2O3 (in grams) is produced?
(c)How much of the excess reactant (in grams) is left at the end of
the reaction?
2. Titanium is a strong & light metal used in rockets & aircrafts. It is
prepared by the reaction between titanium (IV) chloride with moltenmagnesium at around 1000⁰C: TiCl4 + 2Mg -- Ti + 2MgCl2
In a certain industrial operation, 3.54 x 107g of TiCl4 are reacted with
1.13 x 107 g of magnesium.(a)Which is the limiting reactant?
(b)How much Ti (in grams) is produced?
(c)How much of the excess reactant (in grams) is left at the end of