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-- the study of thethe study of the
quantitativequantitative
aspects ofaspects ofchemicalchemical
reactions.reactions.
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Counting AtomsCounting Atoms
Chemistry is a quantitativeChemistry is a quantitative
sciencescience— —we need awe need a
“counting unit.”“counting unit.”
MOLEMOLE
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Lavoisier: The Law of
Conservation of MassEarly 1700’s Lavoisier: Law of
Conservation of Mass
During a chemical change,
matter is neither created nor
destroyed.
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LAW OF CONSERVATION OF MASSLAW OF CONSERVATION OF MASS
In every chemical operation an equalIn every chemical operation an equalquantity of matter exists before and afterquantity of matter exists before and after
the operation.the operation. ThatThat is, the amount ofis, the amount of
matter before a reaction must equal thematter before a reaction must equal the
amount of matter after a reaction. Noamount of matter after a reaction. No
matter is lost.matter is lost.
TheThe total mass of reactants = total mass of productstotal mass of reactants = total mass of products
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LL AW OF AW OF CCONSERVATION OFONSERVATION OF MM ASS ASS
When 0.0976 g of magnesium was heated in air ,
0.1618 g of magnesium oxide (MgO) was
produced. What is the mass of oxygen needed to
produce 0.1618 g MgO?
Using the LCM:Total mass reactants = total mass products
mass of Mg + mass O = mass of MgO
0.0976 g Mg + mass O = 0.1618 g MgO
mass O = 0.1618 g - 0.0976 = 0.0642 g O0.0642 g O
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1799, Proust: Law of Definite
Proportions
A compound always contains the
same elements in certain definite
proportions.
Proust: The Law of
Definite Proportions
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LAW OF DEFINITE PROPORTIONSLAW OF DEFINITE PROPORTIONS Elements combine in specific ratios to form compoundsElements combine in specific ratios to form compounds
Use the Generic equation for percent:Use the Generic equation for percent:
% = ( port ion / total ) 100% = ( port ion / total ) 1001. What is the experimental percent of oxygen in CO2 if42.0 g of carbon reacted completely with 112.0 g ofoxygen?
% O = (mass of O / mass of CO2) 100
% O= [112.0 g O / (42.0 g + 112.0 g) CO2] 100 = 72.7% O72.7% O
2. What is the theoretical percent of aluminum inaluminum oxide?
% Al = (Atomic mass of Al / Formula mass of Al2O3) 100
% Al = (54 amu / 102 amu) 100 = 52.9%52.9%
3. What is the percent composition of sodium chloride?
% Na = 39.3% %% Na = 39.3% % ClCl = 60.7%= 60.7%
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LAW OF MULTIPLE PROPORTIONSLAW OF MULTIPLE PROPORTIONS
When two elements form a series ofcompounds, the masses of the one
element that combine with a fixed
mass of the other element stand to
one another in the ratio of small
integers.
Iron oxide exists in different ratiosIron oxide exists in different ratioswith different propertieswith different properties
FeO and FeFeO and Fe22OO33
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Law of conservation of mass &Law of conservation of mass & LLaw ofaw of ddefiniteefinitepproportionsroportions
When 0.0976 g of magnesium was heated in air,
0.1618 g of magnesium oxide (MgO) wasproduced.
a) what is the percent of Mg in MgO?% Mg = (mass Mg / Mass MgO) 100
= (0.0976g / 0.1618 g) 100 = 60.3 %60.3 %
b) Using only LDP, what mass of oxygen was
needed to combine with the magnesium?% O = 100% MgO - 60.3% Mg = 39.7% O
% O = (mass O / mass MgO) 100
39.7 % = (mass O / 0.1618 g) 100
mass O = 0.397 ( 0.1618 g) = 0.0642 g O0.0642 g O
Same as using the LCM!!
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PRACTICE PROBLEMSPRACTICE PROBLEMSPractic ing Law of conservation of mass:
________1. Aluminum metal combines with oxygen to producealuminum oxide. If 141.0g of aluminum yields 266.7 g of
aluminum oxide, how many grams of oxygen were needed?
________2. Sodium metal reacts with chlor ine gas to produce thesalt, sodium chloride. If 15.0 g of chlorine yields 26.5 g of salt,how much sodium metal is needed?
Practicing the law of definite proportions:
________3. What is the experimental percent of oxygen in a copperoxide if 10.0 g of copper reacted completely wi th 2.52 g ofoxygen?
_______ 4. Based on question #1, what is the experimental percentcomposition of aluminum oxide?
_______ 5. Calculate the theoretical percent composi tion foraluminum chloride and sodium oxide.
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PRACTICE PROBLEMSPRACTICE PROBLEMSPractic ing percents:
________1. Pure gold is too sof t a metal for many uses, so it is
alloyed to give it more mechanical strength. One particular alloy is
made by mixing 29.17 g of gold, 3.81 grams of si lver, and 5.91 g of
copper. What is the percent of gold in this mixture?
________2. If 255 g of a meat sample contains 21.9 g of fat, what
percentage of fat is present?
Using the LAWS:
________3. How many grams of CuO can be obtained from 1.80 g of
copper (use the theoretical percent composi tion)?
4. When aluminum combines with bromine gas, they produce the
substance aluminum bromide, AlBr 3. Write a chemical equationdescribing this reaction.
_______ If 56.88 g of aluminum bromide is formed f rom 5.75 g of
aluminum, how many grams of bromine was needed?
75.0%
8.6%
2.25 g
2Al + 3Br 2 2AlBr 351.13 g
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Connecting Mass to Moles
One of the greatest challenges early chemists
faced was trying to find a way to connect the
mass of a substance to the number of particles
in the sample. It was determined that “elementary particles”
combined in fixed ratios by weight.
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Connecting Mass to Moles
This led Dalton to the “atomic model” of matter
Example: The mass ratio of oxygen to
hydrogen in water is 8:1
This does not tell us how many atoms of eachelement are involved
It could tell us this if we knew the relative mass of
each kind of atom
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Relative Mass
At the time, chemists did not know which was
true and tended to think the latter was more
likely
Example: A bucket of baseballs has fewer ballsthat an identical bucket of golf balls
If this is true in the macroscopic world, why
wouldn’t it be true in the sub-microscopic one?
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Relative Mass
Consider earlier this year when we studied
density: was iron more dense than aluminum
because iron had more particles per given
volume than aluminum or because iron’sindividual particles were more massive than
aluminum’s? Could it be some combination of
both?
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Relative Mass
The truth is, based on the experiments we
conducted earlier in the year, we couldn’t say
which was true.
Dalton did not know what was true during histime either.
Since the mass of individual atoms could not be
determined, a system of atomic masses had to
be determined by comparison.
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Relative Mass
To determine a system of masses by
comparison, one element would have to be
chosen as the basis of comparison for all
others Dalton chose hydrogen and assigned it a mass
of 1.
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Relative Mass
To find the mass of another element like
oxygen:
Compare the masses of equal number of
oxygen and hydrogen atoms ORFind the combining masses of oxygen and
hydrogen in water
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Relative Mass
Dalton thought that the former approach was
invalid because he thought identical volumes of
hydrogen and oxygen gases would have
different numbers of particles He thought the latter was valid but did not take
into account that it is valid ONLY if the ratio of
atomic combination is known
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Reactions of Gases
Research conducted by Gay-Lussac
suggested that equal volumes of gases, at the
same temperature and pressure, contain equal
numbers of particles
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Reactions of Gases
Gay-Lussac noted that gases
appear to react in simple integer
ratios
Example: Two volumes of
hydrogen reacted with one volumeof oxygen to produce two volumes
of water
These findings appeared to
contradict the idea that equalvolumes of gases have equal
numbers of particles
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Reactions of Gases
Why? Well, if water is was H2O,
then two volumes of hydrogen and
one volume of oxygen should
make one volume of water
+ + =
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Avogadro’s Hypothesis
Avogadro assumed
Equal volumes of gases have
equal numbers of molecules
These molecules can be split into
half-molecules during chemical
reactions
That molecules of elemental gases
could contain more than a singleatom
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Avogadro’s Hypothesis
Two volumes of hydrogen react
with one volume of oxygen to
produce two volumes of water
when hydrogen and oxygen can besplit into half-molecules!
+ + = +
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Avogadro’s Hypothesis
If we accept Avogadro’s
Hypothesis, we can compare the
mass of various gases and deduce
the relative mass of the molecules
To do this, we pick a weighable
amount of the lightest element
(how about 1.0 g?) then use mass
ratios to assign atomic masses tothe other elements
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Implications
If two volumes of hydrogencombine with one volume ofoxygen gas, it is reasonable to
assume that two molecules ofhydrogen are reacting with eachmolecule of oxygen
The word chosen to represent thestandard weighable amount of
stuff, the mole, comes from theLatin “mole cula” or little lump
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Counting AtomsCounting Atoms
Chemistry is a quantitativeChemistry is a quantitative
sciencescience— —we need awe need a
“counting unit.”“counting unit.”
1 mole is the amount of substance that1 mole is the amount of substance that
contains as many particles (atoms,contains as many particles (atoms,
molecules) as C atoms in 12.0 g ofmolecules) as C atoms in 12.0 g of1212C.C.
MOLEMOLE
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Molar MassMolar Mass
1 mol of1 mol of 1212CC= 12.00 g of C= 12.00 g of C= 6.022 x 10= 6.022 x 102323 atomsatoms
of Cof C
12.00 g of12.00 g of 1212C is itsC is its
MOLAR MASSMOLAR MASS
Taking into account all ofTaking into account all of
the isotopes of C, thethe isotopes of C, the
molar mass of C ismolar mass of C is
12.011 g/mol12.011 g/mol
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OneOne--mole Amountsmole Amounts
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PROBLEM: What amount of Mg isPROBLEM: What amount of Mg isrepresented by 0.200 g? How manyrepresented by 0.200 g? How many
atoms?atoms?
PROBLEM: What amount of Mg isPROBLEM: What amount of Mg isrepresented by 0.200 g? How manyrepresented by 0.200 g? How many
atoms?atoms?Mg has a molar mass of 24.3050 g/mol.
0.200 g •1 mol
24.31 g = 8.23 x 10
-3 mol
8.23 x 10-3 mol •6.022 x 1023 atoms
1 mol
= 4.95 x 10= 4.95 x 102121 atoms Mgatoms Mg
How many atoms in this piece of Mg?
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MOLECULARMOLECULARWEIGHT ANDWEIGHT AND
MOLAR MASSMOLAR MASSMolecular weightMolecular weight = sum of the= sum of the
atomic weights of all atoms inatomic weights of all atoms in
the molecule.the molecule.
Molar massMolar mass = molecular weight= molecular weight
in grams per mol.in grams per mol.
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What is theWhat is the
molar mass ofmolar mass ofethanol, Cethanol, C22HH66O?O?
1 mol contains1 mol contains
2 moles of C (12.01 g C/1 mol) = 24.02 g C2 moles of C (12.01 g C/1 mol) = 24.02 g C
6 moles of H (1.01 g H/1 mol) = 6.06 g H6 moles of H (1.01 g H/1 mol) = 6.06 g H
1 mol of O (16.00 g O/1 mol) = 16.00 g O1 mol of O (16.00 g O/1 mol) = 16.00 g O
TOTAL =TOTAL = molar mass = 46.08 g/molmolar mass = 46.08 g/mol
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How manyHow many molesmoles of alcohol (Cof alcohol (C22HH66O)O)are there in a “standard” can of beerare there in a “standard” can of beer
if there are 21.3 g of Cif there are 21.3 g of C22HH66O?O?
(a) Molar mass of C2H6O = 46.08 g/mol
(b) Calc. moles of alcohol
21.3 g •1 mol
46.08 g = 0.462 mol
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How manyHow many atoms of Catoms of C are there inare there ina “ standard” can of beer if there are 21.3 ga “ standard” can of beer if there are 21.3 g
of Cof C22HH66O?O?
= 5.57 x 1023 C atoms
There are 2.78 x 1023 molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
2.78 x 1023 molecules •2 C atoms
1 molecule