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Formulas, Compounds,
and The Mole
Calculations
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Counting Atoms
Mg burns in air (O2) toproduce whitemagnesium oxide, MgO.
How can we figure out
how much oxide isproduced from a givenmass of Mg?
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Counting Atoms
Chemistry is a quantitative sciencewe need a counting unit.
The MOLE 1 mole is the amount of substance that
contains as many particles (atoms,molecules) as there are in 12.0 g of12C.
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Particles in a Mole
6.0221367 x 1023
Avogadros Number
Amedeo Avogadro
1776-1856
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Particles in a Mole
6.0221367 x 1023
Avogadros Number
Amedeo Avogadro1776-1856
There is Avogadros number of
particles in a mole of any substance.
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Molar Mass1 mol of12C
= 12.00 g of C= 6.022 x 1023 atoms
of C
12.00 g of12C is its
MOLAR MASS
Taking into account allof the isotopes of C,
the molar mass of C is
12.011 g/mol
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Molar Mass
1 mol of12C = 12.00 g of C= 6.022 x 1023 atoms of C
12.00 g of12C is its MOLAR MASS
Taking into account all of the isotopes ofC, the molar mass of C is 12.011 g/mol
13
Al
26.9815
atomic number
symbol
atomic weight
Find molar
mass fromperiodic
table
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PROBLEM: How many molesare represented by 0.200 g of
Mg?
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PROBLEM: How many molesare represented by 0.200 g of
Mg?
Mg has a molar mass of 24.3050 g/mol.
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PROBLEM: How many molesare represented by 0.200 g of
Mg?
Mg has a molar mass of 24.3050 g/mol.
0.200 g 1 mol
24.31 g= 8.23 x 10-3 mol
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PROBLEM: How many molesare represented by 0.200 g of
Mg?
Mg has a molar mass of 24.3050 g/mol.
How many atoms in this piece of Mg?
0.200 g 1 mol
24.31 g= 8.23 x 10-3 mol
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PROBLEM: How many molesare represented by 0.200 g of
Mg?
Mg has a molar mass of 24.3050 g/mol.
How many atoms in this piece of Mg?
0.200 g 1 mol
24.31 g= 8.23 x 10-3 mol
8.23 x 10-3 mol 6.022 x 1023 atoms1 mol
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PROBLEM: How many molesare represented by 0.200 g of
Mg?
Mg has a molar mass of 24.3050 g/mol.
How many atoms in this piece of Mg?
= 4.95 x 1021 atoms Mg
0.200 g 1 mol
24.31 g= 8.23 x 10-3 mol
8.23 x 10-3 mol 6.022 x 1023 atoms1 mol
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Device for Remembering
How to Do Mass/MoleCalculations
mass
# moles MW
mass
MW# moles# moles =
mass # moles MW=
MW =mass
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MOLECULAR WEIGHT
AND MOLAR MASSMolecular weight is the sum of the
atomic weights of all atoms in the
molecule.
Molar mass = molecular weight in
grams
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What is the molar massof ethanol, C2H6O?
1 mol contains
2 mol C (12.01 g C/1 mol) = 24.02 g C
6 mol H (1.01 g H/1 mol) = 6.06 g H
1 mol O (16.00 g O/1 mol) = 16.00 g O
TOTAL = molar mass = 46.08 g/mol
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Formula = C8H9NO2
Molar mass = 151.2 g/mol
Tylenol
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How many moles of alcohol arethere in a standard can of beer if
there are 21.3 g of C2H6O?
(a) Molar mass of C2H6O = 46.08 g/mol
(b) Calc. moles of alcohol
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How many moles of alcohol arethere in a standard can of beer if
there are 21.3 g of C2H6O?
(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 many moleculesof alcoholare there in a standard can of
beer if there are 21.3 g of C2H6O?
We know there are 0.462 mol of C2H6O.
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How many moleculesof alcoholare there in a standard can of
beer if there are 21.3 g of C2H6O?
= 2.78 x 1023 molecules
We know there are 0.462 mol of C2H6O.
0.462 mol 6.022 x 1023 molecules
1 mol
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How many atoms of C are there ina standard can of beer if there are
21.3 g of C2H6O?
We know there are 2.78 x 1023 molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
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How many atoms of C are there ina standard can of beer if there are
21.3 g of C2H6O?
= 5.57 x 1023 C atoms
We know 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
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Empirical and Molecular
FormulasA pure compound always consists of the
same elements combined in the sameproportions by weight.
Therefore, we can express molecular
composition as PERCENT BYWEIGHT
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Empirical and Molecular
FormulasA pure compound always consists of the
same elements combined in the sameproportions by weight.
Therefore, we can express molecularcomposition as PERCENT BYWEIGHT
Ethanol, C2H6O
52.13% C, 13.15% H,
34.72% O
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Percent CompositionConsider some of the family of nitrogen-
oxygen compounds:
NO2, nitrogen dioxide and closelyrelated, NO, nitrogen monoxide (or
nitric oxide)
Structure of NO2
Chemistry of NO,
nitrogen monoxide
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Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of O?
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Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt. % N =14.0 g N
46.0 g NO2 100% = 30.4 %
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Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt. % O 2 (16 .0 g O per mole )46 .0 g
x 100% 69.6%
Wt. % N =14.0 g N
46.0 g NO2 100% = 30.4 %
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Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt. % O 2 (16 .0 g O per mole )46 .0 g x 100% 69.6%
Wt. % N =14.0 g N
46.0 g NO2 100% = 30.4 %
What are the weight percentages of
N and O in NO?
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Determining Formulas
In chemical analysis we determine the % byweight of each element in a given amount ofpure compound and derive the
EMPIRICALorSIMPLESTformula.
PROBLEM: A compound of B and H is81.10% B. What is its empirical
formula?
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A compound of B and H is 81.10% B.What is its empirical formula?
Because it contains only B and H, it
must contain 18.90% H.
In 100.0 g of the compound there are
81.10 g of B and 18.90 g of H.
Calculate the number of moles of each
constitutent.
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A compound of B and H is 81.10% B.What is its empirical formula?
Calculate the number of moles of each
element in 100.0 g of sample.
81.10 g B
1 mol
10.81 g = 7.502 mol
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A compound of B and H is 81.10% B.What is its empirical formula?
Calculate the number of moles of each
element in 100.0 g of sample.
81.10 g B
1 mol
10.81 g = 7.502 mol
18.90 g H 1 mol
1.008 g
= 18.75 mol
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A compound of B and H is 81.10% B.What is its empirical formula?
Now, recognize that atoms combine in theratio of small whole numbers.
1 atom B + 3 atoms H --> 1 molecule BH3
or
1 mol B atoms + 3 mol H atoms --->
1 mol BH3 molecules
Find the ratio of moles of elements in the
compound.
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A compound of B and H is 81.10% B.What is its empirical formula?
Take the ratio of moles of B and H. Always
divide by the smaller number.
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A compound of B and H is 81.10% B.What is its empirical formula?
Take the ratio of moles of B and H. Always
divide by the smaller number.
18.75 mol H
7.502 mol B =
2.499 mol H
1.000 mol B =
2.5 mol
1.0 mol B
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A compound of B and H is 81.10% B.What is its empirical formula?
But we need a whole number ratio.
2.5 mol H/1.0 mol B = 5 mol H to 2 mol B
EMPIRICAL FORMULA = B2H5
Take the ratio of moles of B and H. Always
divide by the smaller number.
18.75 mol H
7.502 mol B =
2.499 mol H
1.000 mol B =
2.5 mol
1.0 mol B
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A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
itsmolecular formula?Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
B2H6
B2H6 is one example of this class of compounds.
A d f B d H i 81 10% B
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A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula?
We need to do an EXPERIMENT to findthe MOLAR MASS.
Here experiment gives 53.3 g/molCompare with the mass of B2H5
= 26.66 g/unit
Find the ratio of these masses.
A d f B d H i 81 10% B
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A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula?
We need to do an EXPERIMENT to find
the MOLAR MASS.
Here experiment gives 53.3 g/mol
The mass of B2H5 = 26.66 g/unit
Find the ratio of these masses.
53.3 g/mol
26.66 g/unit of B2H5=
2 units of B2H
51 mol
A compound of B and H is 81 10% B
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A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula?
Molecular formula = B4H10
We need to do an EXPERIMENT to find
the MOLAR MASS.
Here experiment gives 53.3 g/mol
The mass of B2H5 = 26.66 g/unit
Find the ratio of these masses.
53.3 g/mol
26.66 g/unit of B2H5=
2 units of B2H
51 mol
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Determine the formula of acompound of Sn and I using the
following data.
Reaction of Sn and I2 is done using excessSn.
Mass of Sn in the beginning = 1.056 g Mass of iodine (I2) used
= 1.947 g
Mass of Sn remaining= 0.601 g
See p. 139
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Find the mass of Sn that combined with1.947 g I2.
Mass of Sn initially = 1.056 g
Mass of Sn recovered = 0.601 gMass of Sn used = 0.455 g
Find moles of Sn used:
Tin and Iodine Compound
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Find the mass of Sn that combined with1.947 g I2.
Mass of Sn initially = 1.056 g
Mass of Sn recovered = 0.601 gMass of Sn used = 0.455 g
Find moles of Sn used:
0.455 g Sn 1 mol118.7 g
= 3.83 x 10-3 mol Sn
Tin and Iodine Compound
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Tin and Iodine Compound
Now find the number of moles of I2 thatcombined with 3.83 x 10-3 mol Sn. Massof I2 used was 1.947 g.
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Tin and Iodine Compound
Now find the number of moles of I2 thatcombined with 3.83 x 10-3 mol Sn. Massof I2 used was 1.947 g.
1.947 g I2 1 mol253.81 g
= 7.671 x 10-3 mol I2
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Tin and Iodine Compound
Now find the number of moles of I2 thatcombined with 3.83 x 10-3 mol Sn. Massof I2 used was 1.947 g.
This is equivalent to 2 x 7.671 x 10-3 or
1.534 x 10-2 mol iodine atoms
1.947 g I2 1 mol253.81 g
= 7.671 x 10-3 mol I2
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Tin and Iodine Compound
Now find the ratio of number of moles ofmoles of I and Sn that combined.
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Tin and Iodine Compound
Now find the ratio of number of moles ofmoles of I and Sn that combined.
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn =
4.01 mol I
1.00 mol Sn
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Tin and Iodine Compound
Now find the ratio of number of moles ofmoles of I and Sn that combined.
Empirical formula is SnI4
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn =
4.01 mol I
1.00 mol Sn
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Hydrated Compounds
Compounds which watermolecules are associated with the
ions of the compound Usually determine experimentally
Weigh the hydrated product dryby heat
weigh the anhydrousproduct convert g to moles
find mole ratio