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```					Unit 8: Chemical Formulas
Chapter 11
Chemistry 1K
Cypress Creek High School
The Mole
•   In the same way a dozen is
worth 12, a mole is worth
6.02 x 1023
•   The mole is a large number
because particles are so small, it
takes many of them make up an
amount we can see and
understand
•   We can use it to count
anything!
Mole

1 gross cars = 144 cars
1 mole of cars = 6.02 X 1023 cars

1 ream Al particles = 500 Al particles
1 mole of Al particles = 6.02 X 1023 particles
Mole Conversions

23
22.4 L                         6.02 X 10 atoms
formula units,
1 MOLE         molecules

Molecular mass in grams
The Mass of a Mole
•   How do we “count” a mole?
•   By measuring mass (just like how they
count aluminum cans for recycling)
•   You can find atomic masses on the
periodic table.
• 1 mole of Mg atoms           =     24.3 g
• 1 mole of Cu atoms           =     63.5 g
Moles to Mass
•   Calculate the mass of 0.625
moles of calcium.
•   According to the periodic
table, the atomic mass of
calcium is 40.078 amu, so the
molar mass (mass of 1 mole) of
calcium is 40.078 g.
X g calcium                      40.078 g calcium
=
0.625 mol calcium                   1 mol calcium
= 25.0 g calcium
Mass to Particles
•   Calculate the number of atoms in
•   First determine how many moles are in
X moles Pb               1 mole Pb
=
4.77 g Pb              2.7.2 g Pb
= 0.0230 mol Pb

•   Now use a second conversion factor to
convert moles to number of particles.
X atoms Pb                   6.02 x 1023 atoms Pb
=
0.0230 mol Pb                         1 mol Pb
= 1.38 x 1022 atom Pb
Chemical Formula
•   A Chemical Formula is the abbreviated way of
naming a compound using numbers and
element symbols.
•   Subscripts denote number of each element in a
compound
• Ex: H2O has 2 hydrogens and 1 oxygen
• Ex: Al2(SO4)3 has 2 aluminums, 3 sulfurs, and 12 oxygens

•   Coefficients act as a scalar (also called a multiplier)
• Ex: 6H2O has 12 hydrogens and 6 oxygens
• Ex: 3Al2(SO4)3 has 6 aluminums, 9 sulfurs, and 36 oxygens
Moles of Compounds
•   If the substance is a molecular compound,
such as ammonia (NH3), a mole is 6.02 x
1023 molecules of ammonia.
•   A mole of a compound contains as many
moles of each element as are indicated by
the subscripts in the formula for the
compound.
•   1 mole of NH3
•1  mole of N
• 3 moles of H
Molar Mass of a Compound
•   The molar mass of a compound is the mass
of a mole of the representative particles of
the compound.
•   In the case of NH3.
Molar mass of NH3 = molar mass of N + 3(molar mass of H)
Molar mass of NH3 = 14.007 g + 3(1.008 g) = 17.031 g/mol
•   You can use the molar mass of a
compound to convert between mass and
moles.
Molar Mass Practice
•   Find the molar mass for:
• SO3
• Na2SO4

• Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake of
serotonin by the brain.
Percent Composition
•   Percent Composition is the percentage by
mass of each element in a compound
•   The percent composition is found by using the
following formula:

Total mass of element
% Mass                         100
molar mass of compound
Calculating Percent Composition
•   What is the percent of each element in sodium sulfite,
Na2SO3?
• 2 mole Na x 22.990 g/mol Na =        45.980 g Na
• 1 mole S x 32.066 g/mol S       =    32.066 g S
• 3 moles O x 15.999 g/mol O = + 47.997 g O
Molar Mass of Na2SO3 = 126.043 g
•   Sodium
• % = (45.980 g / 126.043 g) x 100 = 36.486 % Na
•   Sulfur
• % = (32.066 g / 126.043 g) x 100 = 25.441 % S
•   Oxygen
• % = (47.997 g / 126.043 g) x 100 = 38.080 % O
% Composition Practice

•   What is the percent of
C & H in C2H6?

•   What is the percent of
each element in
Na2SO3?
Law of Definite Proportions
•   Joseph L. Proust was born in
1754 in France and worked
as an apothecary.
carbonate and compared it to
natural copper carbonate.
•   He showed that each had the same
proportion of weights between the
three elements involved.
•   This was called the Law of
Definite Proporitons
Types of Formulas
•   There are 2 types of formulas that can be used
to describe compounds.
•   Empirical Formula
•   Molecular Formula
•   An Empirical Formula is the LOWEST whole
number ratio of the elements in a compound.
•   The empirical formula for C6H6 is CH
•   A molecular formula tells the exact number of
atoms of each element in a molecule or
formula unit of a compound.
•   The actual formula
Examples of Empirical & Molecular
Formulas
•   Example 1: Ethene
• Empirical: CH2
• Molecular: C2H4

•   Example 2: Caffeine
• Empirical: C4N2OH5
• Molecular: C8N4O2H10
Back = Carbon
Blue = Nitrogen
Red = Oxygen
Grey = Hydrogen
Empirical Formulas
•   An Empirical Formula is the LOWEST
whole number ratio of the elements in a
compound.
•   The empirical formula for C6H6 is CH
• C6H6   is a completely different chemical than CH!
•   The subscript cannot be a fraction
•   Ex: There is no CO2H1.5 - it must be C2O4H3
•   What is the empirical formula for each?
• C2H6
• C6H12O6
Empirical Formula Calculations
•   There are 3 types of empirical calculations:
•   Mole Ratio
•   Grams
•   Percent
Empirical Formula from Mole Ratios
•   To calculate the empirical formula from mole
1) Divide each mole quantity by the smallest
mole quantity
2) Assign ratios as subscripts

3) If any of the ratios are not even (ie: .5, 1.25,
1.3), multiply them all by the lowest common
multiple to achieve whole numbers then place
these numbers as the subscripts.
Calculating Empirical Formula from
Mole Ratio
•   Find the empirical formula for a
compound containing 2 mole carbon and
6 mole hydrogen.          LOWEST

•C= 2/2 = 1
•H = 6/2 = 3

The formula is = CH3
Empirical Formula from Grams
•   To calculate the empirical formula from grams, follow
these steps:
1) Convert the grams to moles (divide grams by molar
mass).
2) Divide each mole quantity by the smallest mole
quantity.
3) Assign ratios as subscripts.

4) If any of the ratios are not even (ie: .5, 1.25, 1.3),
multiply them all by the lowest common multiple to
achieve whole numbers then place these numbers as
the subscripts.
Calculating Empirical Formula from
Grams
•Calculatethe empirical formula of a compound
that contains 13.5 grams of calcium,4.05 grams of
carbon, and 16.2 grams of oxygen.
calcium               carbon            oxygen
13.5 g = 0.337        4.05 g  = 0.337 16.2 g   = 1.01
40.08 g/mol mol       12.011 g/mol mol 15.999 g/mol mol
LOWEST
0.337 = 1             0.337             1.01 = 3
=1
0.337                 0.337             0.337
empirical formula = CaCO3
Empirical Formula from Percent Composition
•   To calculate the empirical formula from percent composition,
1) Change % sign to grams. You assume 100 g of the
compound.
2) Convert the grams to moles (Divide grams by molar
mass).
3) Divide each mole quantity by the smallest mole quantity.

4) Assign ratios as subscripts.

5) If any of the ratios are not even (ie: .5, 1.25, 1.3), multiply
them all by the lowest common multiple to achieve
whole numbers then place these numbers as the
subscripts.
Calculating Empirical Formula from % Composition
•   Calculate the empirical formula of a
compound containing 18.8% nickel and 81.2%
iodine.

nickel                     iodine
= 0.320                   = 0.640
18.8 g                       81.2 g
mol                       mol
58.69 g/mol     LOWEST   126.905 g/mol
0.320 = 1                  0.640 = 2
0.320                      0.320
empirical formula = NiI2
Practice Empirical Formula
•   Find the empirical formula given 65.2%
Sc and 34.8 % O
Formula for a Hydrate
•   Many compounds have a certain
number of water molecules that are
associated with them. These are called
hydrates.
•   In the formula for a hydrate the number of
water molecules associated with 1 unit of
the compound is written following a dot.
•   CaCl2  3H2O
Determining the Formula of a Hydrate
2)   Heat the hydrate to remove the water.
3)   Subtract the mass of the compound
without water from the original mass.
1)   This is the mass of water.
4)   Covert the masses to moles by dividing
by the molar mass of each compound.
5)   Calculate the mole ratio by dividing by
the smallest number of moles.
Calculating Molecular Formula
•   A hydrate of BaCl2  XH2O has a mass of 5.00g. After
heating until all the water is lost, the compound has a
mass of 4.26g. What is the formula of the hydrate?
• Mass before heating     5.00g
• Mass after heating      4.26g
• Mass of water        = 0.74g
BaCl2                              H2O
4.26 g                         0.74 g
208.23 g/mol                   18.02 g/mol
0.0205 mol = 1                0.041 mol     = 2
0.0205                        0.0205
•   Formula of Hydrate = BaCl2  2H2O
Practice Molecular Formula
•    A mass of 2.50g of blue, hydrated
CuSO4  XH2O is placed in a crucible and
heated. After heating, 1.59g of white
anhydrous CuSO4 remains. What is the
formula for the hydrate?
•   CuSO4  5H2O
Molecular Formulas
•   For many compounds, the empirical
formula is not the true formula.
•   A molecular formula tells the exact number
of atoms of each element in a molecule or
formula unit of a compound.
• The   actual formula
•   The molecular formula for a compound is
always a whole-number multiple of the
empirical formula.
Determining Molecular Formulas
1)   Determine the empirical formula
2)   Find the empirical formula mass (you
are given the molecular formula mass)
3)   Divide the molecular formula mass by
the empirical formula mass. This is the
scalar (multiplier).
4)   Multiply the subscripts in the empirical
formula by the multiplier.
Calculating Molecular Formula
•   A compound has an empirical formula of Sc2O3.
The molecular mass is 414 g/mol. What is the
molecular formula?
• Empirical Formula = Sc2O3
• Empirical Formula Mass =
•2  mol Sc x 44.956 g/mol Sc =        89.912 g Sc
• 3 mol O x 15.999 g/mol O   =       + 47.997 g O
molar mass =       137.909 g Sc2O3
•   Find the multiplier…
• 414   g / 137.909 g = 3 (the multiplier is 3)
•   Molecular Formula = Sc6O9
Practice Molecular Formula
•   A molecule has 85.6% carbon and
14.5% hydrogen. If the molecule has a
mass of 42.1 grams, what is the
molecular formula?
End of Unit 8
Be Prepared for Unit 8 Test.

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