Download Mole Relationships in chemistry

Mole Relationships in
Chemistry
The Mole Concept and Atomic Masses
O The mole concept and molar mass is historically
based on two laws from Joseph-Louis Proust in 1797
O The Law of Definite Proportions
O This states that compounds always have a definite proportion of
the elements that make it up
O These proportions can be expressed as ratios of atoms, equivalent
mass values, percentage by mass or volumes of gaseous elements
O Example - Water always contains 2 H atoms for every O atom, which is 2 g
H for every 16 g O or 11.1% H and 88.9% O by mass
O The Law of Multiple Proportions
O These laws are based on mass
O John Dalton’s atomic theory used these laws to
formulate his atomic theory
O His atomic theory is based on counts of atoms
O The relationship between mass and count is done
through the mole
Deriving The Mole from
Masses
O The number of atoms in 12.000 grams of 12C can be
calculated:
One atom 12C = 12.000 u = 12 x (1.661 x 10-24 g)
= 1.993 x 10-23 g / atom
# atoms = 12.000 g (1 atom / 1.993 x 10-23 g)
= 6.021 x 1023 atoms
O The number of atoms of any element needed to equal
its atomic mass in grams will always be 6.022 x 1023
atoms
O Called the mole
Why is the Mole So
Important?
O 1 mole of any element = 6.022 x 1023 atoms
O Using moles gives us a practical AND
measurable unit!
O Atoms, ions and molecules are too small to
directly measure in atomic mass units
O So by using the mole, we can relate atoms,
ions and molecules, using an easy to
measure unit -the gram!
The Mole
Molar Mass
O Atoms come in different sizes and masses
O A mole of atoms of one type would have a
different mass than a mole of atoms of another
type
H
1.008 grams / mol
O
16.00 grams / mol
Mo
95.94 grams / mol
Pb
207.2 grams / mol
O We rely on a straight forward system to relate
mass and moles which you learned in first year
chemistry!
Using Molar Mass to Calculate
% Composition (Mass %)
O We can use the molar mass of a compound to determine
the relative masses of the elements in a compound
O Obtained by comparing the MASS OF EACH ELEMENT present
in 1 mole of the compound to the TOTAL MASS of 1 mole of
the compound
O A pure compound should show the same percent mass of
each element consistently
O So given a formula , you should be able to figure out the
percent mass of each element
mass of element in 1 mole of compound
Mass % =
× 100
mass of 1 mole of compound
Elemental Analysis to Determine
Mass Percent of a Compound
O A sample is ‘burned,’ completely converting it to CO2
and H2O
O Each is collected and measured as a weight gain
O By adding other traps elements like oxygen, nitrogen,
sulfur and halogens can also be determined
O2
CO2
trap
furnace
sample
H2O
trap
Elemental Analysis Example
O A compound known to contain only carbon,
hydrogen and nitrogen is examined by
elemental analysis. The following
information is obtained.
Original sample mass
Mass of CO2 collected
Mass of H2O collected
= 0.1156 g
= 0.1638 g
= 0.1676 g
O Determine the % of each element in the
compound
Elemental Analysis
O Mass of carbon
12.01
g
C
= 0.04470 g C
0.1638 g CO2
44.01 g CO2
O Mass of hydrogen
2.016 g H = 0.01875 g H
0.1675 g H2O
18.01 g H2O
O Mass of nitrogen
0.1156 g sample - 0.04470 g C - 0.01875 g H
= 0.05215 g N
Elemental Analysis
O Since we know the total mass of the original sample, we
can calculate the % of each element
0.04470 g
%C=
x 100% = 38.67 %
0.1156 g
% H = 0.01875 g x 100% = 16.22 %
0.1156 g
% N = 0.05215 g x 100% = 45.11 %
0.1156 g
Using % Composition to
Determine Chemical Formulas
O Based on the relative masses of the elements in a
compound, we can directly calculate an empirical
chemical formula
O The lowest whole number ratio of elements in a compound
O CH2
O We need additional information to determine the
molecular formula
O The actual ratio of elements in a compound
O C2H4
O C3H6
O There are instances where empirical and molecular
formula can be the same!
O H2O
Calculating Empirical
Formulas from % Composition
Pretend that you have a 100 gram sample of
the compound
O
O
Change the % to grams
Convert the grams to moles for each element
Write the number of each element as a
subscript in a chemical formula
O
O
O
Keep each number as a decimal at this point!
Divide each subscript by the smallest number
Multiply the result by some integer to get rid of
any fractions
O
O
O
May not be necessary
How to Convert between Empirical
Formulas and Molecular Formulas
O Since the empirical formula is the lowest ratio, the
actual molecule would weigh more
O Molecular formula can always be obtained by
multiplying by some whole number
O To do so, divide the actual molecular molar mass
(usually given in the problem) by the mass of 1 mole
of the empirical formula
O Gives whole number you MUST multiply the
empirical formula by to get the molecular formula
Molecular molar mass
x
Empirical molar mass
Using the Mole to Calculate
Concentration of Solutions
O Concentration is commonly expressed in terms of
molarity
O Defined mathematically as:
moles of 𝒔𝒐𝒍𝒖𝒕𝒆
nsolute
M=
=
volume of 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 (𝐋) Lsolution
O “M” is read as “molar”
O Molarity recognizes that compounds have different
molar masses
O A 1-molar solution of sucrose contains the same
number of molecules as 1-molar solution of ethanol
Other Methods of
Expressing Concentration
O When making different solutions with a specific
molarity, the number of milliliters of solvent
needed to prepare 1 liter of solution will vary
O Sometimes it is necessary to know the exact
proportions of solute to solvent that are in a
particular solution
O Various methods have been devised to express
these proportions
Molality
Molality (m) =
moles solute
kilograms of solvent
mol
=
kg
•Recognizes that the ratio between moles of solute and
kg of solvent can vary
•A 1-molal solution of sucrose contains the same
number of molecules as 1-molal ethanol