Download Mole Relationships in chemistry

Mole Relationships in
Chemistry
Atomic masses
O Atoms are composed of protons, neutrons
and electrons
O Almost all of the mass of an atom comes
from the protons and neutrons
O All atoms of the same element will have the
same number of protons
Atomic Masses
O As a reference point, we use the atomic mass
unit (u), which is equal to 1/12th of the mass of
a 12C atom
O One atomic mass unit (u) = 1.66 x 10-24 gram
O Using this relative system, the mass of all other
atoms can be assigned
O Examples
7Li
14N
29Si
= 7.016 004 u
= 14.003 074 01 u
= 28.976 4947 u
Calculating Average Atomic
Mass
O Most elements occur in nature as a mixture of
isotopes
O Thus, atomic masses on periodic table are usually
average values
Average atomic mass
% natural abundance
=
∙ atomic mass1
100
1
% natural abundance
+
∙ atomic mass2 …
100
2
Average Atomic Masses
O One can calculate the average atomic weight of an
element if the abundance of each isotope for that
element is known
O Example
O Silicon exists as a mixture of three isotopes.
Determine it’s average atomic mass based on the
following data.
Isotope
Mass (u)
Abundance
28Si
29Si
30Si
27.9769265
28.9764947
29.9737702
92.23 %
4.67 %
3.10 %
Average Atomic Masses
28Si
92.23
100
29Si
4.67
100
(28.9764947 u) =
1.35 u
30Si
3.10
100
(29.9737702 u) =
0.929 u
(27.9769265 u) = 25.80 u
Average atomic mass for silicon =
28.08 u
Atomic Masses and the Mole
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
The Mole
O 1 mole of any element = 6.022 x 1023 atoms
O Atoms, ions and molecules are too small to
directly measure in atomic mass units
O Using moles gives us a practical unit!
O We can then relate atoms, ions and
molecules, using an easy to measure unit
O 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
Masses of Atoms and Molecules
O Law of Definite Composition 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
O 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
Mass Percent
O How to Calculate Mass %
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 %
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
× 100
=
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
Determining Chemical
Formulas
O Empirical formula
O The lowest whole number ratio of elements in a
compound
O CH2
O Molecular formula
O The actual ratio of elements in a compound
O C2H4
O C3H6
O Empirical and molecular formula can be the
same!
O H2O
Calculating Empirical
Formulas from Mass %
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
molar mass
x
empirical formula 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
Stoichiometry
Chemical reactions
O A chemical change involves the reorganization of
the atoms in one of more substances
O Chemical equation represents this process with
the reactants on the left side of the arrow and
the products on the right side of the arrow
O Chemical reactions follow the Law of Conservation
of Mass so the equation must be balanced!
O Chemical equation for a reaction gives:
O The nature of the reactants and products
O Physical states!
O The relative number of each
Information conveyed by the balanced equation for
the combustion of methane
Stoichiometric calculations
how to calculate amounts of reactants and products
Stoichiometric calculations
with limiting reactant
Theoretical yield and
Percent yield
O Theoretical yield
O The amount of a product formed when the
limiting reactant is completely consumed
O Determined using stoichiometry!
O Percent yield
O Percentage of theoretical yield that is actually
produced in the laboratory
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑥 100 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑