Download Exercise 4.1 – Masses of Particles Relative Isotopic Mass Chemists

Exercise 4.1 – Masses of Particles
Relative Isotopic Mass
Chemists as early as John Dalton, two centuries ago, used
experimental data to determine the ____________________________ of
different atoms relative to one another. Dalton estimated relative
atomic weights based on a value of one unit for the
____________________________ atom. In 1961, it was decided that the most
common ____________________________ of 12C would be used as the
reference standard. On this scale, the 12C isotope is given a relative
mass of exactly 12 units.
“The relative ____________________________ mass (Ir) of an isotope is the
mass of an atom of the that isotope relative to the mass of an
____________________________ of 12C taken as 12 units exactly.”
Chlorine has ____________________________ isotopes. These have different
____________________________ as they have different amounts of
____________________________ . Using the 12C isotope as a standard, the
relative isotopic masses of these two isotopes are 34.969 (35Cl) and
36.966 (37Cl). Naturally occurring ____________________________ is made of
75.80% of the lighter isotope and 24.20% of the
____________________________ isotope.
Isotopic Composition of some Common Elements
ELEMENT
ISOTOPES
RELATIVE
ISOTOPIC MASS
ABUNDANCE
(%)
____________________
1
1.008
99.986
2
2.014
0.014
3
3.016
0.001
12 exactly
98.888
H
H
H
Carbon
12
C
13
13.003
1.112
14
14.003
Approx 10
16
O
15.995
99.76
17
O
16.999
0.04
17.999
0.20
106.9
51.8
108.9
48.2
C
C
Oxygen
18
O
_____________________
107
Ag
108
Ag
-10
Relative isotopic masses of elements can be obtained using an
instrument called a mass ____________________________ . This separates
the individual ____________________________ in a sample of the element
and determines the ____________________________ of each isotope. The
information is presented graphically and is known as a mass
____________________________ . In a mass spectrum showing the isotopes of
an element:
–The number of ____________________________ indicates the number of
isotopes,
–The position of each peak on the ____________________________ axis
indicates the relative ____________________________ mass,
–The relative heights of the peaks correspond to the relative
____________________________ of the isotopes.
Relative Atomic Mass
A naturally occurring sample of an element contains the same
____________________________ in the same proportions regardless of its
____________________________ . It is sufficient to imagine that the
hypothetical average atoms of the element are being used. This
average is known as the relative ____________________________ mass. Its
symbol is Ar. The relative atomic mass of an element is actually a
____________________________ average of the relative isotopic masses
because it takes into account the relative amounts or
____________________________ of each isotope in natural samples of the
element.
Atomic Mass Example
1.Imagine taking 100 atoms from a sample of chlorine of chlorine –
there will be 75.80 atoms of 35Cl and 24.20 of 37Cl. Find the relative
atomic mass…
Equation to use:
((relative isotopic mass1 x % abundance1) + (relative isotopic mass2 x
% abundance2)) /100 OR
Ar = ∑(relative isotopic mass x %abundance) / 100
34.969  75.80  36.966  24.20
100
Ar =
Ar =
2650.65  894.58
100

 Ar = 35.452
Equations with Mass Spectrums
A simplified mass spectrum of ____________________________ is shown on
page 58 (of textbook). From the mass spectrum calculate the
percentage abundances of each of its three isotopes.
-The heights of the peaks for the graph are:
-24Mg = 7.9 cm
-25Mg = 1.0 cm
-26Mg = 1.1 cm
•So, the percentage of the 24Mg isotope
=
7.9
7.9 1.0 1.1
100%
= 79%

•The percentage
=
of 25Mg isotope
1.0
100
7.9 1.0 1.1

= 10%
•The percentage of 26Mg isotope
=
1.1
7.9 1.0 1.1
100
=11%
Percentage Abundance Example
Copper has two isotopes. 63Cu has a relative isotopic mass of 62.95
and 65Cu has a relative ____________________________ mass of 64.95. The
relative atomic mass of copper is 63.54. Calculate the percentage
____________________________ of the two isotopes.
1.Let x be the percentage abundance of 63Cu
2.So, 100-x is the percentage abundance of 65Cu
3. Ar(Cu) = ∑(relative isotopic mass x %abundance)
100
So 63.54 =
62.95x  64.95(100  x)
100
6354
= 62.95x + 6495 – 64.95x
6354 = 6495 – 2x
2x = 6495 – 6354
2x = 141
x = 70.5
Relative Molecular Mass
The relative molecular mass (Mr) of a ____________________________ is the
mass of one ____________________________ of that substance
____________________________ to the mass of a 12C atom. The Mr is
calculated by taking the ____________________________ of the relative
atomic masses of the ____________________________ in the molecular
formula.
Calculate the Mr of CO2:
Mr(CO2) = Ar(C) + 2 x Ar (O)
= 12.0 + 2 x 16.0
= 44.0
For non-molecular compounds the term relative
____________________________ mass is used. This is solved exactly the same
way.
Exercise 4.2 – The Mole
A ____________________________ is defined as the amount of substance that
contains the same ____________________________ of specified particles as
there are atoms in 12g of carbon-12. The mole has been defined in a
convenient way:
–1 atom of 12C has a relative ____________________________ mass of 12
exactly
–1 mol of atoms of 12C has a mass of 12g exactly.
The number of particles in 1 mole is given by the symbol ____________.
Because atoms and molecules are so small, NA, needs to be a very
large number if 1 mole of a substance is to be an amount that is
convenient to work with. It is for this reason that
____________________________ have selected 12g of Carbon-12 atoms to
define the mole.
The symbol for the ____________________________ of substance is n. The
unit of ____________________________ for the amount of substance is mol.
NA, also known as ____________________________ Constant, = 6.02 x 1023
Therefore 1 mole of particles equals 6.02 x 1023. If given the amount
of substance, we can calculate the number of particles.
____________________________
N = number of particles
n = amount of substance
NA = Avagadro’s Constant
Find the number of O2 molecules in 2.5 mol of oxygen.
–N = n x NA
–N = 2.5 x 6.02 x 1023
–N = 15.05 x 1023
–N = 1.5 x 1024
You can also use this equation to find the amount of substance, in
mol, by rearranging the equation.
____________________________
Exercise 4.3 - Molar Mass
The mass of 1 mole of a particular element or compound is known as
the molar mass. Because the particles of different
____________________________ or compounds have different
____________________________ , the masses of 1 mol samples of different
particles will be different. In general
–The molar mass of an element is the ____________________________ atomic
mass of the relative atomic mass expressed in
____________________________ .
–The molar mass of a ____________________________ is the relative
molecular or relative ____________________________ mass of the compound
expressed in grams.
The symbol for Molar Mass is ____________________________ and its unit of
measurment is gmol-1.
Calculate the molar mass of table sugar, sucrose (C12H22O11)
1.What are the mass numbers of the individual elements?
C = 12.0
H=1
O = 16
2. Find the molar mass of each element.
C = 12 x 12
H = 22 x 1
O = 11 x 16
3. Add them together.
Mr = (12x12) + (22x1) + (11x16)
Mr = 342g mol-1
____________________________ or ____________________________
m = mass
n = amount of substance
M = Mr = molar mass
This means that is we have the mass of substance we can find the
____________________________ in mol by substituting the
____________________________ in for m and working out the molar
____________________________ . Or, we can find the mass if the amount of
substance is known.
Exercise 4.4 – Formulas of Compounds
Percentage Composition
The values for molar masses of elements in ____________________________
can be used to calculate the ____________________________ composition of a
compound once the formula is known. For example if a company is
producing ____________________________ from alumina (Al2O3), the
management would want to know the mass of aluminium that can be
____________________________ from a given quantity of alumina.
Use the space below to complete an example for percentage
composition.
Formula for percentage composition:
Mass of element in 1 mole of compound
Mass of 1 mole of compound
x 100
Empirical Formulas
The empirical formula of a compound is the formula that gives the
simplest whole ____________________________ ratio, by number of
____________________________ , of each element in the compound.
____________________________ formulas are determined experimentally
usually by determining the ____________________________ of each element
present in a given mass of ____________________________ .
1.Measure the ____________________________ (m) of each element in the
compound.
2.Calculate the ____________________________ in mol (n) of each element in
the compound.
3.Calculate the simplest whole number ____________________________ of
moles of each element in compound.
4.____________________________ formula of compound.
Use the space below to complete an example for empirical formula
Molecular Formulas
While an ____________________________ formula give the simplest whole
number ratio of an element in a compound, a
____________________________ formula gives the ____________________________
number of atoms in one molecule of the compound. The empirical
and molecular formula can be the ____________________________ or
different. The molecular formula is always a whole number
____________________________ of the empirical formula. A molecular
formula can be obtained from the ____________________________ formula if
the ____________________________ mass of a compound is known.
Use the space below to complete an example for finding the molecular
formula.