Download File - Evergreen Tutor Zone

A spectrometer (from G 10) is used to measure the
masses of small particles like atoms.
It has been established that the mass of an
H atom is = 1,66 x 10-27 kg. i.e.
0, 00000000000000000000000000166 kg
This really a very small number!
Spectrometer
1
Basic Concepts
Atomic and mass numbers
• Atomic number (Z):
– number of protons (also equal to number ē).
• Mass number:
– number of nucleons (particles in nucleus).
• Element:
– a substance of which all atoms have the
same atomic number.
• Isotope:
– different atoms of the same element with
different mass numbers. Isotopes of hydrogen
35
17
Cl
37
17
Cl
Are isotopes of chlorine.
2
Basic Concepts
• Relative atomic mass Ar:
– A number that compares the average
mass of an atom with the mass of the
C-12 atom taken to be 12 units.
– Take note that it is a rough, average
idea of the number of nucleons an
atom has. Because we are looking at
a mixture of isotopes, we get fractions
though. No real atom of Carbon has
12,011 nucleons for instance.
Relative atomic mass
3
The mole
• The mole
– The number of elementary particles that is
equal to the number of carbon atoms in 12g
of carbon-12.
Mole in chemistry
i.e. there are 6,02 x 1023 atoms of carbon in
12 g (RAM) of carbon.
1 mole of any element thus contains
6,02 x 1023 atoms of that element
Also: the RAM of a substance expressed in
grams will contain 1 mole of particles of that 4
substance.
Calculating RAM for Cl
35
Mass of 17
Cl
= 75,4 x 35 = 2639
Mass of 37 Cl = 24,6 x 37 = 9102
17
Mass of 100 Cl atoms: = 3549,2
∴RAM of Cl = 3549,2/100 = 35,492
This is the average mass of a Cl atom –
as they occur in nature (Cl only has 2
isotopes)
5
Relative molecular/formula mass
Relative formula mass:
Mass of a molecule or formula unit of a
compound, calculated by adding relative
atomic masses (RAM) together in their ratios.
Mr [Ca(NO3)2] = 40 +(2x14) +(6x16)
= 164
i.e. Mass of one formula unit is 164 – relative to C-12.
C-12 is the international standard used today & masses
of all other atoms are determined from this particular
isotope of C.
6
Relative molecular/formula mass
Now find the relative molecular mass of each of the
following elements:
HCl, H2O, CO2, H2SO4, & K2Cr2O7
Answers: 36,5, 18, 44, 98, & 294 respectively.
Ensure that you understand how each of these
values is obtained.
Relative atomic mass
7
Counting in Chemistry
You have come across the words: pairs, dozens &
century.
In chemistry we use huge numbers and we count
numbers of atoms, molecules, ions, protons &
electrons etc in moles.
1 mole = 6,02 x 1023 particles. i.e.
602 000 000 000 000 000 000 000.0 particles
This really huge!
The mole is also called the
Avogadro number or Avogadro
constant.
There are only about
1 x 1014 people on
earth!
8
The mole
Consider each of the following masses:
1gH
16 g of O 35,5 g of Cl 36,5 g HCl
Each contain 1 mole of
atoms
These are the relative
atomic masses expressed
in gram.
44 g CO2
Both contain 1 mole
of molecules
These are the relative
molecular masses
expressed in gram.
Each contain 6,02 x 1023 particles – either
atoms or molecules.
9
The mole
1 g of H contains 6,02 x 1023 atoms or
1 g of H contains ½ (6,02 x 1023 ) molecules
(Since H is diatomic: H2)
2 g of H contains 6,02 x 1023 molecules or
2 g of H contains 2(6,02 x 1023) atoms
Find the following:
The number of molecules in 16 g of oxygen.
The number of molecules in 142 g of chlorine.
The number of atoms in 64 g of oxygen
10
Molar mass Mr
Whereas the mass of 1 molecule is called the
relative molecular mass and the mass of 1
formula unit is called the relative formula mass,
the mass of 1 mole of molecules is called the
molar mass and is expressed in g∙mol-1.
Consider CO2.
12 + 2(16) = 44 = mass of 1 molecule of CO2.
But 44 g∙mol-1 is the molar mass (Mr) = mass of 1
mole of CO2 molecules i.e. 6,02 x 1023 molecules.
Molar mass
11
Finding the number of moles
• One mole of any substance is the relative atomic
mass or the relative formula mass expressed in
g.mol-1.
• One mol of Na is Mr = 23 g.mol-1, while one mol of
water is Mr = 18 g.mol-1
• The number of moles is determined as follows:
• n= m/Mr
• Where n = no. of moles, m = mass, Mr = molar mass
• How many moles are there in 40 g of carbon?
• n = m/Mr n = 40 g/ 12 g.mol-1 = 3.33 mol of C.
12
Molar concentration
Concentrations of solutions can easily be
established by means of the following
equations:
m
m
n
n
c
c
Mr
M rV
V
Where n = no. of moles, m = mass, Mr = molar
mass, c = concentration & V = volume solution.
Solutions with specific concentrations
are made up in volumetric flasks. The
unit is mol∙dm-3.
13
Molar gas volume
This is the volume of 1 mole of any gas at STP.
Using densities, we can find the volume of any gas
at STP.
32 g
3
m
=
22,4
dm
V = D = 1,43 g.dm-3
For oxygen:
Using the masses and densities of other gases, we
find that 1 mole of any/every gas occupies 22,4 dm3
at STP.
This value is known as molar gas volume – the
volume of 1 mole of every gas at STP.
Molar volumes
14
Molar gas volume
1 mole of every gas
occupies 22,4 dm3 at
STP.
1mole
H2
1mole
CH4
22,4 dm3
at STP.
1mole
NH3
1mole
O2
1mole
N2
This phenomenon
only applies to gases
– not solids & liquids.
Experiment to find molar gas volume
Avogadro’s hypothesis:
Equal volumes of gases under the same conditions of
temperature & pressure have equal numbers of
molecules i.e.
15
10 cm3 of H2 & O2 have the same number of molecules at STP.
Avogadro’s hypothesis
He said: equal volumes of different gases (at the
same temp & pressure) contain the same number
of molecules.
The ratio of masses of equal volumes of gases
must be the same ratio as their relative molecular
masses.
You can thus also determine the relative molecular
mass of a gas by using volumes & the known
relative molecular mass of another gas.
Avogadro's law
16
Calculation
Fertilizers need to be applied to lawns &
ground where crops are to be grown.
Most fertilizers have a proportion of some
of nitrogen, phosphorus, potassium,
calcium, magnesium, hydrogen, chlorine
& oxygen.
Common fertilizers are: NH4NO3, Ca(H2PO4)2 & KCl
We need to be able to correctly calculate the
fertilizer required for certain crops in certain soils
lacking particular elements.
17
Percentage composition
Find the percentage composition of potassium
chloride.
Write down formula & find
formula mass
KCl
= 39 + 35,5
= 74,5 g.mol-1
% K in KCl is:
39 X 100 = 52,3%
74,5
% Cl in KCl is:
35,5 X 100 = 47,7%
74,5
Percentage composition
We can use this method to determine which
fertilizers to use and what % of each element they 18
should contain.
Formula of compound from %
If you know the percentages of the elements
present in a compound, you can determine its
formula or empirical formula.
If a certain compound contains 11,1% H and
88,9% O, find the formula of the compound.
If you convert to gram: 11,1 g of H & 88,9 g of O
No. of mole of H = 11,1 = 11,1 mole
No. of mole of O =
1
88,9 = 5,55 mole
16
Ratio of H : O is 11,1 : 5,55 i.e. 2: 1
∴Formula of compound is H2O
Empirical formula
19
A compound consists of 26,67 % carbon,
71,11 % oxygen and 2,22 % hydrogen by
mass. If the relative formula mass of the
compound is 90, calculate its true formula.
20
Calculate the empirical formula first:
n (C) =
26,67/
n (O) =
n (H) =
71,11/
2,22/
mole ratio:
12
16
1
= 2,22 mol
= 4,44 mol
= 2,22 mol
2,2 : 4,44 : 2,22
1 : 2 : 1
empirical formula: CO2H
21
Empirical & true formula
Calculate the relative empirical formula mass:
Mr (CO2H) = 12 + 32 + 1 = 45
Determine the ratio of the true formula mass
to the empirical formula mass:
Mr (given) : Mr (CO2H) Mass of true
90
: 45
2
:
1
Trueisformula:
True formula
∴ C2O4HC2 2O4H2
formula is
twice that of
empirical
formula. 22
Chemical equations
Chemical equations are an integral part of Chemistry
and give us the following information about the
reacting substances:
Reactants reacting
Products formed
Numbers of moles of reactants & products involved
Proportional of masses of reactants & products
Proportional volume of gases that react or are formed
The physical sate (g, ℓ, aq, or s) of reactants &
products
23
Chemical equations
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g) + energy
The above balanced equation tells us:
1 atom of Mg reacts with 2 molecules of HCl to form
1 molecule of MgCl2 & 1 molecule of H2. OR
1 mole of solid Mg reacts with 2 mole of HCl in
solution (water), to produce 1 mole of MgCl2 in
solution & 1 mole of H gas in proportion.
24 g of Mg reacts with 2(1 + 35,5) = 73 g of HCl to
form (24 + 71) = 95 g of MgCl2 & 2 g of H2.
Stoichiometry is the study of the amounts of
substances that react in chemical reactions.
24
Mass-mass chemical reactions
Crucible
Mg ribbon
Gauze
Find mass of crucible without Mg,
with Mg & after reaction
completed after heating.
Tripod
Bunsen
2Mg + O2
2MgO
2(24) of Mg reacts with (16)2 O2 to form 2(24+16) MgO
∴ 48 g Mg + 32 g O2 forms 80 g of MgO
∴ 2g Mg reacts with 2 x 32 g O2 to form 2 X 80 g of MgO
48
48
2g Mg reacts with 1,33 g of O2 to form 3,33 g of MgO
Relate these to the numbers you got.
25
Mass - mass calculation
Mass-volume chemical reactions
If you are required to find the volume of a gas
formed, first find the number of moles of gas formed
& then use the equation pV = nRT to find the
Mass - volume calculations
volume of the gas.
2,4 g of Mg reacts completely to form MgCl2 & H2. Find
the volume of the gas formed at 20 0C & 100 kPa.
Mg + 2HCl
MgCl2 + H2
24 g of Mg forms
∴2,4 g of Mg forms
2 g of H2
0.2 g of H2 (0,1 mole)
Now use pV = nRT to find the volume of the gas.
26
Mass-volume chemical reactions
pV = nRT
100 x 103 x V = 0,1 x 8,31 293
∴V = 0,00243 m3 = 2,43 dm3
If the gas is at STP then you can use molar gas volume
of 22,4 dm3 to find the volume of the gas.
If this problem had been at STP then:
1 mole of H2 at STP occupies 22,4 dm3
∴ 0,1 mole at STP occupies 0,1 x 22,4 = 2,24 dm3
The ChemCollective:
Stoichiometry Applet
27
25 cm3 of hydrogen and 20 cm3 of oxygen are
mixed and an electric spark passed through the
mixture. Steam is formed.
Calculate the volume of the remaining gases if
all volumes are measured at the same
temperature and pressure and the temperature
is above 1000 C.
Volume - volume calculations
28
2H2(g)
2 vol
+
O2 (g)
1 vol

2H2O (g)
2 vol
25 cm3 reacts with 12,5 cm3 and yields 25 cm3
Volume of O2 left = 20 - 12,5 = 7,5 cm3
Total volume of gas left = 7,5 + 25 = 32,5 cm3
Here the 25 cm3 of H2 is called the limiting
reagent, since it is consumed completely, leaving
29
excess O2 + H2O product.
Certain chemical reactions are explosive and
produce a very large number of gaseous product
molecules.
Mining: 2NH4NO3 2N2(g) + 4H2O(g) + O2(g)
Car engine:2C8H18 +25O2 16CO2(g) +18H2O(g)
2 STROKE PETROL ENGINE
Sodium azide in car airbags:
2NaN3(s)  2Na(s) + 3N2(g)
Airbag operation
30