1. Week 3 : Summary Isotopes Cathode rays Mass spectrometer Calculations
Amount of a substance
Mass ,Molar mass, Mole : Equation 1
Avogadro constant – Number of particles :Equation 2
Molecular formula weight
Calculation of molecular formula

2. ISOTOPES

3. ISOTOPES: Almost all atoms have "Isotopes"
Elements with the same number of protons (atomic number) but differing number of neutrons – isotopes are the same elements (atoms) with different masses
isotopes will have slightly different chemical and physical properties due to the difference in mass, which can be very helpful in characterizing substances

4. Isotopes of Carbon C below are three isotopes of carbon, C: # p 6 6 6
            C6               C6                 C6
        # p          6                         6                          6
        # e          6                         6                          6
        # n          6                         7                          8
notice that the sum of number of protons and number of neutrons is equal to the mass number

5. Natural isotopes of C
Carbon has two main isotopes which occur naturally (C-14 is present in very small quantities and is radioactive) These are shown below with their relative percent abundance C amu 98.9% abundant C amu 1.1 % abundant

6. Calculation of mean RAM
To determine the mass of naturally occurring carbon, we calculate the weighted average of the two isotopes by summing the individual mass of each isotope:
mass C = (0.989)( ) + (0.011)( ) = amu

7. because nearly all elements have one or more isotopes, the mass of a naturally occurring element will be a "weighted average" of all the isotopes which occur naturally, for example:
ISOTOPES OF Neon (Ne)

8. Relative atomic mass unit
amu – atomic mass unit – is one such relative mass scale--one amu equals exactly 1/12 the mass of an atom of carbon-12 isotope
rounded masses:       C         12 amu /g                                     Mg       24 amu /g                                     Al         27 amu /g                                      F          19 amu /g

9. Periodic table

10. PERIODIC TABLE

11. Cathode rays

12. Cathode ray tube

13. CRT ....1
Cathode rays are streams of electrons observed in vacuum tubes, i.e. evacuated glass tubes that are equipped with at least two electrodes, a cathode (negative electrode) and an anode (positive electrode).

14. CRT
When the cathode is heated, electrons are emitted from the metal. These are attracted to the anode. If the inner glass walls behind the anode are coated with a fluorescent material called a phosphor the incident electrons induce a glow

17. J J Thompson - CRT

18. First T V s

19. Old T V

20. T V

21. CRT --- T V

22. T V
Almost all T V s in use today rely on a device known as the CRT to display their images.
It is even possible to make a television screen out of thousands of ordinary 60-watt light bulbs!
CRTs are the most common way of displaying images today.

23. T V
Electrons are negative. The anode is positive, so it attracts the electrons pouring off the cathode. In a T V , the stream of electrons is focused by a focusing anode into a tight beam and then accelerated by an accelerating anode. This tight, high-speed beam of electrons flies through the vacuum in the tube and hits the flat screen at the other end of the tube. This screen is coated with phosphor, which glows when struck by the beam.

24. MASS SPECTROMETER

25. MASS SPECTROMETER
masses can be accurately calculated using a mass spectrometer.
All of our atomic masses we will use will come from published values on a periodic table of the elements to be discussed next.

26. Diagram of a Mass spectrometer

27. Mass spectrometer

28. Principle of MS
Atoms can be deflected by magnetic fields - provided the atom is first turned into an ion. Electrically charged particles are affected by a magnetic field although electrically neutral ones aren't.

29. Ionisation
The atom is ionised by knocking one or more electrons off to give a positive ion. This is true even for things which you would normally expect to form negative ions (chlorine, for example) or never form ions at all (argon, for example). Mass spectrometers always work with positive ions.

30. Acceleration
The ions are accelerated so that they all have the same kinetic energy.

31. Deflection
The ions are then deflected by a magnetic field according to their masses. The lighter they are, the more they are deflected.
The amount of deflection also depends on the number of positive charges on the ion - in other words, on how many electrons were knocked off in the first stage. The more the ion is charged, the more it gets deflected.

32. Detection
The beam of ions passing through the machine is detected electrically.

33. Mass spectrometer

34. Mass spectrometer of Neon

35. Isotopes of Cl
Isotopes are atoms of the same element with different numbers of neutrons in the nucleus The RAM of an element is the weighted (to take account of relative abundance) average of the RAM of all of the isotopes of that element. Chlorine has two isotopes with mass numbers 35 and 37 and abundancies 75% and 25%. Calculate RAM.

36. Calculations

37. Calculations: Amount
Equation 1: Amount of a substance= mass over molar mass n = amount of a substance (moles) m = mass (g) M= molar mass (g) (from molecular formula) Definition: 1 mole of any substance contains the same number of particles as there are atoms in 12 g of C-12 (you must learn this definition) Mole counting unit for atoms, molecules, ions

38. Molecules and moles
Number of molecules in any macroscopic sample is huge
Use moles as a unit of quantity
1 mole= x 1023
This is an experimentally measured number named Avogadro`s constant
Remember 6.0 x 1023

39. Avogadro`s number Defined by setting 1 mole 12C = 12 g
I mole of 12C atoms contains 6,0x 1023 and weighs 12 grams (g)

40. Avogadro`s constant
Definition: 1 mole of any substance contains 6.02 x 1023 particles (atoms, molecules, ions) per mole Equation 2: N= n x L N = number of particles L = 6.02 x 1023 (Avogadro constant) n= number of moles

41. Simple mole calculation
A bar of iron weights 10 kgs
How many moles of iron is that?
How many iron atoms

42. Mole problem 10 kgs Fe= ? Moles Fe What we know?
atomic weight of Fe= 56
1 mole Fe= 56 g
1kg= 1000g
Now just calculate the units
Kgs-> gs-> moles

43. Calculating formulae
Remember that we have two kinds of formulae
Empirical formula
Molecular formula

44. Composition and formulas: Example
Nicotine contains 74.0% C, 8.65% H and 17.35% N.
If the molar mass of nicotine is 162, what is the chemical formula of nicotine?
Atomic weights of C, H, and N are 12, 1 and 14 respectively

45. Empirical vs. Molecular formulas
Nicotine shows the difference between these ideas
Empirical formula: simplest possible formula with correct ratios of atoms
Molecular formula: formula showing actual composition of a molecule

46. Formula and percentage composition
Given a chemical formula, its easy to find mass percentage
Find molar mass and the mass from each element
As an example, calculate mass percentage of each element in the previous molecule

47. Composition and formulas
Reverse process is useful in analyzing unknowns
Instruments can tell us the elemental composition of a substance ( elemental analysis)
We can convert information into a molecular formula

48. Problem 1
Calculate the empirical formula from the masses An 18.2 g sample of a hydrated compound contained 4.0 of calcium, 7.1 of chlorine and 7.2g of water. Calculate its empirical formula

49. Solution
list the mass of each component and its molar mass. Although water is a molecule, in the calculation we treat it in the same way as we do atoms.
2.From this information calculate the amount of each substance present using the expression:
3. Calculate the relative amount of each substance by dividing each amount by the smallest one
The relative amounts are in the simple ration 1:2:4
From this result you can see that the empirical formula is CaCl2.4H20 Ca Cl
H2O
Mass/g
4.0
7.1
7.2
Molar mass/g mol -1
40.0
35.5
18.0 Ca Cl
H2O
Amount/ mol
4.0/40.0=0.10
7.1/35.5=0.20
7.2/18.0=0.40 Ca Cl
H20
Amount/ smallest amount= relative amount
0.10/0.10=1.0
0.20/0.10=2.0
0.40/0.10=4.0

50. Problem 2
Calculate empirical formula from percentage composition by mass An organic compound was analyzed and was found to have the following percentage composition by mass: 48.8% carbon, 13.5 % hydrogen and 37.7% nitrogen. Calculate empirical formula of the compound.

51. Empirical formula= C3H10N2
Solution
If we assume the mass of the sample is 100,0 g we can write immediately the mass of each substance: 48.8% carbon, 13.5 % hydrogen and 37.7% nitrogen. Then we set up a table as before. The instructions between each step are omitted this time but you should check your calculation
Empirical formula= C3H10N2 C H N
Mass/ g
48.8
13.5
37.7
Molar mass/g mol-1
12.0
1.00
14.0
Amount/mol
4.07
2.69
Amount/ smallest amount
4.07/2.69=1.51
13.5/2.69=5.02
2.69/2.69=1.00
Simplest ratio of relative amounts 3 10 2

52. Problem 3
Calculate empirical formula from quantitative analysis An 1.0 g sample of a compound was burnt in an excess oxygen. This reaction gave the following products: 2.52g of carbon dioxide and 0.443g of water. Determine the empirical formula of the compound, assuming that it contains carbon, hydrogen and oxygen only.

53. Solution
Calculate the amount of carbon in the sample from the mass of carbon dioxide absorbed.
Amount of CO2=mass/molar mass=2.52 g/44.0 g mol-1= mol
2. Calculate the amount of hydrogen in the sample from the mass of water absorbed
Amount of H20= mass/molar mass=0.443/18.0 gmol-1= mol
3. Now calculate the mass of carbon and mass of hydrogen
Mass of C=amount x molar mass= x 12.0g mol-1=0.688g
Mass of H=amount x molar mass=0.0492x1.00g mol-1=0.0492g
4.The mass of oxygen=mass of sample-(mass of C+ mass of H)
=1.00g-(0.688g g)=0.263g
5.Now construct a table as in the previous worked examples.

54. C H O
Mass/ g
0.688g
0.0492g
0.263g
Molar mass/g mol-1
12.0g
1.0g
16.0g
Amount/mol
0.0573
0.0492
0.0164
Amount/ smallest amount
0.0573/0.0164=3.49
0.0492/0.0164=3.00
0.0164/0.0164=1.00
Simplest ration of relative amounts 7 6 2

55. Formulas and Relative Masses:
Atoms combine to form molecules. If we know the relative masses of substances which combine, we can determine their relative combining ratio:
Now, if we also know the relative atomic masses of the elements, and the masses of the elements that combine, we can calculate the ratio in which they combine:

56. Ca cation and O anion are chemically combined in a 1:1 ratio mass Ca combined                           8.02 g mass CaO after combination        11.22 g mass O combined                            3.20 g therefore the relative mass of Ca : O is 8.02 /3.20 = 2.51 so Ca is 2.51 times heavier than O

57. A compound containing only nitrogen and oxygen is analyzed
A compound containing only nitrogen and oxygen is analyzed. A sample weighing 2.20 g is found to contain 1.40 g of nitrogen. What is the formula of the nitrogen oxide? mass N combined                1.40 g mass O combined                0.80 g tot. mass of sample              2.20 g

58. Ratio of N : O is 1. 40/0. 80 = 1. 75 so in this compound N is 1
Ratio of N : O is 1.40/0.80 = 1.75 so in this compound N is 1.75 heavier than O We know that the mass of N is amu and the mass of O is amu and we also know that in our sample, the ratio of N/O must=1.75
           2 x mass of N      =      =1.75              mass of O                   and the ratio or chemical formula is N2O