1. Isotopes Atoms with the same number of protons, but different numbers of neutrons. Atoms of the same element (same atomic number) with different mass numbers Isotopes of chlorine 35 Cl 37 Cl17 chlorine - 35 chlorine - 37

2. Natural abundances of isotopes Natural chlorine contains : 75 % 35 Cland 25 % 37 Cl. 17 17 These percentages are known as the natural abundances of the isotopes and determined by “mass spectrometry.”

3. Average atomic mass/average mass/atomic mass/ atomic weight Is the weighted mean.

4. Exercise 1 The atomic mass of iridium is 192.2 g / mol. What are the naturally occurring percentages of the two isotopes of Ir-191 and Ir-193?

5. solution Iridium is a mixture of 40% 191 Ir and 60 % 193 Ir.

6. Isotopes of Hydrogen element has the biggest abundance in nature. Tritium Deuterium

7. A mass spectrometer is an instrument which separates particles according to their masses, records the relative proportions of these, and determine natural abundances of the isotopes of an element. Therefore, it also allows us to calculate the atomic mass of an element. The most accurate way for determining atomic and molecular weights is provided by mass spectrometer. Mass spectrometry

8. mass spectrometer, invented by the English physicist Francis William Aston (1877-1945) when he was working in Cambridge with J. J. Thomson. It was in his use of this instrument that the existence of isotopes of elements was discovered. Aston eventually discovered many of the naturally occurring isotopes of non-radioactive elements. He was awarded the Nobel Prize for Chemistry in 1922. Mass spectrometry

9. A B C D E accelerating F

10. A: a gaseous sample is very slowly introduced to the mass spectrometer. B: atoms/molecules are bombarded by a stream of high energy electrons to produce positive ions, mostly w/ a 1+ charge. These electrons collide w/ electrons in the particle knocking them out and leaving a positive ion. C: positively charged ions are accelerated high enough to make the particles pass through the slits and magnetic field by high electrical voltage on the negatively charged grid. With the slits, the ions were made a beam of ions.

11. D: Fast moving ions enter a magnetic field produced by an electromagnet. Ions are deflected by a magnetic field into a curved path. The deflection of the ions depends on “charge to mass ratio(q/m). ”The more massive the ion, the less the deflection. The ions w/ equal mass and charge will deflect the same. E: By changing the strength of the magnetic field or the accelerating voltage on the negatively charged grid, ions of varying masses can be made to enter the detector at the end of the instrument.

12. On the detector, ions are collected on a metal plate and the current flows through the metal plate to neutralise the ions and this current is recorded. In this way, the relative abundances of ions of different masses in the sample can be determined and put into a graph called “mass spectrum.”

13. F: The mass spectrometer must be at a high vacuum for its correct operation and its correct operation depends on particles being able to pass through it w/o colliding with any other particles.

14. A: vapourised sample introduced B: ionization by electron bombardment C: Positive ions accelerated by electrical field D: ions deflected by a magnetic field E: detector records ions of a particular mass F: vacuum prevents molecules colliding

15. Atomic weight measurements How was the atomic weight measured? By mass spectrometry –This also measures % natural abundance for a given isotope -The graph is called as “mass spectrum.”

16. Atomic weight calculation There are three naturally occuring isotopes of neon (Ne): 20 Ne isotopic mass = 19.99244018 amu 21 Ne isotopic mass = 20.9938467 amu 22 Ne isotopic mass = 21.9913855 amu the atomic weight is reported in text as: 20.1797 amu

17. Masses of Atoms A scale designed for atoms gives their small atomic masses in atomic mass units (amu) An atom of 12 C was assigned an exact mass of 12.00 amu Relative atomic masses of all other atoms was determined by comparing each to the mass of 12 C An atom twice as heavy has a mass of 24.00 amu (Mg=24.00amu). An atom 12 times lighter than 1 C-12 atom is 1.00 amu(H=1.00amu).

18. Atomic mass unit(amu or u or Da) 1/12 th of the mass of one C-12 atom.

19. Average atomic mass(atomic weight/average mass/atomic mass) “atomic weight or mass” = average mass of an atom calculated from the masses and natural abundances of all isotopes

20. Atomic Mass Average atomic mass is based on all the isotopes and their abundance % Atomic mass is not a whole number!!!! Na 22.99

21. Calculating Atomic Weight or Mass Percent(%) abundance of isotopes Mass of each isotope of that element Weighted average = mass isotope 1 (%) + mass isotope 2 (%) + … 100 100

22. Naturally occurring C is composed of 98.93 % 12 C and 1.07 % 13 C. The masses of these nuclides are 12 amu (exactly) and 13.00335 amu, respectively. Average atomic mass(atomic mass) or atomic weight

23. Atomic Mass of Magnesium Isotopes Mass of Isotope Abundance 24 Mg =24.0 amu 78.70% 25 Mg = 25.0 amu 10.13% 26 Mg = 26.0 amu 11.17% Atomic mass (average mass) Mg = 24.3 amu Mg 24.3

24. Atomic mass calculation How was the atomic mass calculated? multiply each isotopic mass by the reported natural abundance for the isotope, then: add these individual contributions for each isotope to get the average atomic mass for the element

25. Atomic mass calculation There are three naturally occuring isotopes of neon (Ne): 20 Ne mass # = 19.99244018 amu (90.51%) 21 Ne mass # = 20.9938467 amu (0.27%) 22 Ne mass # = 21.9913855 amu (9.22%) the atomic mass is reported in text as: 20.1797 amu 18.10 + 0.057 + 2.03 = 20.19 amu

26. Learning Check 5 Gallium is a metallic element found in small lasers used in compact disc players. In a sample of gallium, there is 60.2% of gallium-69 (68.9 amu) atoms and 39.8% of gallium-71 (70.9 amu) atoms. What is the atomic mass of gallium ?

27. Solution Ga-69 68.9 amu x 60.2 = 41.5 amu for 69 Ga 100 Ga-71 (%/100) 70.9 amu x 39.8 = 28.2 amu for 71 Ga 100 Atomic mass Ga = 69.7 amu

28. Finding An Isotopic Mass A sample of boron consists of 10 B (mass 10.0 amu) and 11 B (mass 11.0 amu). If the average atomic mass of B is 10.8 amu, what is the % abundance of each boron isotope?

29. Assign X and Y values: X = % 10 B Y = % 11 B Determine Y in terms of X X + Y = 100 Y = 100 - X Solve for X: X (10.0) + (100 - X )(11.0) = 10.8 100 100 Multiply through by 100 10.0 X + 1100 - 11.0X = 1080

30. Collect X terms 10.0 X - 11.0 X = 1080 - 1100 - 1.0 X = -20 X = -20 = 20 % 10 B - 1.0 Y = 100 - X % 11 B = 100 - 20% = 80% 11 B

31. Learning Check 6 Copper has two isotopes 63 Cu (62.9 amu) and 65 Cu (64.9 amu). What is the % abundance of each isotope? (Hint: Check Zumdahl or any other chemistry text for atomic mass) 1) 30%2) 70%3) 100%

32. Solution 2) 70% Solution 62.9X + 6490 = 64.9X = 6350 -2.0 X = -140 X = 70%

33. Atomic Masses 13 C 12 C 13.00335 amu (1.11%) 12.0000 amu (98.89%) atomic weight of C = 12.01115 amu WHY?

34. Calculating masses of atoms relative to 12 C (mass of 12 C atom) * 1.58320 = mass of F atom = 18.99840 reported atomic weight of F = 18.9984