1. 4.2 – NOTES The Mass of the Atom

2. III. How atoms differ A. Atomic Number
Equal to the number of protons – never changes; equals the number of electrons if AND ONLY IF the atom is neutral; Moseley discovered each atom has a unique number of protons in the nucleus;

3. B. Isotopes and mass number
1. Definition of isotopes: atoms of the same element with differing numbers of neutrons;
H has 3 isotopes –
protium (1 proton)
deuterium (1 proton and 1 neutron)
tritium (1 proton and 2 neutrons);
usually have similar properties, but can have specific uses like C-14 (carbon dating); most elements are found as a mixture of their isotopes

5. 2. Mass number: represents the total number of protons and neutrons; specific to each isotope
(U-238 vs U-235);
#n0 = mass # - atomic #;
does not include mass of electron b/c mass of electron is ~2000x less than that of proton and/or neutron

6. 3. Nuclear symbols and nuclides: nuclide – specific atom with a mass number attached (Li-7);
Mass # Atomic symbol atomic # = nuclear symbol;
Examples: 1H1 = protium (H isotope); 2H1 = deuterium; 3H1 = tritium
H – 1; H – 2; H – 3;

7. C. Mass of individual atoms
1. Atomic mass units (amu) – the mass of a single proton or neutron is too small to use easily
- scientists developed a method to measure the mass of an atom relative to an atomic standard
– carbon 12; 1 amu = 1/12 the mass of a carbon-12 atom; the value of 1 amu and that of a proton are slightly different; explains why Si-30 is really

9. 2. Atomic weights on the periodic table are weighted averages.

10. Mg is a good example: 78. 99% is Mg-24 has a mass of 23. 985amu, 10
Mg is a good example: 78.99% is Mg-24 has a mass of amu, 10.00% is Mg – 25 with a mass of and 11.01% is Mg-26 with a mass of What is the average weight? Any element’s average mass can be calculated if the # of isotopes, mass of each isotope and relative abundance of each isotope is known. **% should be put in decimal form.
Average mass = (%)(mass) + ( %)(mass) + …
Mg – 24: (0.7899)(23.985) = 18.95
Mg – 25: (0.1000)(24.986) =
Mg – 26: (0.1101)(25.982) =
= amu

11. Sample Problems:
1. The abundance of B-10 is 19.78% and has a mass of amu. The abundance of B-11 is 80.22% with a mass of amu. Determine the atomic weight of boron:
B – 10: (0.1978)(10.011) = 1.980
B – 11: (0.8022)(11.009) = 8.831
amu

12. 2. Chlorine has 2 isotopes: Cl-35 and Cl-37
2.Chlorine has 2 isotopes: Cl-35 and Cl-37. Since the atomic weight of Cl is about 35.5, what are the approximate relative abundances of these isotopes?
Easiest way to solve this problem is to make a number line.
KEEP IN MIND – this is a weighted average! What that means is since the average (35.5) is closer to Cl-35, Cl-35 is more abundant (= larger percentage); in this problem, the percentages are 25% and 75% so Cl- 35 must be 75% of the abundance