4.2 – NOTES The Mass of the Atom

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Presentation transcript:

4.2 – NOTES The Mass of the Atom

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;  

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

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

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;

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 29.974

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

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 23.985amu, 10.00% is Mg – 25 with a mass of 24.986 and 11.01% is Mg-26 with a mass of 25.982. 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) = 2.499 Mg – 26: (0.1101)(25.982) = 2.861 24. 310 = 24.31 amu

Sample Problems: 1. The abundance of B-10 is 19.78% and has a mass of 10.011amu. The abundance of B-11 is 80.22% with a mass of 11.009amu. Determine the atomic weight of boron: B – 10: (0.1978)(10.011) = 1.980 B – 11: (0.8022)(11.009) = 8.831 10.811 amu

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