1. Atomic Structure and Isotopes

2. Recap: Atomic Structure An atom is the simplest form of an element that can take part in a chemical reaction Three subatomic particles make up an atom. ▫Electron : outside nucleus (shells)  charge: -1  mass is negligible ▫Proton: inside nucleus  charge: 1+  mass is 1 amu (atomic mass unit) ▫Neutron: inside nucleus  Charge 0 (neutral)  mass is 1 amu

3. Atomic numbers, Mass numbers Elements are often symbolized with their mass number (A) and atomic number (Z) E.g. Oxygen: O 16 8 Z = # of protons = # of electrons A - Z = # of neutrons Calculate # of e –, n 0, p + for Ca, Ar, and Br

4. 3545358035Br 1822184018Ar 20 4020Ca e–e– n0n0 p+p+ Mass # Atomic # Practice!

5. Li atom Li 1+ ion Ions An element that has lost or gained an electron An element which has lost an electron is called a CATION ▫Ex: Na 1+ and Fe 2+ An element which has gained an electron is called an ANION ▫Ex: Cl 1- and O 2- nucleus

6. Ion Practice How has the number of electrons changed from Ca to Ca 2+ ? ▫2 less electrons Determine the number of each subatomic particle for P 3- ▫Protons: 15 ▫Electrons: 18 ▫Neutrons: 16

7. Isotopes Isotopes are atoms of the same element having different masses due to varying numbers of neutrons. IsotopeProtonsElectronsNeutronsNucleus Hydrogen–1 (protium) 110 Hydrogen-2 (deuterium) 111 Hydrogen-3 (tritium) 112

8. Atomic Masses IsotopeSymbolComposition of the nucleus % in nature Carbon-12 12 C6 protons 6 neutrons 98.89% Carbon-13 13 C6 protons 7 neutrons 1.11% Carbon-14 14 C6 protons 8 neutrons <0.01% Atomic mass is the average of all the naturally isotopes of that element… This means that due to isotopes, mass numbers are not simple whole numbers Carbon = 12.011

9. Isotopic Abundance For most elements, the isotopic composition remains constant E.g. any sample of naturally occurring carbon has the same percentage of C-12 and C-13 Because this is constant, we can use an average value for the atomic mass of carbon The percentage of an isotope in a sample is called its isotopic abundance

10. E.g. Given Isotopic Abundance of C-12 = 98.89% Isotopic Abundance of C-13 = 1.110% Calculate the average atomic mass of carbon. m avg = [m C-12 x % C-12 ] + [m C-13 x % C-13 ] = [12 amu x 0.9889] + [13 amu x 0.0110] = 11.8668 u + 0.1443 u = 12.01 u

11. E.g. Chlorine has 2 isotopes, Isotopic Abundance of Cl-35 = 75.53% Isotopic Abundance of Cl-37 = 24.47% Calculate the average atomic mass of chlorine. m avg = [m Cl-35 x % Cl-35 ] + [m Cl-37 x % Cl-37 ] = [35 amu x 0.7553] + [37 amu x 0.2447] = 26.4355 u + 9.0539 u = 35.49 u

12. You practice… Naturally occurring magnesium exists as a mixture of three isotopes. These are Mg-23.985 amu: 78.70% Mg-24.985 amu: 10.13% Mg-25.983 amu; 11.17% Calculate the average atomic mass of magnesium. Answer: 24.31 amu

13. More Practice – Homework! P. 39 # 1, 3 P. 167 # 1-4