1. Unit 3 Atomic Structure and Periodicity

2. Dalton’s Atomic Theory 1.All matter is composed of _____________. 2. Atoms of the same element are _______________. Atoms of different elements are _______________. 3.Atoms of different elements can _______________ mix together or can _______________ combine with one another in simple whole-number ratios to form _______________. 4. Chemical change involves a ____________________ of atoms. ATOMS IDENTICAL DIFFERENT PHYSICALLY CHEMICALLY COMPOUNDS REARRANGEMENT

3. Charges and relative masses of the three main subatomic particles. PROTON NEUTRON ELECTRON charge = +1; mass = 1 amu; no charge; mass = 1 amu; charge = -1; mass = 1/1846 amu

4. IN A NEUTRAL ATOM The atomic number is the number of __________ and the number of _____________. (in a charged atom, # p = #e) The mass number is the total number of protons __________ neutrons. To find the number of neutrons, __________ the atomic number from the mass number. PROTONS ELECTRONS PLUS SUBTRACT

5. An atom is identified as platinum – 195. (a)What does the number represent? (a)Symbolize this atom using superscripts and subscripts. Mass number 195 Pt 78 Mass number = p + n Atomic number = # protons

6. ISOTOPES Isotopes of the same element identical number of protons different number of neutrons, therefore… different masses and mass numbers

7. List the number of protons, neutrons, and electrons in each pair of isotopes. (1) Li-6, Li-7 Li-6: 3 p +, 3 e -, 3 n O Li-7: 3 p +, 3 e -, 4 n O (2) Ca-42, Ca-44 Ca-42: 20 p +, 20 e -, 22 n O Ca-44: 20 p +, 20 e -, 24 n O

8. AVERAGE ATOMIC MASS Each isotope exists in nature in different abundances (the abundance of Li – 6 is 7.5%; the abundance of Li – 7 is 92.5%) The average atomic mass on the periodic table = the weighted average of all the isotopes for that element

9. to calculate the average atomic mass of an element: 1.Multiply the abundance (in decimal form, ex. 92.5% =.925) by the mass of the isotope 2.repeat step #1 for each isotope 3.Add all calculations together (do not round)

10. Example…Cesium has three known isotopes: Cs – 133, Cs – 132, and Cs – 134. Their abundances in nature are 75%, 20%, and 5% respectively. What is the average atomic mass of cesium? Steps #1, #2 and #3 can be performed together: (.75)(133) + (.20)(132) + (.05)(134) 99.75 + 26.4 + 6.7 132.85 amu

11. Using the data for nitrogen listed in Table 4, p. 82, calculate the average atomic mass of nitrogen. N – 14 and N – 15 Abundances, 99.63% and.37%, respectively Avg. Atomic Mass = (.9963)(14) + (0.0037)(15) = 13.9482 +.0555 = 14.0037 amu