2. Lesson 17 Atomic Number and Mass  Mr.Feiock Mr.Reed 

3. Standard deviants Tape 1 III B Stoichiometry: atomic number and weights

4. Objectives: 1.The student will explain the relationship between atomic number and number of protons. 2. The student will calculate numbers of protons, neutrons, and electrons, as well as atomic mass and mass number. - 3. The student will calculate the average atomic mass for an element, given composition data.

5. I. Atomic Number and Mass a. Each element has an atomic number i. The number of protons found in the nucleus of an atom is known as its atomic number (Z) 1. Ex – Hydrogen – 1 2. Ex – Oxygen – 8 ii. If you know the atomic number of an element, you also know the number of electrons in the element – they are the same.

6. b.Isotopes of the same element have different mass numbers i. Atoms of the same element, with the same number of protons, but different numbers of neutrons, are known as isotopes. ii. Examples: 1. Hydrogen or protium – 1 proton, 0 neutrons, 1 electron 2. Deuterium – 1 proton, 1 neutron, 1 electron 3. Tritium – 1 proton, 2 neutrons, 1 electron 4. Further Examples: Carbon –12 (C 12 ) 6p+, 6n, 6e- Carbon –14 (C 14 ) 6p+, 8n, 6e- iii.Isotopes have very similar properties

7. iv. The mass number (A) for any element or isotope is the number of protons and neutrons that it possesses Mass number = (p + + n) v. Isotopes can be represented in two different ways 1. By placing the mass number to the upper left of the symbol and the atomic number to the lower left of the symbol, for that isotope 2. By writing the name of the element followed by a dash and the mass number. Ex – Hydrogen-2 is the same as deuterium.

8. Elements Name  Atomic  number Symbol  Mass number  i.First letter of symbol must always be capitalized, second letter must always be lowercase ii.Atomic number is always a whole number iii.Mass number or Atomic weight may contain decimals

9. d.The number of neutrons in an element is easy to find. i. If you know the mass number of an element, all you need to do is subtract the atomic number (protons) from the mass number (protons + neutrons) to determine the number of neutrons. ii. Mass number is found on the periodic table by rounding the atomic mass to the nearest whole number. iii. Examples: number of protons = atomic # number of electrons = atomic # number of neutrons = mass # - atomic #

10. e.Ions: when an atom gains or loses an electron an ion is formed. i. For each e- lost a positive charge is gained. ii. For each e- gained a negative charge is gained. iii. Examples H loses 1 electron forming a H + ion Mg loses 2 e-, forming a Mg 2+ ion O gains 2 e-, forming an O 2- ion iv. Presence of a charge will then affect your calculation of how many electrons are in the atom from being equal to the atomic # to (atomic # - charge) v. Example calculations: Mg 2+ 12p+, 10e-, 12n O 2- 8p+, 10e-, 8n

11. II. Atomic Mass is expressed in Atomic Mass Units a. The actual masses of atoms are too small to be worked with easily. b. To simplify calculations, atomic masses are expressed in arbitrary units. c. For the scale we use, carbon-12 is the standard d. The mass of one carbon-12 atom is defined as exactly 12.00 atomic mass units. e. This is similar to defining the weight of an egg as 1/12 the mass of a carton of eggs. f. Based on this, the definition of the atomic mass unit is one-twelfth of the mass of the carbon-12 isotope. g. The symbol is amu. h. Atomic Mass is defined as the mass of an atom in atomic mass units. It is what the masses we have been working with so far are expressed in.

12. Actual Masses of Subatomic Particles in amu Mass of a single proton:1.0073μ Mass of a single neutron:1.0087μ Mass of a single electron:0.000549μ

13. III. The periodic table lists average atomic mass a. Atomic masses listed on the periodic table are not rounded numbers – they are all decimals of some sort. b. This is because the mass on the periodic table is a weighted mass, based on the percentage of the naturally occurring isotopes of that atom. c. Weighted averages are used for your grades – here is how they work:

14. i. Example: Copper has two naturally occurring isotopes: copper-63 and copper-65. The relative abundance of copper-63 is 69.17%; the atomic mass of copper-63 is 62.94 amu. The relative abundance of copper-65 is 30.83%; its atomic mass is 64.93 amu. Determine the average atomic mass for copper. Multiply each mass by the decimal equivalent of the percent for each isotope. Add these together to get the final answer. (.6917) (62.94) = 43.535598 (.3083) (64.93) = 20.017919 _________________ 63.553517  rounds to 63.55 amu Remember not to round off your answer until you have the final answer. Each multiplication problem had 4 significant figures, so your final answer has four significant figures.