Section 4.3

Standard: 1a, 11c, 3b Mastering Concept: 112(47-52) Terms: 98 Practice Problems: 99(11-13), 101(14), 104(15-17) homework: Cornell Notes: 4.3 Section Assessment: 104(18-20) Mastering Problems: 113 (59-68)

Explain why atomic masses are not whole numbers. Section 4-3 Section 4.3 How Atoms Differ Explain the role of atomic number in determining the identity of an atom. Define an isotope. Explain why atomic masses are not whole numbers. Calculate the number of electrons, protons, and neutrons in an atom given its mass number and atomic number.

The number of protons and the mass number define the type of atom. Section 4-3 Section 4.3 How Atoms Differ (cont.) periodic table: a chart that organizes all known elements into a grid of horizontal rows (periods) and vertical columns (groups or families) arranged by increasing atomic number atomic number isotopes mass number atomic mass unit (amu) atomic mass The number of protons and the mass number define the type of atom.

Each element contains a unique positive charge in their nucleus. Section 4-3 Atomic Number Each element contains a unique positive charge in their nucleus. The number of protons in the nucleus of an atom identifies the element and is known as the element’s atomic number.

Isotopes and Mass Number Section 4-3 All atoms of a particular element have the same number of protons and electrons but the number of neutrons in the nucleus can differ.

Atoms with the same number of protons but different numbers of neutrons are called isotopes.

The relative abundance of each isotope is usually constant. Isotopes and Mass Number (cont.) Section 4-3 The relative abundance of each isotope is usually constant. Isotopes containing more neutrons have a greater mass. Isotopes have the same chemical behavior. The mass number is the sum of the protons and neutrons in the nucleus.

Section 4-3 Isotopes and Mass Number (cont.)

Section 4-3 Mass of Atoms One atomic mass unit (amu) is defined as 1/12th the mass of a carbon-12 atom. One amu is nearly, but not exactly, equal to one proton and one neutron.

Mass of Atoms (cont.) Section 4-3 The _____________ of an element is the weighted average mass of the isotopes of that element. atomic mass

Mastering Concept: 112(47-52) 47. Does the existence of isotopes contradict part of Dalton’s original atomic theory? Explain. (4.3) Yes; not all atoms of an element are identical in mass.

Mastering Concept: 112(47-52) 48. How do isotopes of a given element differ? How are they similar? (4.3) Different number of neutrons, masses; same: chemical properties, number of protons and electrons

Mastering Concept: 112(47-52) 49. How is the mass number related to the number of protons and neutrons an atom has? (4.3) Mass number = number of p+ + number of n0

Mastering Concept: 112(47-52) 50. What do the superscript and subscript in the notation 40 K represent? (4.3) 19 The superscript represents the mass number (40) and the subscript represents the atomic number (19).

Mastering Concept: 112(47-52) 51. Explain how to determine the number of neutrons an atom contains if you know its mass number and its atomic number. (4.3) Number of n0 = mass number – atomic number

Mastering Concept: 112(47-52) 52. Define the atomic mass unit. What were the benefits of developing the atomic mass unit as a standard unit of mass? (4.3) amu = 1/12 of the mass of a C-12 atom; Scientists defined the atomic mass unit as a relative standard that was closer in size to atomic and subatomic masses.

Calculating Average Atomic Mass x Sum of the products = average atomic mass Example: The isotopes of element “Bob” are found below: Bob-18, 25% Bob-19, 60% Bob-20, 15% What is the average atomic mass of naturally occurring Bob?  x  x  1amu = 1.66x10-27kg

Isotopes, Ions, and Allotropes (Oh my) A review… Isotopes, Ions, and Allotropes (Oh my) Isotopes atoms of the same element with different number of neutrons. Ions atoms of the same element with different number of electrons. Ions are easy to create; adding or removing electrons can be done with electric current. Allotropes forms of the same element, bonded in different structures. Diamond and pencil graphite are allotropes. They are both pure carbon, but in different structures.

Practice Problems: 99(11-13) 11. How many protons and electrons are in each of the following atoms? Boron c. Platinum Radon d. Magnesium 12. An atom of an element contains 66 electrons. What element is it? 13. An atom of an element contains 14 protons. What element is it?

Practice Problems: 99(11-13) 11. How many protons and electrons are in each of the following atoms? Boron a. boron, 5

Practice Problems: 99(11-13) 11. How many protons and electrons are in each of the following atoms? b. Radon Radon, 86

Practice Problems: 99(11-13) 11. How many protons and electrons are in each of the following atoms? c. Platinum platinum, 78 78 Pt Platinum 195.084

Practice Problems: 99(11-13) 11. How many protons and electrons are in each of the following atoms? d. Magnesium magnesium, 12

Practice Problems: 99(11-13) 12. An atom of an element contains 66 electrons. What element is it? 12. dysprosium word=dysprosium

Practice Problems: 99(11-13) 13. An atom of an element contains 14 protons. What element is it? 13. silicon

Isotope Composition Data Practice Problems: 101 (14) 14. Determine the number of protons, electrons, and neutrons, Name each isotope, and write its symbol. Isotope Composition Data Element Atomic Number Mass Number P/ e n isotope symbol b. Calcium 20 46 26 Calcium-46 c. Oxygen 8 17 9 Oxygen-17 d. Iron 57 31 Iron-57 e. Zinc 30 64 34 Zinc-64 f. Mercury 80 204 124 Mercury-204

Practice Problems: 104 (15-17) 15. Boron has two naturally occurring isotopes: boron-10 (abundance = 19.8%, mass = 10.013 amu), boron-11 (abundance = 80.2%, mass = 11.009 amu). Calculate the atomic mass of boron. For isotope boron-10 : Mass: 10.013 amu abundance 19.8% = 0.198 Mass contribution = (mass)(percent abundance) = 10.013 amu (0.198) = 1.98257

Practice Problems: 104 (15-17) 15. Boron has two naturally occurring isotopes: boron-10 (abundance = 19.8%, mass = 10.013 amu), boron-11 (abundance = 80.2%, mass = 11.009 amu). Calculate the atomic mass of boron. For isotope boron-11 : Mass: 11.009 amu abundance 80.2% = 0.802 Mass contribution = (mass)(percent abundance) = 11.009 amu (0.802) = 8.829 amu Sum the mass contributions to find the atomic mass. 1.98257 amu + 8.829 amu = 10.8116 amu

Alternative Isotope % Abundance Mass Mass contribution boron-10 19.8% 0.198 10.013 1.98257 boron-11 80.2% 0.802 11.009 8.829 10.8116

Practice Problems: 104 (15-17) 16. Helium has two naturally occurring isotopes, helium-3 and helium-4. the atomic mass of helium is 4.003 amu. Which isotope is more abundant in nature? Explain. 16. Helium-4 is more abundant in nature because the atomic mass of naturally occurring helium is closer to the mass of helium-4 (~4 amu) than to the mass of helium-3 (~3 amu).

Practice Problems: 104 (15-17) Calculate the atomic mass of magnesium. The three magnesium isotopes have atomic masses and relative abundances of 23.985 amu (78.99%), 24.986 amu (10.00%), and 25.982 amu (11.01%) 23.985 amu (0.7899) = 18.94575 24.986 amu (0.1000) = 2.4986 25.982 amu (0.1101) = 2.8606 24.31 amu