Section 4.3

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Standard: 1a, 11c, 3bMastering Concept: 112(47-52)Terms: 98Practice Problems: 99(11-13), 101(14), 104(15-17)

homework: Cornell Notes: 4.3Section Assessment: 104(18-20)Mastering Problems: 113 (59-68)

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Section 4-3Section 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.

Section 4-3Section 4.3 How Atoms Differ (cont.)

atomic numberisotopesmass number

The number of protons and the mass number define the type of atom.

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 mass unit (amu)atomic mass

Section 4-3Atomic 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.

Section 4-3Isotopes and Mass Number

• 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.

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

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

Section 4-3Mass 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.

Section 4-3Mass of Atoms (cont.)

• 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

• 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?

2015.01960.01825.0 xxx

amu90.18

1amu = 1.66x10-27kg

(%)abundancerelative )(amumassisotope

(%)abundancerelative

(%)abundancerelative

)(amumassisotope

)(amumassisotope

:1isotope

:2isotope

:3isotope x

xx

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.

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A review…

Practice Problems: 99(11-13)11. How many protons and electrons are in each of

the following atoms?a. Boron c. Platinumb. 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?

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Practice Problems: 99(11-13)11. How many protons and electrons are in each of

the following atoms?

a. Boron

a. boron, 5

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Practice Problems: 99(11-13)11. How many protons and electrons are in each of

the following atoms?b. Radon

Radon, 86

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Practice Problems: 99(11-13)11. How many protons and electrons are in each of

the following atoms? c. Platinum

platinum, 78

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78Pt

Platinum195.084

Practice Problems: 99(11-13)11. How many protons and electrons are in each of

the following atoms?d. Magnesium

magnesium, 12

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Practice Problems: 99(11-13)

12. An atom of an element contains 66 electrons. What element is it?

12. dysprosium

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word=dysprosium

Practice Problems: 99(11-13)

13. An atom of an element contains 14 protons. What element is it?

13. silicon

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Practice Problems: 101 (14)14. Determine the number of protons, electrons, and

neutrons, Name each isotope, and write its symbol.

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Isotope Composition DataElement Atomic

NumberMass

NumberP/ e

n isotope symbol

b. Calcium 20 46 20 26 Calcium-46c. Oxygen 8 17 8 9 Oxygen-17d. Iron 26 57 26 31 Iron-57e. Zinc 30 64 30 34 Zinc-64f. Mercury 80 204 80 124 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% = 0.198Mass contribution = (mass)(percent abundance)

= 10.013 amu (0.198) = 1.98257

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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% = 0.802Mass 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 amu32

AlternativeIsotope % Abunda

nceMass 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).

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Practice Problems: 104 (15-17)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) = 18.9457524.986 amu (0.1000) = 2.498625.982 amu (0.1101) = 2.8606 24.31 amu

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