C H E M I S T R Y

Chapter 2Chapter 2Atoms, Molecules, and IonsAtoms, Molecules, and Ions

Conservation of Mass and the Conservation of Mass and the Law of Definite ProportionsLaw of Definite Proportions

Law of Conservation of Mass: Mass is neither created nor destroyed in chemical reactions.

Conservation of Mass and the Conservation of Mass and the Law of Definite ProportionsLaw of Definite Proportions

HgI2(s) + 2KNO3(aq)Hg(NO3)2(aq) + 2KI(aq)

4.55 g + 2.02 g = 6.57 g

3.25 g + 3.32 g = 6.57 g

Conservation of Mass and Conservation of Mass and the Law of Definite the Law of Definite ProportionsProportions

Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass.

By mass, water is: 88.8 % oxygen11.2 % hydrogen

The Law of Multiple The Law of Multiple Proportions and Dalton’s Proportions and Dalton’s Atomic TheoryAtomic Theory

Law of Multiple Proportions: Elements can combine in different ways to form different substances, whose mass ratios are small whole-number multiples of each other.

Insert Figure 2.2 p37

7 grams nitrogen per 8 grams oxygen

7 grams nitrogen per 16 grams oxygen

nitrogen monoxide (NO):

nitrogen dioxide (NO2):

The Law of Multiple The Law of Multiple Proportions and Dalton’s Proportions and Dalton’s Atomic TheoryAtomic Theory

• Elements are made up of tiny particles called atoms.

• Each element is characterized by the mass of its atoms. Atoms of the same element have the same mass, but atoms of different elements have different masses.

• The chemical combination of elements to make different chemical compounds occurs when atoms join in small whole-number ratios.

• Chemical reactions only rearrange how atoms are combined in chemical compounds; the atoms themselves don’t change.

Atomic Structure: ElectronsAtomic Structure: ElectronsCathode-Ray Tubes: J. J. Thomson (1856-1940) proposed that cathode rays must consist of tiny negatively charged particles. We now call them electrons.

His first experiment was to build a cathode ray tube with a metal cylinder on the end. This cylinder had two slits in it, leading to electrometers, which could measure small electric charges.

He found that by applying a magnetic field across the tube, there was no activity recorded by the electrometers and so the charge had been bent away by the magnet. This proved that the negative charge and the ray were inseparable and intertwined.

he constructed a slightly different cathode ray tube, with a fluorescent coating at one end and a near perfect vacuum. Halfway down the tube were two electric plates, producing a positive anode and a negative cathode, which he hoped would deflect the rays.

the rays were deflected by the electric charge, proving beyond doubt that the rays were made up of charged particles carrying a negative charge

Atomic Structure: Atomic Structure: ElectronsElectrons

Atomic structure: ElectronsAtomic structure: Electrons The strength of deflecting magnetic electric field

The size of the negative charge of electron

The mass of the electron

Atomic Structure: ElectronsAtomic Structure: Electrons

Millikan’s oil drop experiment

Millikan's experiment involved measuring the force on oil droplets in a glass chamber sandwiched between two electrodes, one above and one below. With the electrical field calculated, he could measure the droplet's charge, the charge on a single electron being (1.592×10−19  C).

Atomic Structure: Protons and Atomic Structure: Protons and NeutronsNeutrons

The results of this experiment gave Rutherford the means to arrive at two conclusions: one, an atom was much more than just empty space and scattered electrons (J.J. Thomson model argued), and two, an atom must have a positively charged center that contains most of its mass (which Rutherford termed as the nucleus).

Atomic Structure: Protons and Atomic Structure: Protons and NeutronsNeutrons

The mass of the atom is primarily in the nucleus.

The charge of the proton is opposite in sign but equal to that of the electron.

Atomic NumbersAtomic Numbers

Atomic Number (Z): Number of protons in an atom’s nucleus. Equivalent to the number of electrons around an atom’s nucleus

Mass Number (A): The sum of the number of protons and the number of neutrons in an atom’s nucleus

Isotope: Atoms with identical atomic numbers but different mass numbers

Atomic NumbersAtomic Numbers

Atomic NumbersAtomic Numbers

carbon-13 or C-13

C13

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atomic number

mass number

carbon-12 or C-12

C12

6

atomic number

mass number

6 protons6 electrons7 neutrons

6 protons6 electrons6 neutrons

Isotopic symbolsIsotopic symbols

XAZ

or X - A

ExamplesExamples

1. How many protons, electrons and neutrons are present in an atom of

2. Write isotopic symbols in both forms for Selenium isotope with 40 neutrons

3. An atom has 32 electrons and 38 neutrons. What is its mass number and what is the element?

Ca46

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Atomic Masses and the MoleAtomic Masses and the Mole

The mass of 1 atom of carbon-12 is defined to be 12 amu.

Atomic Mass: weighed according to the natural abundance of each isotope. Sometimes called average atomic mass or relative atomic mass.

Atomic Masses and the MoleAtomic Masses and the Mole

carbon-12: 98.89 % natural abundance 12 amu

carbon-13: 1.11 % natural abundance 13.0034 amu

Why is the atomic mass of the element carbon 12.01 amu?

= 12.01 amu

mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111)

= 11.87 amu + 0.144 amu

Atomic Masses and the Atomic Masses and the MoleMoleAtomic mass unit (amu) is the mass in

grams of a single atom◦1 amu = 1.6605 x 10-24g◦E.g 1 H atom = 1.01 amu

Calculating Atomic MassCalculating Atomic Mass

The calculation for atomic mass requires the

• percent(%) abundance of each isotope.

• atomic mass of each isotope of that element.

• sum of the weighted averages.

Atomic mass = isotopic mass 1 x % abundance + isotopic mass 2 x % abundance

Calculating Atomic Mass for Calculating Atomic Mass for CopperCopper

Copper has two naturally occurring isotopes: Cu-63 with mass 62.9396 amu and a natural abundance of 69.17% and Cu-65 with mass 64.9278 amu and a natural abundance of 30.83%. Calculate the atomic mass of copper

• Use atomic mass and percent of each isotope to calculate the contribution of each isotope to the weighted average. Atomic mass Cu-63 x % abundance =Atomic mass Cu-65 x % abundance =

• Sum is atomic mass of Cu is

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ExampleExampleBromine has two naturally

occurring isotopes (Br-79 and Br-81) and has an atomic mass of 79.904 amu. The mass of Br-81 is 80.9163 amu, and its natural abundance is 49.31%. Calculate the mass and natural abundance of Br-79

A Mole of AtomsA Mole of Atoms

A mole is

• a unit of measurement used in chemistry to express amounts of a chemical substance, the same number of particles as there are carbon atoms in 12.0 g of carbon.

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Collection TermsCollection Terms

A collection term statesa specific number of items.

• 1 dozen donuts

= 12 donuts

• 1 ream of paper

= 500 sheets

• 1 case = 24 cans

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Atomic Masses and the MoleAtomic Masses and the Mole

Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units.

1 mole of anything = 6.02 x1023

Molar Mass: The mass in grams of one mole of any element. It is numerically equivalent to its atomic mass.

E.g 1 H atom = 1.01 amu 1 mol H = 1.01 g

Samples of 1 Mole QuantitiesSamples of 1 Mole Quantities

1 mole of C atoms = 6.02 x 1023 C atoms

1 mole of Al atoms = 6.02 x 1023 Al atoms

1 mole of S atoms = 6.02 x 1023 S atoms

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Avogadro’s Number and the Avogadro’s Number and the MoleMole

Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 3/28

Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units.

One mole of any substance is equivalent to its molar mass.

1 mole = 12.01 gC:

6.022 x 1023 molecules = 12.01 g

1 mole = 1.08 g

6.022 x 1023 molecules = 1.08 g

H

Molar Mass from Periodic Molar Mass from Periodic TableTable

Molar mass is the atomic mass expressed in grams.

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1 mole of Ag 1 mole of C 1 mole of S = 107.9 g = 12.01 g = 32.07 g

ExamplesExamples

Give the molar mass for each

A. 1 mole of Li atoms = ________

B. 1 mole of Co atoms = ________

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Avogadro’s NumberAvogadro’s Number

Avogadro’s number, 6.02 x 1023, can be written as an

equality and two conversion factors.

Equality:1 mole = 6.02 x 1023 particles

Conversion Factors:6.02 x 1023 particles and 1 mole

1 mole 6.02 x 1023 particles

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ExamplesExamples

1. Calculate the number of copper atoms in 2.45 mol of copper

2. The number of moles of S in 1.8 x 1024

atoms of S

3. The number of atoms in 2.0 g of Al atoms is

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