Modern Chemistry Chapter 3 Atoms: The Building Blocks of Matter law of conservation of mass- mass is neither created nor destroyed during ordinary chemical.

Modern Chemistry Chapter 3Modern Chemistry Chapter 3Atoms: The Building Blocks of Atoms: The Building Blocks of

MatterMatter

law of conservation of mass-law of conservation of mass- mass is neither created nor mass is neither created nor destroyed during ordinary chemical destroyed during ordinary chemical reactions or physical changesreactions or physical changes

e.g. 20 g A + 20 g B e.g. 20 g A + 20 g B 40 g AB 40 g AB

law of definite proportions-law of definite proportions- a a chemical compound contains the same chemical compound contains the same elements in exactly the same proportions elements in exactly the same proportions by mass regardless of the size of the by mass regardless of the size of the sample or the source of the compoundsample or the source of the compound

e.g. If 10 grams of A combine with 20 grams e.g. If 10 grams of A combine with 20 grams of B to form compound AB, how many grams of of B to form compound AB, how many grams of B will be necessary to combine with 20 grams B will be necessary to combine with 20 grams of A to form AB? Answer 40: grams of B.of A to form AB? Answer 40: grams of B.

law of multiple proportions-law of multiple proportions- if two if two or more different compounds are or more different compounds are composed of the same two elements, composed of the same two elements, then the ratio of the masses of the then the ratio of the masses of the second element combined with a second element combined with a certain mass of the first element is certain mass of the first element is always a ratio of small whole numbersalways a ratio of small whole numbers

e.g. He.g. H22O & HO & H22OO22 or NO or NO22 & N & N22OO55

Dalton’s Atomic TheoryDalton’s Atomic Theory

In 1808, an English In 1808, an English school teacher school teacher named John Dalton named John Dalton proposed an proposed an explanation for the explanation for the law of conservation law of conservation of mass, the law of of mass, the law of definite proportions, definite proportions, and the law of and the law of multiple multiple proportions.proportions.

Dalton’s Atomic Theory:Dalton’s Atomic Theory:1- All matter is composed of extremely 1- All matter is composed of extremely

small particles called atoms.small particles called atoms.2- Atoms of a given element are 2- Atoms of a given element are identical identical in size, mass and other in size, mass and other properties.properties.3- Atoms cannot be subdivided, created 3- Atoms cannot be subdivided, created or or destroyed.destroyed.4- Atoms of different elements combine 4- Atoms of different elements combine in in simple whole-number ratios to form simple whole-number ratios to form

chemical compounds.chemical compounds.5- In chemical reactions, atoms are 5- In chemical reactions, atoms are combined, separated, or rearranged.combined, separated, or rearranged.

Modern Atomic TheoryModern Atomic Theory

Dalton’s theory was a good one, but Dalton’s theory was a good one, but it has since been modified.it has since been modified.– Atoms are divisible into even Atoms are divisible into even

smaller particles.smaller particles.– Atoms of a given element can have Atoms of a given element can have

different masses.different masses.– Atoms can be destroyed.Atoms can be destroyed.

Do Section Review #3 on page 71.Do Section Review #3 on page 71.

Section review #3 page 71Section review #3 page 71

IF each compound contains 1.0 g oxygen IF each compound contains 1.0 g oxygen and the three samples contain:and the three samples contain:

compound A:compound A: K = 1.22 g / 1.22 = 1.0K = 1.22 g / 1.22 = 1.0

compound B:compound B: K = 2.44 g / 1.22 = 2.0K = 2.44 g / 1.22 = 2.0

compound C:compound C: K = 4.89 g / 1.22 = 4.0K = 4.89 g / 1.22 = 4.0

1:2:4 ratios of potassium 1:2:4 ratios of potassium multiple multiple proportionsproportions

Chapter 3 Section 2Chapter 3 Section 2The Structure of the AtomThe Structure of the Atom

atomatom- the - the smallest particle smallest particle of an element of an element that retains the that retains the chemical chemical properties of properties of that elementthat element

atomic nucleus-atomic nucleus- the small, densely the small, densely packed, positively charged central portion packed, positively charged central portion of the atom that contains nearly all of its of the atom that contains nearly all of its mass but nearly none of its volumemass but nearly none of its volume– neutron-neutron- The neutral particle of the The neutral particle of the

nucleus of an atom.nucleus of an atom.– proton-proton- The positively charged particle The positively charged particle

of the nucleus of an atom.of the nucleus of an atom.

electron cloud-electron cloud- The large area The large area surrounding the nucleus of an atom in surrounding the nucleus of an atom in which the electrons are located.which the electrons are located.– electron-electron- the negatively charged the negatively charged

particles of an atomparticles of an atom

The Discovery of ElectronsThe Discovery of Electrons In the late 1800’s, electric current was In the late 1800’s, electric current was

passed through cathode ray tubes. It was passed through cathode ray tubes. It was discovered that the cathode ray was discovered that the cathode ray was attracted to the positive pole of a magnet attracted to the positive pole of a magnet and repelled by the negative pole.and repelled by the negative pole.– This led to the discovery of electrons.This led to the discovery of electrons.– In 1909, Robert Millikan In 1909, Robert Millikan measured the measured the

negative charge of the electronnegative charge of the electron..– From this, it was found that the mass of From this, it was found that the mass of

an electron is 9.109 x 10an electron is 9.109 x 10-31 -31 kg.kg.– The mass of an electron is 1/1837The mass of an electron is 1/1837thth the the

mass of the simplest hydrogen atom.mass of the simplest hydrogen atom.– The negative charges of the electrons The negative charges of the electrons

equal the positive charges of an atom equal the positive charges of an atom (protons).(protons).

The Discovery of the Atomic The Discovery of the Atomic NucleusNucleus

In 1911, Ernest Rutherford, Hans In 1911, Ernest Rutherford, Hans Geiger, & Ernest Marsden Geiger, & Ernest Marsden bombarded a thin foil of gold with bombarded a thin foil of gold with positively charged particles called positively charged particles called alpha particles.alpha particles. They were surprised They were surprised to find that a few of the particles (1 to find that a few of the particles (1 in 8000) were reflected from the foil in 8000) were reflected from the foil straight back toward their source. straight back toward their source. They reasoned that this must mean They reasoned that this must mean that there was that there was a small positively a small positively charged portion of each gold atomcharged portion of each gold atom. .

Composition of the NucleusComposition of the Nucleus protons-protons- are positively charged are positively charged

particles in the nucleus (their + charge particles in the nucleus (their + charge = - charge of an electron)= - charge of an electron)– mass = 1.673 x 10mass = 1.673 x 10-27 -27 kg or 1836/1837 the kg or 1836/1837 the

mass of a protium atommass of a protium atom

neutrons-neutrons- neutral particles of the neutral particles of the nucleusnucleus– mass = 1.675 x 10mass = 1.675 x 10-27-27 kg = mass of 1 kg = mass of 1

electron electron + 1 proton + 1 proton

Composition of the NucleusComposition of the Nucleus

How can the numerous positively How can the numerous positively charged protons exist packed into the charged protons exist packed into the nucleus without flying apart due to nucleus without flying apart due to their like charges repelling one their like charges repelling one another?another?

nuclear forces-nuclear forces- are short range are short range forces (proton-proton, proton-neutron, forces (proton-proton, proton-neutron, & neutron-neutron) that hold the & neutron-neutron) that hold the nuclear particles together.nuclear particles together.

The Size of AtomsThe Size of Atoms

Atomic radius ranges between 40 & Atomic radius ranges between 40 &

270 pm (10270 pm (10-12-12 m) m)Discuss how small a picometer would be.Discuss how small a picometer would be.

Nuclear radius = 0.001 pm.Nuclear radius = 0.001 pm.The size of the nucleus to the entire atom The size of the nucleus to the entire atom

would be about the same as if you placed a would be about the same as if you placed a dime at the center of the football stadium.dime at the center of the football stadium.

Section 3.3Section 3.3Counting AtomsCounting Atoms

atomic number-atomic number- is equal to the is equal to the number of protons in the nucleus of number of protons in the nucleus of each atom of an elementeach atom of an element

mass number-mass number- is the number of is the number of protons plus neutrons in a single protons plus neutrons in a single atom of an element.atom of an element.

Counting AtomsCounting Atoms isotopes-isotopes- are atoms of the same element that are atoms of the same element that

have different masses due to different numbers of have different masses due to different numbers of neutronsneutrons

nuclide-nuclide- is a general term for a specific isotope of is a general term for a specific isotope of an elementan element

– We designate isotopes using one of two different We designate isotopes using one of two different designations.designations.

– Hydrogen has three isotopes; protium (m# = 1), Hydrogen has three isotopes; protium (m# = 1), deuterium (m# = 2), and tritium (m# = 3).deuterium (m# = 2), and tritium (m# = 3).

hyphen notationhyphen notation H-1 H-2 H-1 H-2 H-3H-3 1 2 3 1 2 3

nuclear symbol notationnuclear symbol notation H H H H H H 11 11

11

IsotopesIsotopes

Do practice Do practice problems #1, 2, & problems #1, 2, & 3 on page 80.3 on page 80.

Practice problems page 80Practice problems page 80

1-1- bromine-80bromine-80 35 protons35 protons

35 electrons35 electrons

80-35 = 45 neutrons80-35 = 45 neutrons

2-2- carbon-13 carbon-13 1313

CC 66

3-3- 15 electrons, so 15 protons & 15 electrons, so 15 protons & element is element is phosphorus. 15 + 15 phosphorus. 15 + 15 phosphorus-30phosphorus-30

Identifying IsotopesIdentifying Isotopes

elementelementsymbolsymbol atomic #atomic # mass #mass # pp++ n n00 e e--

calciumcalcium 40 40 ________

uraniumuranium 92 92 235 235 ____ ____

uraniumuranium 238 238 ________

1313 27 27 ________

I-127I-127 ________

Relative Atomic MassesRelative Atomic Masses

atomic mass unit (amu)-atomic mass unit (amu)- exactly exactly 1/121/12thth of the mass of a carbon-12 atom of the mass of a carbon-12 atom

average atomic mass-average atomic mass- is the is the weighted average of the atomic masses weighted average of the atomic masses of the naturally occurring isotopes of an of the naturally occurring isotopes of an elementelement

– see table 4 on page 82see table 4 on page 82

Calculating Average Atomic MassCalculating Average Atomic Mass

We will be calculating the “average mass” of the science We will be calculating the “average mass” of the science textbooks in the classroom. This is the approximate textbooks in the classroom. This is the approximate method used to determine average atomic mass of the method used to determine average atomic mass of the isotopes of an element.isotopes of an element.

1- Using a bathroom scale, find the weight of the physics 1- Using a bathroom scale, find the weight of the physics textbooks and count their number. What is the textbooks and count their number. What is the

average average weight of each book?weight of each book?

2- Find the weight of the chemistry textbooks and count 2- Find the weight of the chemistry textbooks and count their their number. What is the average weight of each number. What is the average weight of each book?book?

3- Add the weights of the textbooks and add the numbers 3- Add the weights of the textbooks and add the numbers of of books.books.

4- Divide the total weight by the total number of books to 4- Divide the total weight by the total number of books to find find the average weight. How does this compare to the average weight. How does this compare to the the average weight of each book?average weight of each book?

Relating Mass to Numbers of Relating Mass to Numbers of AtomsAtoms

mole (mol)-mole (mol)- the the amount of a amount of a substance that substance that contains the same contains the same number of particles number of particles as there are in 12 as there are in 12 grams of carbon-12grams of carbon-12

The concept is The concept is similar to that of a similar to that of a dozen.dozen.

Avogadro’s Avogadro’s number-number- is equal is equal to the number of to the number of particles in one particles in one mole of a mole of a substance and is substance and is equal to 6.022 x equal to 6.022 x 10102323

Molar MassMolar Mass molar mass-molar mass- the the

mass of one mole of mass of one mole of a pure substancea pure substance

– We can find molar We can find molar mass by using the mass by using the average atomic average atomic mass found on mass found on the periodic table the periodic table and changing the and changing the units from amu to units from amu to grams.grams.

Mole-molar mass Mole-molar mass conversionsconversions

#mole x molar mass = mass (#grams)#mole x molar mass = mass (#grams)

mass (#grams) mass (#grams) ÷ molar mass = #moles÷ molar mass = #moles

Mole HillMole Hill

#moles#moles

(mol)(mol)

÷ molar mass ÷ molar mass x molar massx molar mass

(g/mol)(g/mol) (g/mol) (g/mol)

mass in grams mass in grams mass in mass in gramsgrams

(g)(g) (g)(g)

Gram to Mole ConversionsGram to Mole Conversions We can use conversion factors to We can use conversion factors to convert convert

between grams and molesbetween grams and moles..

2.00 mol He x 2.00 mol He x 4.00 g He4.00 g He = 8.00 g He = 8.00 g He 1 mol He1 mol He

8.00 g He x 8.00 g He x 1 mol He 1 mol He = 2.00 mol He = 2.00 mol He 4.00 g He4.00 g He

Do practice problems #1-4 and #1-3 on page 85Do practice problems #1-4 and #1-3 on page 85

Mole to Gram Conversions pg Mole to Gram Conversions pg 8585

1)1) 2.25 mol Fe x 55.85 g/mol = 126 g Fe2.25 mol Fe x 55.85 g/mol = 126 g Fe

2)2) 0.375 mol K x 39.10 g/mol = 14.7 g K0.375 mol K x 39.10 g/mol = 14.7 g K

3)3) .0135 mol Na x 22.99 g/mol = 0.310 .0135 mol Na x 22.99 g/mol = 0.310 g Nag Na

4)4) 16.3 mol Ni x 58.69 g/mol = 957 g Ni16.3 mol Ni x 58.69 g/mol = 957 g Ni

Gram to Mole Conversions pg Gram to Mole Conversions pg 8585

1)1) 5.00 g Ca 5.00 g Ca ÷ 40.08 g/mol = 0.125 mol Ca÷ 40.08 g/mol = 0.125 mol Ca

2)2) 3.60 x 103.60 x 10-5-5 g Au ÷ 196.97 g/mol = 1.83 x 10 g Au ÷ 196.97 g/mol = 1.83 x 10-7-7 mol Aumol Au

3)3) 0.535 g Zn ÷ 65.39 g/mol = 8.18 x 100.535 g Zn ÷ 65.39 g/mol = 8.18 x 10-3-3 mol Zn mol Zn

Conversions Using Avogadro’s NumberConversions Using Avogadro’s Number

3.01 x 103.01 x 102323 Ag atoms x Ag atoms x 1 mole Ag atoms 1 mole Ag atoms 6.022 x 10 6.022 x 102323 Ag atoms Ag atoms

= 0.500 moles of Ag= 0.500 moles of Ag

1.20 x 101.20 x 1088 atoms Cu x atoms Cu x 1 mol Cu atoms 1 mol Cu atoms x x 63.55 g63.55 g

6.022 x 106.022 x 102323 Cu atoms Cu atoms

= Cu = 1.27 x 10= Cu = 1.27 x 10-14-14 g Cu g Cu

Do section review problems #2-6 on page 87.Do section review problems #2-6 on page 87.

Section review page 87Section review page 872a-2a- sodium-23 sodium-23 11 protons, 11 electrons, 12 neutrons 11 protons, 11 electrons, 12 neutrons

b-b- calcium-40 calcium-40 20 protons, 20 electrons, 20 neutrons 20 protons, 20 electrons, 20 neutrons

c-c- copper-64 copper-64 29 protons, 29 electrons, 35 neutrons 29 protons, 29 electrons, 35 neutrons

d-d- silver-108 silver-108 47 protons, 47 electrons, 61 neutrons 47 protons, 47 electrons, 61 neutrons

3a-3a- silicon-28silicon-28

b-b- iron-56iron-56

4-4- potassium potassium 39.10 amu & 39.10 g/mol 39.10 amu & 39.10 g/mol

5a-5a- 2.00 mol x 14.0 g/mol = 28.0 g N2.00 mol x 14.0 g/mol = 28.0 g N

6a-6a- 12.15 g / 24.3 g/mol = 0.50 mol Mg12.15 g / 24.3 g/mol = 0.50 mol Mg

Section Review page 87Section Review page 87

#7-#7- 2.06 mol Cu2.06 mol Cu 222 g Ag222 g Ag

Which has the larger mass?Which has the larger mass?

2.06 mol Cu x 63.5 g/mol = 130.8 g Cu2.06 mol Cu x 63.5 g/mol = 130.8 g Cu

222 g Ag > 130.8 g Cu222 g Ag > 130.8 g Cu

Which beaker has the larger number of atoms?Which beaker has the larger number of atoms?

222 g Ag ÷ 107.9 g/mol = 2.06 mol Ag222 g Ag ÷ 107.9 g/mol = 2.06 mol Ag

SINCE both beakers have the same number of SINCE both beakers have the same number of moles, they have equal numbers of atoms.moles, they have equal numbers of atoms.

Chapter 3- Practice problemsChapter 3- Practice problems

1-1- U-235U-235

92 electrons92 electrons & & 92 protons92 protons

235-92 = 143 neutrons235-92 = 143 neutrons

2-2- U-238U-23892 electrons & 92 protons92 electrons & 92 protons

238-92 = 146 neutrons238-92 = 146 neutrons

3-3- C-14C-14

6 electrons & 6 protons6 electrons & 6 protons

14-6 = 8 neutrons14-6 = 8 neutrons

4-4- I-127I-127

53 electrons & 53 protons53 electrons & 53 protons

127-53 = 74 neutrons127-53 = 74 neutrons

5-5- K-41K-41 19 electrons & 19 protons; 41 – 19 = 22 19 electrons & 19 protons; 41 – 19 = 22 neutronsneutrons

#6)#6) 4.25 mol Na 4.25 mol Na

x 22.99 g/mol = 97.7 g Nax 22.99 g/mol = 97.7 g Na

#7)#7) 0.0013 mol Au 0.0013 mol Au

x 196.97 g/mol = 0.26 g Aux 196.97 g/mol = 0.26 g Au

#8)#8) 111.5 mol Ca 111.5 mol Ca

x 40.08 g/mol = 4469 g Cax 40.08 g/mol = 4469 g Ca

#9)#9) 2.5 mol C 2.5 mol C

x 12.01 g/mol = 30 g Cx 12.01 g/mol = 30 g C

#10)#10) 0.025 mol Ag 0.025 mol Ag

x 107.87 g/mol = 2.7 g Agx 107.87 g/mol = 2.7 g Ag

#11)#11) 100.3 g Ca 100.3 g Ca

÷ 40.07 g/mol = 2.503 mol Ca÷ 40.07 g/mol = 2.503 mol Ca

#12)#12) 72.0 g O 72.0 g O

÷ 16.00 g/mol = 4.50 mol O÷ 16.00 g/mol = 4.50 mol O

#13)#13) 0.06 g C 0.06 g C

÷ 12.01 g/mol = 0.005 mol C÷ 12.01 g/mol = 0.005 mol C

#14)#14) 5.4 g Au 5.4 g Au

÷ 196.97 g/mol = 0.027 mol Au÷ 196.97 g/mol = 0.027 mol Au

#15)#15) 3.449 x 103.449 x 101111 g He g He

÷ 4.00 g/mol = 8.62 x 10÷ 4.00 g/mol = 8.62 x 101010 mol mol HeHe

Chapter 3 ReviewChapter 3 Review

Do problems #2, 6-11, Do problems #2, 6-11, 17-19, 21-24, & 28 on 17-19, 21-24, & 28 on pages 89 & 90 of the pages 89 & 90 of the textbook.textbook.

Do the Math Tutor Do the Math Tutor problems #1 & 2 on problems #1 & 2 on page 92.page 92.

Do the Standardized Do the Standardized Test Prep on page 93.Test Prep on page 93.

Chapter 3 vocabularyChapter 3 vocabulary

law of conservation of masslaw of conservation of mass law of definite law of definite proportions- proportions-

law of multiple proportions- law of multiple proportions- atom- atom-

atomic nucleus- atomic nucleus- neutron- neutron-

******** ******** electron cloud- electron cloud-

electron- electron- protons- protons-

neutrons- neutrons- nuclear forces- nuclear forces-

atomic number- atomic number- mass number-.mass number-.

isotopes- isotopes- atomic mass unit atomic mass unit

average atomic mass-average atomic mass- mole mole

Avogadro’s number- Avogadro’s number- molar mass- molar mass-

Chemistry Chapter 3 TestChemistry Chapter 3 Test30 multiple choice Questions:30 multiple choice Questions: definitions & uses of the Laws of Conservation of Mass, definitions & uses of the Laws of Conservation of Mass,

Definite Proportions, & Multiple ProportionsDefinite Proportions, & Multiple Proportions Dalton’s Atomic Theory: its 5 points & modificationsDalton’s Atomic Theory: its 5 points & modifications the cathode ray experiment & the discovery of electronsthe cathode ray experiment & the discovery of electrons Rutherford’s experiment & the discovery of the atomic Rutherford’s experiment & the discovery of the atomic

nucleusnucleus definitions of proton, neutron, electron, atomic nucleus definitions of proton, neutron, electron, atomic nucleus

(& its characteristics), nuclear forces, atomic number, (& its characteristics), nuclear forces, atomic number, mass number, isotopes, average atomic mass, mole, mass number, isotopes, average atomic mass, mole, molar mass, & Avogadro’s numbermolar mass, & Avogadro’s number

determine the number of protons, electrons, & neutrons determine the number of protons, electrons, & neutrons of an element from its atomic and mass numbersof an element from its atomic and mass numbers

mass to mole & mole to mass calculationsmass to mole & mole to mass calculations

Honors Chemistry Chapter 3 TestHonors Chemistry Chapter 3 Test

50 Multiple Choice questions50 Multiple Choice questions definitions & implications of the Laws of Conservation of Mass, definitions & implications of the Laws of Conservation of Mass,

Definite Proportions, Multiple Proportions, Dalton’s Atomic Definite Proportions, Multiple Proportions, Dalton’s Atomic Theory (and its modifications)Theory (and its modifications)

implications of the cathode ray experiment & the discovery of implications of the cathode ray experiment & the discovery of the electronthe electron

Rutherford’s experiment & the discovery of the atomic Rutherford’s experiment & the discovery of the atomic nucleusnucleus

description of an atom, atomic nucleus, & electron clouddescription of an atom, atomic nucleus, & electron cloud definitions & implications of atom, proton, neutron, electron, definitions & implications of atom, proton, neutron, electron,

nuclear forces, isotopes, average atomic mass, atomic nuclear forces, isotopes, average atomic mass, atomic number, mass number, Avogadro’s number, mole, molar massnumber, mass number, Avogadro’s number, mole, molar mass

describe the isotopes of hydrogen (protium, deuterium, describe the isotopes of hydrogen (protium, deuterium, tritium)tritium)

determine numbers of protons, electrons & neutrons from determine numbers of protons, electrons & neutrons from atomic & mass numbersatomic & mass numbers

mass to mole & mole to mass conversionsmass to mole & mole to mass conversions