chemistry 161 chapter 2

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CHEMISTRY 161 Chapter 2

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CHEMISTRY 161 Chapter 2. MATTER. ATOMS. John Dalton matter is composed of ‘building blocks’. MOLECULES ELEMENTS. Dalton’s postulates Elements are composed of small particles called atoms; atoms of an element are identical with same properties (mass) - PowerPoint PPT Presentation

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Page 1: CHEMISTRY 161  Chapter 2

CHEMISTRY 161

Chapter 2

Page 2: CHEMISTRY 161  Chapter 2

MATTER

ATOMS

MOLECULES

ELEMENTS

John Dalton

matter is composed of

‘building blocks’

Page 3: CHEMISTRY 161  Chapter 2

Dalton’s postulates

1. Elements are composed of small particles called atoms;atoms of an element are identical with same properties (mass)

2. Compounds are composed of atoms of more than one element; in a compound, the ratio of the number of atoms is an integer or fraction

LAW OF DEFINITE PROPORTIONS

3. A chemical reaction involves separation and combination of atoms; atoms are neither created nor destroyed

an atom is the basic, undividable unit of any element

Page 4: CHEMISTRY 161  Chapter 2

1. Law of definite proportions

2. Law of conservation of mass

in a chemical reaction

no gain or loss of mass is observed

unit of mass is g (gram) or kilogram (kg)

each atom has a constant mass

atomic mass – atomic weight

Page 5: CHEMISTRY 161  Chapter 2

Structure of an Atom

subatomic particles

electrons

protons neutrons

Page 6: CHEMISTRY 161  Chapter 2

proton,p 1.67262 × 10-27 kg +1.6022 × 10-19 C +1

electron,e 9.10939 × 10-31 kg -1.6022 × 10-19 C -1

m(p) / m(e) ≈ 1836

the atom’s positive charge is located in a small, dense central core ‘nucleus’

protons are a constituent of the nucleus

Page 7: CHEMISTRY 161  Chapter 2

+ ++ +

+

1. mass of the nucleus constitutes most of the mass of the atom

2. the nucleus is positively charged and contains protons

3. the nucleus constitutes only about 1/1013 space of an atom

4. electrons are as clouds around the nucleus

Page 8: CHEMISTRY 161  Chapter 2

proton, p 1.67262 × 10-27 kg +1.6022 × 10-19 C +1

electron, e 9.10939 × 10-31 kg -1.6022 × 10-19 C -1

neutron, n 1.67493 × 10-27 kg 0 0

Neutron

m(n) / m(e) ≈ 1838

m(n) > m(p)

Page 9: CHEMISTRY 161  Chapter 2

Structure of an Atom

subatomic particles

electrons(‘cloud’)

protons(nucleus)

neutrons(nucleus)

neutrons are the ‘glue’ of the nucleus

Page 10: CHEMISTRY 161  Chapter 2

Atomic Number and Mass Number(PSE)

AZXatomic number

(number of protons)

(number of electrons)

mass number

(number of protons plus neutrons)

in an atom, the number of electrons and protons are identical (charge neutrality)

Page 11: CHEMISTRY 161  Chapter 2

136C

6 electrons 6 protons

12 protons plus neutrons6 neutrons 126C

13 protons plus neutrons7 neutrons

6 electrons 6 protons

elements with same number of electrons and protons but different number of neutrons

ISOTOPES

Page 12: CHEMISTRY 161  Chapter 2

11H

31H

21H

Page 13: CHEMISTRY 161  Chapter 2

168O

11H

126C

Page 14: CHEMISTRY 161  Chapter 2

Periodic Table of the Elements

period

group

Page 15: CHEMISTRY 161  Chapter 2

Periodic Table of the Elements

8 main groupsmetals

nonmetalsmetalloids (semi metals)

metals – shine, conduct electricitynonmetals – do not shine, do not conduct electricitymetalloids – properties between metals and nonmetals

10 transition metal groups

lanthanides/actinides

metals

metals

Page 16: CHEMISTRY 161  Chapter 2

HOW HEAVY ARE ATOMS?

CALCULATION

Page 17: CHEMISTRY 161  Chapter 2

CHEMICAL MASS SCALE

standard / calibration

atomic mass unit (amu, u)

one atom of carbon-12 12 u (exactly)

we have to correlate u with kg

Page 18: CHEMISTRY 161  Chapter 2

MOLE

Page 19: CHEMISTRY 161  Chapter 2

one mole of a compound contains the same number of molecules/atoms as the number of atoms

in exactly 12 g of 12C

Avogadro’s numberNa

6.023 x 1023

Page 20: CHEMISTRY 161  Chapter 2

1 mole of H2O 6.023 x 1023

1 mole of 12C

6.023 x 1023

6.023 x 10231 mole of NaCl

1 mole of Na

6.023 x 1023

Avogadro’s numberlinks micro and

macroscopic world

molecules

molecules

atoms

atoms

Page 21: CHEMISTRY 161  Chapter 2

H2O

O: 15.999 u

H: 1.008 u

H: 1.008 u

H2O: 18.015 u

PSE

formula mass: weight of one moleculecorrelation between u and kg

Page 22: CHEMISTRY 161  Chapter 2

H2OO: 15.999 u

H: 1.008 u

H: 1.008 u

H2O: 18.015 u

1 mole of H2O – 18.015 g

15.999 g/mol

1.008 g/mol

1.008 g/mol

Na

Na

Na

Page 23: CHEMISTRY 161  Chapter 2

CaO

O: 15.999 u

Ca: 40.08 u

CaO: 56.08 u

1 mole of CaO – 56.08 g

Page 24: CHEMISTRY 161  Chapter 2

2 H2 + O2 2 H2O

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

4.03176g 31.9988g 36.03g

STOICHIOMETRY

x g y g 70.0g

Page 25: CHEMISTRY 161  Chapter 2

2 H2 + O2 2 H2O

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

4.03176g 31.9988g 36.03g

STOICHIOMETRY

x g y g 70.0g

Page 26: CHEMISTRY 161  Chapter 2

Example I: How many grams of iron are in a 15.0 g

sample of iron(III) oxide?

1. molecular formula

3. weight of one mole Fe2O3

2. weight of one molecule

Fe2O3

159.7 u

159.7 g4. 1 molecule Fe2O3 contains 2 atoms of Fe

5. 1 mole Fe2O3 contains 2 moles of Fe

159.7 g 111.69 g

15.0 g x gx = 10.5 g

Page 27: CHEMISTRY 161  Chapter 2

Example II:

How many atoms are in 10 kg of sodium?

1 mole sodium = 22.98977 g

6.023 x 1023 atoms = 22.98977 g

x atoms = 10,000 g

x = 2.6 x 1026 atoms

Page 28: CHEMISTRY 161  Chapter 2

Example III

How heavy are 1 million gold atoms?

1 mole gold = 196.96654 g

6.023 x 1023 atoms = 196.96654 g

1,000,000 atoms = x g

x = 3.2 x 10-16 g = 0.32 fg

Page 29: CHEMISTRY 161  Chapter 2

MOLEAvogadro’s number

Na

6.023 x 1023