SURVEY OF CHEMISTRY I

CHEM 1151

CHAPTER 2

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 2

ATOMS AND MOLECULES

THE ATOMIC THEORY OF MATTER

Law of Constant Composition- The relative numbers and kinds of atoms are constant in a given compound

- All samples of a given chemical compound have the same elemental composition

Example- Water (H2O) always contains 1 g of H for every 8 g of O

Law of Conservation of Mass (Matter)- The total mass of materials after a chemical reaction is equal to the total mass

before the chemical reaction

Law of Multiple Proportions- When two or more elements combine to form a compound, their masses in that

compound are in a fixed and definite ratio

- Elements combine in a ratio of small whole numbers

- If two elements form more than one compound, the ratios of the masses of the second element combined with a fixed mass of the first element will be in ratios

of small whole numbers

THE ATOMIC THEORY OF MATTER

Law of Multiple Proportions

- C and O can combine to form CO and CO2

CO1.33 g O combine with 1.00 g C

CO2

2.66 g O combine with 1.00 g C

- Ratio of O is 2.66 g : 1.33 g = 2 : 1

THE ATOMIC THEORY OF MATTER

Dalton’s Atomic Theory

1. All matter (every element) is made up of very small particles called atoms

- Atoms are indivisible and indestructible

2. All atoms of a given element are identical in mass and properties

- Atoms of a given element are different from atoms of all other elements

THE ATOMIC THEORY OF MATTER

Dalton’s Atomic Theory

3. Compounds are formed from a combination of two or more different kinds of atoms

- A given compound always has the same relative number and kind of atoms

4. A chemical reaction is a rearrangement of atoms - Atoms are neither created nor destroyed in a chemical reaction

THE ATOMIC THEORY OF MATTER

Modern atomic theory is more involved but based on Dalton’s theory

- Atoms can be destroyed by nuclear reactions but not by chemical reactions

- There are different kinds of atoms within an element (isotopes - different masses, same properties)

THE ATOMIC THEORY OF MATTER

THE ATOMIC STRUCTURE

Atom - Is the smallest particle of an element that retains the chemical

identity of the element- Is the basic building block of ordinary matter

- Made up of smaller particles (the building blocks of an atom) called subatomic particles

Three Types of Subatomic ParticlesElectron: possesses a negative (-) electrical chargeProton: possesses a positive (+) electrical charge

Neutron: has no charge (it is neutral)

THE ATOMIC STRUCTURE

Electronic Charge equals 1.602177 x 10-19 C(C = coulombs)

- Charges are usually expressed as multiples of the electronic charge

Charge of an electron = -1.602177 x 10-19 C = -1

Charge of a proton = +1.602177 x 10-19 C = +1

Atoms have no net electrical charge since they have equal number of electrons and protons

THE ATOMIC STRUCTURE

- Protons and neutrons have very large masses (about 2000 x) as compared to electrons

- Atoms generally have extremely small masses- Atomic Mass Unit (u) is used to express such small masses

1 u = 1.66054 x 10-24 g or 1 g = 6.02212 x 1023 u

Charge

Negative (-1)Positive (+1)Neutral (0)

Particle

ElectronProtonNeutron

Mass (g)

9.109 x 10-28

1.673 x 10-24

1.675 x 10-24

Mass (u)

5.486 x 10-4

1.00731.0087

Relative Mass

118371839

THE ATOMIC STRUCTURE

- The center of an atom is small, dense, and positively chargedcalled the nucleus

- The nucleus contains all protons and neutrons and are referred to as necleons

- The nucleus is, therefore, positively charged and contributes about 99.9% of the mass of an atom

THE ATOMIC STRUCTURE

- The electrons move rapidly around the nucleus

- Outer region called the extranuclear region

- Account for most of the volume of an atom

Electron Cloud - Volume occupied by electrons

- Negatively charged

ATOMIC NUMBER (Z)

- The number of protons in the nucleus of an atom- determines the identity of the element

- Since atoms have no net electrical chargenumber of protons = number of electrons

Z = number of protons = number of electrons

MASS NUMBER (A)

- The sum of the number of protons and the number of neutrons in the nucleus of an atom

-The total number of subatomic particles in the nucleus of an atom

- The number of nucleons of an atom

A = number of protons + number of neutrons

number of neutrons = mass number - atomic number = A - Z

ATOMIC AND MASS NUMBERSMASS NUMBER

ATOMIC NUMBERCHEMICAL SYMBOL

SYMBOLA

Z

C12

6O

16

8Ca

40

20

Mass number is the superscript to the leftAtomic number is the subscript to the left

An atom has an atomic number of 56 and a mass number of 138.What are the numbers of protons, electrons, and neutrons presentin the atom? What is the number of subatomic particles present

in the nucleus of the atom?

Number of protons = atomic number = 56Number of electrons = atomic number = 56

Number of neutrons = mass number – atomic number = 138-56 = 82Number of subatomic particles in the nucleus = mass number = 138

ATOMIC AND MASS NUMBERS

CHEMICAL PROPERTIES OF ATOMS

- The number of protons (the atomic number) characterizes an atom

- Electrons determine the chemical properties of an atom

- Atoms with the same atomic number have the same chemical properties

- Atoms with the same atomic number are atoms of the same element

Chapter 1 definition of An Element - Is a pure substance that cannot be reduced to a simpler substance

by normal chemical means

Chapter 2 definition of An Element- Is a pure substance in which all atoms present have the same

atomic number

CHEMICAL PROPERTIES OF ATOMS

ISOTOPES

- Atoms of an element with the same atomic number but different mass numbers

- Atoms of an element with the same number of protons and the same number of electrons but different numbers of neutrons

- Isotopes of an element have the same chemical propertiesbut slightly different physical properties

- The atomic number is usually omitted since it is the same for isotopes of a given element

ISOTOPES

12

6C

6 66

11 13 14

H1 2 3

1 1 1

Si28 29 30

14 14 14

Most abundant is carbon-12

Most abundant is silicon-28

Most abundant is hydrogen-1

C C C

H H

Si Si

AVERAGE ATOMIC MASS

- Determined by using the masses of an element’s various isotopes and their respective natural abundances

Units 1 u = 1.66054 x 10-24 g or 1 g = 6.02214 x 1023 u

u (amu): atomic mass unit

u is defined by assigning a mass of exactly 12 u to an atom of carbin-12 (reference point)

For an element with n isotopes which have atomic masses in u (m1, m2, m3,….., mn) and

natural abundances expressed as fractions (x1, x2, x3,……,xn)

Average Atomic Mass = m1x1 + m2x2 + m3x3 +….+ mnxn

The natural abundance is usually expressed as a percentage

Divide by 100 to convert to the decimal form (fractional abundance)

AVERAGE ATOMIC MASS

The mass spectrometer is an instrument used to measurethe masses and relative (natural) abundances of the

isotopes present in a sample of an element

HomeworkDescribe the operation and uses of the mass spectrometer

One page maximum and double spaced

AVERAGE ATOMIC MASS

Naturally occurring copper is 69.09% 63Cu, which has a relative mass of 62.93 u, and 30.91% 65Cu, which has a relative mass of

64.93 u. Calculate the average atomic mass of copper.

63Cunatural abundance = 69.09%

fractional abundance = 69.09/100 = 0.6909

65Cunatural abundance = 30.91%

fractional abundance = 30.91/100 = 0.3091

Average Atomic Mass = (62.93)(0.6909) + (64.93)(0.3091) = 63.5478 = 63.55 u

AVERAGE ATOMIC MASS

FORMULA MASS

- The sum of atomic masses of all the atoms present in the chemical formula of a substance

- Relative mass based on the carbon-12 relative-mass scale

-It is advisable to use two decimal places for atomic masses

Calculate the formula mass of H2SO4

H: 2 x 1.01 u = 2.02 uS: 1 x 32.06 u = 32.06 uO: 4 x 16.00 u = 64.00 u

Formula mass = (2.02 + 32.06 + 64.00) u = 98.08 u

FORMULA MASS

Calculate the formula mass of H2OH: 2 x 1.01 u = 2.02 u

O: 1 x 16.00 u = 16.00 uFormula mass = (2.02 + 16.00) u = 18.02 u

Calculate the formula mass of Fe2(SO4)3

Fe: 2 x 55.85 u = 111.70 uS: 3 x 32.07 u = 96.21 u

O: 12 x 16.00 u = 192.00 uFormula mass = (111.70 + 96.21 + 192.00) u = 399.91 u

FORMULA MASS

Calculate the formula mass of CaCO3

Ca: 1 x 40.08 u = 40.08 uC: 1 x 12.01 u = 12.01 uO: 3 x 16.00 u = 48.00 u

Formula mass = (40.08 + 12.01 + 48.00) u = 100.09 u

THE MOLE

The amount of substance of a system, which contains as manyelementary entities as there are atoms in 12 grams of carbon-12

- abbreviated mol

1 mole (mol) = 6.02214179 x 1023 entities

- known as the Avogadro’s number (after Amedeo Avogadro)

- usually rounded to 6.022 x 1023

THE MOLE

The number of entities (or objects) can be atoms or molecules

1 mol C = 6.022 x 1023 atoms C

1 mol CO2 = 6.022 x 1023 molecules CO2

2 conversion factors can be derived from each

THE MOLE

How many atoms are there in 0.40 mole nitrogen?

= 2.4 x 1023 nitrogen atoms

How many molecules are there in 1.2 moles water?

= 7.2 x 1023 water molecules

nitrogenmol1

atomsnitrogen10x6.022xnitrogenmol0.40

23

watermol1

moleculeswater10x6.022xwatermol1.2

23

How many H atoms are there in 1.2 moles water?

= 1.4 x 1024 H atoms

molecule)water(1

atoms)H(2x

water)mol(1

molecules)water10x(6.022xwatermol1.2

23

THE MOLE

MOLAR MASS

- The mass of a substance in grams that is numerically equal tothe formula mass of that substance

- Add atomic masses to get the formula mass (in u) = molar mass (in g/mol)

- The mass, in grams, of 1 mole of the substance

MOLAR MASS

Consider the following

Sodium (Na) has an atomic mass of 22.99 uThis implies that the mass of 1 mole of Na = 22.99 g

Molar mass of Na = 22.99 g/mol

Formula mass of NaCl = 58.44 uThe mass of 1 mole of NaCl = 58.44 g

Molar mass of NaCl = 58.88 g/mol

Formula mass of CaCO3 = 100.09 uThe mass of 1 mole of CaCO3 = 100.09 g

Molar mass of CaCO3 = 100.09 g/mol

MOLAR MASS

Calculate the mass of 2.4 moles of NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)

= 85.00 g /mol NaNO3

= 204 g NaNO3

= 2.0 x 102 g NaNO3

3

333 NaNOmol1

NaNOg85.00xNaNOmol2.4NaNOg

MOLAR MASS

How many moles are present in 2.4 g NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)

= 85.00 g /mol NaNO3

= 0.028 mol NaNO3

= 2.8 x 10-2 mol NaNO3

3

333 NaNOg85.00

NaNOmol 1xNaNOg2.4NaNOmol

CHEMICAL FORMULA

Subscripts represent both atomic and molar amounts

Consider Na2S2O3:

- Two atoms of sodium, two atoms of sulfur, and three atoms ofoxygen are present in one molecule of Na2S2O3

- Two moles of sodium, two moles of sulfur, and three moles ofoxygen are present in one mole of Na2S2O3

CHEMICAL FORMULA

How many moles of sodium atoms, sulfur atoms, and oxygenatoms are present in 1.8 moles of a sample of Na2S2O3?

I mol Na2S2O3 contains 2 mol Na, 2 mol S, and 3 mol O

Namol3.6OSNamol1

Namol2xOSNamol1.8Namol

322322

Smol3.6OSNamol1

Smol2xOSNamol1.8Smol

322322

Omol5.4OSNamol1

Omol3xOSNamol1.8Omol

322322

CHEMICAL CALCULATIONS

Calculate the number of molecules present in 0.075 g of urea,(NH2)2CO

Given mass of urea: - Convert to moles of urea using molar mass

- Convert to molecules of urea using Avogadro’s number

= 7.5 x 1020 molecules (NH2)2CO

CO)(NHmole1

CO)NH(molecules10x6.022x

CO)(NHg60.07

CO)(NHmole1xCO)(NHg0.075

22

2223

22

2222

CHEMICAL CALCULATIONS

How many grams of carbon are present in a 0.125 g of vitamin C,C6H8O6?

Given mass of vitamin C: - Convert to moles of vitamin C using molar mass

- Convert to moles of C (1 mole C6H8O6 contains 6 moles C)- Convert moles carbon to g carbon using molar mass

= 0.0511 g carbon

Cmol1

Cg12.01x

OHCmol1

Cmol6x

OHCg176.14

OHCmol1xOHCg0.125

686686

686686

PERCENTAGE COMPOSITION

- Percentage by mass contributed by individual elements in a compound

100%xcompoundofmass

element of masselement%

100%xcompoundofmassformula

element)ofatomsofumberelement)(nofmass(atomicelement%

PERCENTAGE COMPOSITION

Calculate the percentage of carbon, hydrogen, and oxygen, inethanol (C2H5OH)

% 13.13100%xu 46.07

u)(6) (1.01H%

% 73.34100%xu 46.07

u)(1) (16.00O%

% 52.14100%xu 46.07

u)(2) (12.01C%

PERCENTAGE COMPOSITION

Calculate the percent composition by mass of each elementin the following compounds

C9H8O4

(NH4)2PtCl4

C2H2F4

C8H10N4O2

Pt(NH3)2Cl2

EMPIRICAL FORMULA

Given mass % elements:

- Convert to g elements assuming 100.0 g sample

- Convert to mole elements using molar mass

- Calculate mole ratio (divide each by the smallest number of moles)

- Round each to the nearest integer

- Multiply through by a suitable factor if necessary( __.5 x 2 or __.33 x 3 or __ .25 x 4)

EMPIRICAL FORMULA

Determine the empirical formula for a compound that gives the following percentages upon analysis (in mass percents):71.65 % Cl 24.27 % C 4.07 % H

- Assume 100.0 g of sample and convert grams to moles

Clmol2.021Clg35.45

Clmol1xClg71.65

Cmol2.021Cg12.01

Cmol1xCg24.27

Hmol04.4 Hg 1.01

Hmol1xHg07.4

71.65 g Cl

24.27 g C

4.07 g H

EMPIRICAL FORMULA

1.0002.021

2.021:Cl

1.0002.021

2.021:C

- Calculate mol ratios

99.12.021

4.04:H

- Round to nearest integers and write empirical formula

Cl: 1, C: 1, H: 2 giving CH2Cl