Chapter 1~ The Atomic Theory and Moles ~

Miss Faridah Bt. Abu Bakarhttp://www.formspring.me/MissFaridah

Classification of Matter

Atom• The smallest object that retain the properties of an element. Composed of electrons and a nucleus (containing protons and neutrons).

Elements & Compound

Separation by

Physical method

Separation by chemical method

The Structure of the Atom

Electrically neutral particles having a mass

slightly greater than that of proton

Positively charged particles in the nucleus

Electron

Nucleus

Proton

Neutron

Atomic Number, Mass Number and Isotopes

All atom can be identified by the number of protons and neutron they contain

XAZ

Mass Number (A)The total number of neutrons and protons present in the nucleus of an atom of an element

Atomic Number (Z)The number of proton in the nucleus of each atom of an element

• Atomic Number (Z) = proton (p)• Proton (p) = Electron (e)

The atomic number also indicates the number of electrons present in the atom

H H H1

1

2

1

3

1

Hydrogen Deuterium Tritium

Isotopes

Molecule and Ion

Example Example

Loosing electron Accepting electron

Average Atomic Mass

6C

12.01

Atomic Number

Atomic Mass

Example

The atomic masses of copper of its two stable isotopes are as below:

Cu6329 Cu65

29(69.09 %) (30.91 %)

62.93 amu 64.9278 amu

Calculate the average atomic mass of copper.

Answer

Percent are converted to fraction

69.09/100 = 0.6909 30.91/100 = 0.3091

[ (0.6090) (62.93 amu) ] + [ (0.3091) (64.9278) ] = 63.55 amu

Avogadro’s Number & the Molar Mass of an Element

Mass of element (m)

Number of moles of

element (n)

Number of atoms of

element (N)

m/M nNA

N/NAnM

Avogadro’s Number (NA) = 6.0221367 X 1023

Where…1 mole carbon-12 atom = 6.0221367 X 1023 carbon-12 atom

1 g = 6.0221367 X 1023 amu1 amu = 1.661 X 10-24 g

Number of mole

Number of mole = mass of X (g)Relative Molar Mass of X (gmol-1)

Example

How many moles of He atoms are in 6.46 g of He?

4.003 g He 1 mole He1 g He 0.2498 mole He6.46 g He (0.2498) X (6.46)

= 1.61 mol He

Example

Sulfur (S) is a nonmetallic element that is present in coal. When coal is burned, sulfur is converted to sulfur dioxide and eventually to sulfuric acid that gives rise to acid rain phenomenon. How many atoms are in 16.3 g of sulfur?

Where 1 mol of S = 32.07 g S

Answer

32.07 g 1 mol16.3g [(1/32.07) X (16.3)]

= 0.5082 mol

1 mol NA (6.022 X 1023) S atom0.5082 mol 3.060 X 1023 S atoms

Molecular Mass

Example

H2O = 2 (atomic mass of H) + atomic mass of O

= 2 (1.008 amu) + 16.00 amu

= 18.02 amu

Percent Composition of Compounds

Mass spectrometer

n X molar mass of elementMolar mass of compound X 100 %

Empirical Formular

Procedure for calculating the empirical formular of a compound from it’s percent composition

Example

Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, tooth-paste and carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P and O in this compound.

Atomic mass

H = 1.008 P = 30.97 O= 16.00

Answer

The molar mass of H2PO4 is 97.99 g. The percent by mass of each of the elements are as follows;

% H = 3 (1.008 g) H 97.99 g H2PO4

% P = 30.97 g P 97.99 g H2PO4

% O = 4 (16.00 g) O 97.99 g H2PO4

X 100 % = 3.086 %

X 100 % = 31.61 %

X 100 % = 65.31 %

Example

Chalcopyrite (CuFeS2) is a principle mineral of copper. Calculate the number of kilograms of Cu in 3.71 X 103 kg of chalcopyrite.

Relative Molecular mass

Cu = 63.55Fe = 55.85S = 32.07

Answer

Calculate the molar mass

Cu = 63.55 g

CuFeS2 = (63.55) + (55.85) + [2(32.07)]= 183.5 g

Therefore

% Cu = molar mass of Cu molar mass of CuFeS2

X 100 %

% Cu = molar mass of Cu molar mass of CuFeS2

= 63.55 g 183.5 g

= 34.63%

Therefore

Mass of Cu in CuFeS2 = [(34.63/100) X (3.71 X 103)]= 1.28 X 103 kg

X 100%

X 100%

Chemical FormulaChemist used the chemical formula to express the composition of

molecules and ionic composition in terms of chemical symbol

Example

Benzene

Empirical Formula : CH

Molecular Formula : C6H6

Structural Formula :

Example

Write the empirical formula for glucose (C6H12O6)

Answer

From the molecular formula there are 6 carbon atoms, 12 hydrogen atom and 6 oxygen atom. Dividing the subscript by 6, we will obtain the formula CH2O.

Eventhough we might divide the subscript by 3 and gain C2H4O2, but this is not the empirical formula because its subscript are not in the smallest whole-number ratio

Empirical Formula

Determination of Empirical and Molecular Formulas

•The formula calculated from percent composition by mass is always the empirical formula because the coefficients in the formula are always reduced to the smallest whole number.

•Therefore, to calculate the actual molecular formula we must know the approximate molar mass of the compound in the addition to its empirical formula.

Divided by molar mass

Multiply by stoichiometric ratio

Multiply by molar mass

Example

A sample of a compound contains 1.52 g of nitrogen (N) and 3.47 g of oxygen (O). The molar mass of this compound is between 90 g to 95 g. Determine the molecular formula of the compound

Relative Molecular Mass

N = 14.01 gO = 16.00 g

Answer

N

14.01 g N 1 mole of N1.00 g N [(1/14.01)]

= 0.07 mole of N

Therefore

1.52 g N 0.07 X 1.52= 0.106 mol of N

All the subscript will be divided by the smallest number to gain the smallest

whole-number ratio

O

16.00 g O 1 mole of O1 g O (1/16.00)

= 0.06 mole of O

Therefore

3.47 g O 0.06 X 3.47= 0.208 mole of O

We may write the formula as

= N0.106O0.208

= NO2

From the empirical formula, we may calculate the empirical molecular mass

NO2

= 14.01 g + [2(16.00)]= 46.01 g

Given in the question that the molar mass of the compound is around 90 to 95 g. Therefore the molecular formula is

Molar mass = 90 gEmpirical molar mass 46.01 g

= 2

The molecular formula = (NO2 ) 2

Formula of Ionic Compounds•The formula of ionic compounds are usually the same as their empirical formulas because ionic compounds do not consist of discrete molecular units.

•Basically, in order for ionic compounds to be electrically neutral, the sum of the charges on the cation and anion in each formula unit must be zero.

•The subscript of the cation is numerically equal to the charge on the anion and the subscript of the anion is numerically equal to the charge on the cation

Potassium Bromide

The potassium cation K+ and the bromine anion Br- combine to form the ionic compound potassium bromide. The sum of the charges is [(+1)+(-1)] = 0. Therefore, no subscript is necessary.

The formula is KBr

Zinc Iodide

The zinc cation Zn2+ and the iodine anion I- combine to form zinc iodide. The sum of charges of 1 Zn2+ ion and one I- ion is as below

[(+2) + (-1)] = +1

Therefore to make the charges add up to zero, we multiply the -1 charge of the anion by 2 and add the subscript “2” to the symbol for Iodine.

The formula for zinc iodide is ZnI2

Aluminum Oxide

The cation is Al3+ and oxygen anion O2- . The following diagram helps to determine the subscript for the compound formed by the cation an the anion

Al3+ O2-

Al2O3

The sum of the charge is [2(+3)] + [3(-2)] = 0. Therefore the formula for Aluminum oxide is Al2O3

Oxidation State (Oxidation Number)For element and monoatomic ion, the oxidation number is the same as the charge.

There are rules that have to be followed as below.

Rule 1

The oxidation number of the atom in an element is zero.

For example, in the reaction below

3 CuCl2 + 2 Al 2 AlCl3 + 3 Cu

Oxidation number = 0

Rule 2

The oxidation number in the monoatomic ion equals to the charge on the ion.

The Cu2+ and Cl- in copper (II) chloride have oxidation number of 2+ and -1 respectively.

Cu2+ Cl- Cl-

3 CuCl2 + 2 Al 2 AlCl3 + 3 Cu

Therefore: [(+2) + (-1) + (-1)] = 0

Rule 3

The oxidation number of oxygen is -2 except for peroxide which is -1

1.Sodium oxide, Na2O contains O2- ions. Therefore the oxidation number will be -2.

2.Magnesium oxide, MgO contains O2- ions. Therefore the oxidation number will be -2.

3.Iron (III) oxide, Fe2O3 contains O2- ions. Therefore the oxidation number will be -2.

4.Cesium peroxide, Cs2O2 contains O22- ions. Therefore the oxidation

number will be -1.

Rule 4

The oxidation number of Hydrogen is +1 in most of its compound (except in binary compound with metal such as NaH, where the oxidation number of -1.

1.Hydrochloric acid, HCl contains H+ ions. Here the oxidation number is +1.

2.Sulfuric acid, H2SO4 contains H+ ions. Here the oxidation number is +1.

3.Calcium hydrate, CaH2 contains H- ions. Here the oxidation number is -1.

Rule 5

The oxidation number of Fluorine is always -1. All other halogens (Cl, Br, I) have the oxidation number of -1 in binary compound with some exception.

1.Calcium Floride, CaF2 contains F- ions. Here the oxidation number of F is equal to -1.

2.Nickel (II) bromide, NiBr2 contains Br- ions. Here the oxidation number of Br is equal to -1.

Rule 6

The sum of oxidation numbers of the atom in a compound is zero.

The sum of oxidation number of atom in the polyatomic ion equals to the charge on the ion

-1-2-3+3+1 +2

Chemical Reaction and Chemical Equations

• The substance formed as a result of a chemical reactionProduct

Balancing Chemical EquationsIn general, we can balance a chemical equation by following step:

1. Identify all reactants and products and write their correct formulas on the left and the right side of the equation respectively

Reactant Product

2. Begin balancing the equation by trying different coefficient to make the number of atoms of each element the same on both sides of the equation.

Note that we can change the coefficient (the numbers preceding the formulas) but not the subscript as it will

change the identity of the substance!

3. Look for elements that appear only once on each side of the equation with the same number of atoms on each side. The formulas containing these elements must have the same coefficient.

4. Next, look for elements that appear only once on each side of the equation but in unequal numbers of atom. Balance these elements.

5. Finally, balance the elements that appear in two or more formulas on the same side of the equation.

Example

When potassium chlorate was heated , it will produce oxygen and potassium chloride. The chemical equation for this reaction can be written as follows.

The chemical formula for each compound as as follows

Oxygen : O2

Potassium Chloride : KClPotassium Chlorate : KClO3

Therefore…Reactant Product

KClO3 O2 + KCl

KClO3 O2 + KCl

1. Because there are 3 O atoms on the left and 2 on the right of the equation, we can balance it by placing a 2 in front of KClO3 and a 3 in front of O2

2. Finally, the K and Cl atom is balanced by placing 2 in front of the KCl

2 3 2

Reactant ProductK (2) K (2)Cl (2) Cl (2)O (6) O (6)

Example

When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. Write a balance equation for the formation of Al2O3.

The chemical formula for each compound are as follows;

Aluminum : AlAir : O2

Aluminum Oxide : Al2O3

Therefore… Reactant Product

Al + O2 Al2O3

Al + O2 Al2O3 2 2

32

4 Al + 3 O2 2 Al2O3

Reactant ProductAl (4) Al (4)O (6) O (6)

Amount of Reactant & Product

Example

All alkali metal react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is between lithium and water

2 Li + 2 H2O 2 LiOH + H2

(a) How many moles of H2 will be formed by the complete reaction of 6.23 moles of Li with water? (b) How many grams of H2 will be formed by the complete reaction of 80.57 g of Li in the water

Answer (a)

From the given chemical equation

2 mol of Li 1 mol of H2

Therefore..

1 mol of Li 0.5 mol of H2

6.23 mol of Li (0.5 X 6.23) mol of H2

= 3.115 mol of H2

Answer (b)

6.941 g of Li 1 mol of Li1 g of Li 0.144071459 mol of Li

Therefore

80.57 g of Li (0.144071459 X 80.57)= 11.60783749 mol of Li

From the equation

1 mol of Li 0.5 mol of H2

11.60783749 mol of Li 5.803918744 mol of H2

1 mol of H2 (2 x 1.008) g of H2

= 2.016 g of H2

Therefore

5.803918744 mol of H2 (2.016 X 5.803918744) g of H2

= 11.70 g of H2

Types of Stoichiometric Calculation

Limiting Reagent

Example

Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide

2 NH3 + CO2 (NH2)2CO + H2O

In one process, 637.2 g of NH3 are treated with 1142 g of CO2.

(a)Which of the two reactant is the limiting reagent? (b)Calculate the mass of (NH2)2CO formed. (c)How much excess reagent (in grams) is left at the end of the reaction?

Answer (a)

2 NH3 + CO2 (NH2)2CO + H2O

Calculate the number of moles for NH3 and CO2

Number of mole of NH3 =

=

= 37.4075 moles

massmolarrelative

mass

17.034

637.2

Number of mole of CO2 =

=

= 25.9486 moles

Therefore….

1 mole of NH3 0.5 mole of (NH2)2CO 37.4075 moles of NH3 (0.5 X 37.4075)

= 18.7037 moles of (NH2)2CO

1 mole of CO2 1 mole of (NH2)2CO 25.9486 mole of CO2 (1 X 25.9486)

= 25.9486 moles of (NH2)2CO

massmolarrelative

mass

01.44

1142

From the calculation we may see that the NH3 is the limiting reagent because it produce a smaller amount of (NH2)2CO.

Answer (b)

Since the ammonia is the limiting reagent, therefore, the number amount of (NH2)2CO produce are highly dependent on it.

From the calculation 18.7037 moles of (NH2)2CO will be produce.

Therefore….

Number of mole =

18.7037 mole =

Mass = 18.7037 X 60.062

= 1123.3816 g

massmolarrelative

mass

062.60

mass

Answer (c)

Since only 18.7037 moles of (NH2)2CO were produce during the reaction, therefore the number of mole of CO2 used in the reaction is 18.7037 moles .

Therefore…

Number of mole CO2 =

18.7037 =

mass = 823.1498 g

massmolarrelative

mass

01.44

mass

Given in the question, the initial amount of CO2 used in the reaction is 1142 g.

From the previous calculation, the final mass of CO2 used in the reaction is 823.1498 g.

Therefore, the mass of excess CO2 is

= (1142 – 823.1498)g= 318.8502 g of excess CO2

Reaction Yield

%100xyieldltheoretica

yieldactual

Example

Titanium is a strong, lightweight, corrosion-resistant metal that is used in rocket, aircraft, jet engines and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950 °C and 1150 °C.

TiCl4 + 2 Mg Ti + 2 MgCl2

In a certain industrial operation 3.54 X 107 g of TiCl4 are reacted with 1.13 X 107 g of Mg.

(a)Calculate the theoretical yield of Ti in grams(b)Calculate the percent yield if 7.91 X 106 g of Ti are actually obtained.

Answer (a)

Number of mole of TiCl4 =

=

= 1.02 x 105 moles of TiCl4

Number of mole of Mg =

=

= 4.65 x 105 moles of Mg

massmolarrelative

mass

2.346

1054.3 7x

massmolarrelative

mass

31.24

1013.1 7x

1 mole TiCl4 1 mole Ti1.02 x 105 moles of TiCl4 1.02 x 105 moles of Ti

2 moles of Mg 1 mole of Ti1 mole of Mg 0.5 moles of Ti4.65 x 105 moles of Mg 2.33 x 105 moles of Ti

Therefore the limiting reactant is TiCl4

Mass of Ti produce

Number of mole of Ti

1.02 x 105 moles of Ti

Mass of Ti 2.08 X 107 g of Ti

massmolarrelative

mass

4.204

mass

Answer (b)

% of yield =

=

= 38.03 %

%100xyieldltheoretica

yieldactual

%10010 x 2.08

1091.77

6

xx

Solution & Their Concentrations

Molar concentration (Cx)

Molar concentration (Cx) =

OR

Analytical Molarity

~ The total number of moles of a solute in 1 L of solution (or the total number of milimoles in 1 mL).

solutionLno

solutemolno

.

.

solutionmLno

solutemmolno

.

.

Example

Calculate the molar concentration of ethanol in an aqueous solution that contains 2.30 g of C2H5OH (46.07 g/mol) in 3.50 L of solution.

Answer

The number of mole of C2H5OH

=

=

= 0.04992 mol C2H5OH

RMM

mass

07.46

30.2

Molarity

C C2H5OH =

=

= 0.014 M

solutionofLofno

soluteofmoleofno

.

.

50.3

04992.0

Example

Describe the preparation of 2.00 L of 0.108 M of BaCl2 from BaCl2· 2 H2O (244.3 g/mol).

Answer

Molarity =

0.108 M =

Number of mole = 0.216 mole of BaCl2· 2 H2O

Lofnumber

moleofnumber

L

moleofnumber

2

Therfore…

Number of mole =

0.216 mole =

mass = 52.8g of BaCl2· 2 H2O

The preparation..

Dissolve 52.8 g of BaCl2· 2 H2O in water and dilute to 2 L

RMM

mass

3.244

mass

Dilution of Solution

Because molarity is define as moles of solute in one liter of solution, we see that the number of moles of solute is given by

soluteofmoleslitersinsolutionofvolumexsolutionofliters

soluteofmole)(

M V

soluteofmoleslitersinsolutionofvolumexsolutionofliters

soluteofmole)(

The equation above can also be written as

MV = moles of solute

Because all the solute comes from the original stock solution, therefore we may conclude that

M1V1 = M2V2

Moles of solute before

dilution

Moles of solute after

dilution

Example

What is the volume of H2SO4 needed to prepair 5.00 X 102 mL of a 1.75 M H2SO4 solution, starting with an 8.61 M stock of H2SO4

Answer

M1= 8.61 M V1 = ????M2= 1.75 M V2 = 5.00 X 102

M1V1 = M2V2

(8.61) (V) = (1.75) (5.00 X 102 )

8.61 V = 875

V = 1.02 X 102 mL/ 102 mL

Acid-Base Titration

Example

In a titration experiment, a student finds that 23.48 mL of NaOH solution are needed to neutralized 0.5468 g of KHC8H4O4. What is the concentration (in molarity) of NaOH ssolution?

solutionofL

NaOHofmolNaOHofmolarity

Need to calculate

Need to calculate

Given

The reaction will be as follows

KHC8H4O4 + NaOH KNaC8H4O4 + H2O

From the equation, 1 mole of KHC8H4O4 neutralized 1 mole of NaOH

Number of mole of KHC8H4O4 =

=

= 2.68 X 10-3 mol KHC8H4O4

RMM

mass

204

5468.0

Since 1 mole of KHC8H4O4 = 1 mol of NaOH…

Therefore…

2.68 X 10-3 mol KHC8H4O4 = 2.68 X 10-3 mol NaOH

Molarity of NaOH =

=

= 0.1141 mol NaOH/L

= 0.1141 M

solutionL

moleofno

L0.02348

NaOH mol 10 X 2.68 -3

Example

How many milimeters (mL) of a 0.610 M NaOH solution are needed to neutralized 20.0 mL of a 0.245 M H2SO4 solution?

Answer

The chemical equation for the reaction are as follows

2 NaOH + H2SO4 Na2SO4 + 2 H2O

From the equation, 1 mole of H2SO4 is ≈ 2 mole NaOH.

The number of mole for H2SO4 is

The number of mole of H2SO4 = 0.245 X 0.02

= 4.9 X 10-3 mol H2SO4

Therefore, the number of mole of NaOH is

= (4.9 X 10-3 mol) X 2= 9.8 X 10-3 mol of NaOH

L

molemolarity

Molarity of NaOH

L

molemolarity

NaOHmL

NaOHLL

x

1.16

01606.0

1080.9610.0

3

Percent concentration

The 3 common methods are:

%100)/( x

solutionweight

soluteweightwwpercentweight

%100)/( xsolutionvolume

solutevolumevvpercentvolume

%100)(

)()/(/ x

mLsolutionvolume

gsoluteweightvwpercentvolumeweight

Part per Million and Parts per Billion

For very dilute solutions, parts per million (ppm) is a convenient way to express concentration:

• Where Cppm is the concentration in parts per million.

• For even more dilute solutions, 109 ppb is employed to give the result in parts per billion (ppb)

ppmxgsolutionofmass

gsoluteofmassC ppm

610)(

)(

A handy rule in calculating parts per million is to remember that for dilute aqueous solution whose densities are approximately 1.00 g/mL therefore 1 ppm is equal to 1 mg/L.

910)(

)(

)(

)(

xgsolutionofmass

gsoluteofmassC

Lsolutionofmass

mgsoluteofmassC

ppb

ppm

Mole Fraction

The mole fraction of a component in a solution is simply the number of moles of that component divided by the total moles of all the components. The mole fraction of component i is written as Xi.

For a solution consisting of nA moles of component A, nB moles of component B, nC moles of component C, etc., then the mole fraction of component A is given by

Mole fractions are strictly additive. The sum of the mole fractions of all components is equal to one.

cBA

AA nnn

nX