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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