Chapter 1 - Mole and Atomic Concepts

DAS 12102/12203 Chapter 1 - Mole and Atomic Concepts

MOLE AND ATOMIC CONCEPT 1-1 OBJECTIVES After completing this chapter, students will be able to:

distinguish the matter based on their classification and composition write the symbol for atom and the formula of chemical compound calculate the atomic mass from percent abundance and mass of atom

produced from mass spectrometer apply the concept of mole to determine the empirical and molecular formula apply the mole concept to calculate the quantities of reactant and products in

reaction stoichiometry. apply the mole concept in stoichiometry reaction for volumetric analysis in

titration reaction. 1-2 INTRODUCTION

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Matter can be described as anything that has mass and it must take up space. Depending on its temperature, matter can be a solid, liquid or gas.

Figure 1.1: Physical States of Matter.


1-3 CLASSIFICATION OF MATTER The classification of matter is based on:

physical properties such as solid, liquid or gases. the composition including atom, molecule, substance, element, compound or

mixtures. Atom

An atom is the smallest particle differentiable as a certain chemical element. When an atom of an element is divided, it ceases to be that element. The example of atoms are Sodium (Na), Iron (Fe), Gold (Au), Oxygen (O), Hydrogen (H), Chlorine (Cl) and Carbon (C).

Molecule A molecule is the smallest indivisible portion of a pure compound that retains a set of unique chemical properties. A molecule consists of two or more atoms bonded together. The examples of molecules are Oxygen (O2), Hydrogen (H2), Chlorine (Cl2), Ozone (O3), Water (H2O). Pure Substances Any sample of matter can be classified as a pure substance or a mixture. A pure substance can be either an element or a compound. The composition of a pure substance is definite and fixed. The examples of pure substances are pure water (compound) and pure gold (element). Element A chemical element often called simply element is a substance that cannot be divided or changed into different substances by ordinary chemical methods. The smallest particle of such an element is an atom, which consist of electrons centered around a nucleus of protons and neutrons. The examples of elements are: Metals – Iron (Fe), Gold (Au), Silver (Ag) and Mercury (Hg). Gases - Oxygen (O2), Chlorine (Cl2), Nitrogen (N2), Helium (He) and Neon (Ne). Non Metals - Carbon (C). Compound A chemical compound is a chemical substance formed from two or more elements, with a fixed ratio determining the composition. For example dihydrogen monoxide (water) is a compound composed of two hydrogen atoms for every oxygen atoms. Mixtures

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A mixture is a combination of two or more substances often chemicals, in which the substances remain chemically distinct, retaining their particular composition and properties. There are two types of mixtures:

Heterogeneous mixtures – does not have uniform properties throughout: the composition of one part (or phase). Example: • a mixture of oil and water. • a mixture of nuts and cake

Homogenous mixture – is the same throughout. It has uniform composition

and appearance throughout. Example:

• a cup of coffee. • a mixture of alcohol and water. • a solution

Figure 1.2: Classification of Matter

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1-4 ATOM AND IONS An atom consists of three sub-atomic particles which are electron, proton and neutron. All atoms can be identified by the number of protons and neutrons they contain:

The atomic number (Z) is the number of protons in the nucleus of each atom of an element (Z = p). In a neutral atom the number of protons is equal to the number of electrons (p = e).

The mass number (A) is the total number of neutrons and protons present in the nucleus of an atom of an element. In general the mass number is given by:

Mass number = number of protons + number of neutrons = atomic number + number of neutrons

Figure 1.3: Structure of an Atom

The accepted way to denote the atomic number and mass of an atom of an element (X) is as follow:

Mass number

X Atomic number

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Thus, the denotations of three isotopes of hydrogen atom are

11 hydrogen),H ( 2

1 (deutrium) ;H 31 (tritium)H

Figure 1.4: Isotopes of Hydrogen EXAMPLE 1 Indicate the number of protons, neutrons, electrons and charge of ion in each of the following table, assuming the unknown atom as X and Y.

Symbol +35626Fe −Br80

35

Proton 23

Neutron 28 14

Electron 21 10

Charge 4+

SOLUTION:

Symbol +356

26 Fe −Br8035

+25123 X +428

14Y

Proton 35 23 14

Neutron 45 28 14

Electron 36 21 10

Charge -1 2+ 4+

EXERCISE 1 Fill in the blanks in Example 1.

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An ion is an atom or group of atoms with a net electric charge.

Na+ H+ H+

H+

Cl – O2 –

Figure 1.5: Ionization Process A negative charge is known as an anion and a positively charged is known as a cation.

Figure 1.6: Cation and Anion

Cations are positively charged ions. Cations are the opposite of anions, since cations has fewer electrons than protons. Anion are negatively charged ion, which has more electrons in its electron shell than it has proton in its nuclei. For example:

Na (p = 11, e- = 11) → Na+ (cation, p = 11, e- = 10) + e- Cl (p = 17, e- = 17) + e- → Cl- (anion, p = 17, e- = 18)

Combination of positive and negative ion produced ionic compound with zero charge. For example:

Na → Na+ (cation) + e-

Cl + e- → Cl- (anion) Na+ + Cl- → NaCl (ionic compound)

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1-5 ATOMIC AND MOLECULAR MASS Atomic mass is the mass of one atom expressed in units (atomic mass unit, amu or u) where 1 amu is equal to 1/12 of the actual mass of carbon-12 (C-12). Atomic mass is also called as relative atomic mass. Most elements have several naturally occuring isotopes with different abundance. The atomic mass shown in the periodic table for an element is actually a weighted average of the masses of all isotopes of the element and is referred to as the Average Atomic Mass. EXAMPLE 2 Calculate the average atomic mass of vanadium. The relative abundances and mass of atom are given :

Isotopes Relative Abundance (%) Mass of atom (amu) 50V 0.25 49.9472 51V 99.75 50.9440

SOLUTION: Average atomic mass = [ ( 49.9472 100

0.25 × ) x ( 50.9440) 10099.75 × ] amu

= (0.125 + 50.817) amu = 50.942 amu EXERCISE 2 (i) The atomic masses of 69Ga (60.4 %) and 71Ga (39.60 %) are 68.94 amu and 70.92

amu respectively. Calculate the relative atomic mass for gallium.

Solution : 69.72 amu

(ii) The atomic masses of 10B and 11B are 10.0129 amu and 11.0093 amu. Calculate the natural abundances of these two isotopes. The average atomic mass of B is 10.81amu.

Solution : 10B = 20% ; 11B = 80%

Relative Atomic Mass

Relative Atomic Mass is unitless. It is the atomic mass (amu) divided by 121

the

mass of one C-12 atom (amu).

Relative atomic mass = atomic mass (amu) x 12 mass of 12C (12)

Relative Molecular Mass Relative Molecular Mass is also unitless. It is the mass of a molecule (amu) divided

by 121

the mass of one C-12 atom (amu).

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The most direct and most accurate method for determining atomic and molecular masses is mass spectrometry. EXAMPLE 3 Calculate the relative molecular mass of H2SO4, given relative atomic mass H = 1, S = 32, O = 16 SOLUTION: Relative Molecular Mass H2SO4= 2(atom H) + 1(atom S) + 4(atom O) = 2(1) + 1(32) + 4(16) = 98 EXERCISE 3 Calculate the relative molecular mass of Fe2(SO4)3. The relative atomic mass refers at periodic table in text book.

Solution : 400

1-6 MOLE CONCEPT 1 mole is the amount of a substance that contains as many elementary entities (atom, molecules or other particles) as there are atoms in exactly 12 g of the carbon-12 isotopes. The actual number of atoms in 12 g of carbon-12 is determined experimentally. This number is called Avogadro Number (NA). The accepted value is 6.022 x 1023. EXAMPLE 4 How many atoms are in: (i) 2 moles of Na atom, (ii) 0.1115 moles of Ca atoms. SOLUTION: (i) 1 mole Na atom ≡ 6.022 x 1023 Na atoms 2 mole Na atom ≡ 6.022 x 1023 Na atoms x 2 = 12.04 x 1023 Na atoms (ii) 1 mole Ca atom ≡ 6.022 x 1023 Ca atoms 0.1115 mole Ca atom ≡ 6.022 x 1023 x 0.1115 = 6.7 x 1022 Ca atoms EXERCISE 4 (i) Calculate how many moles of Sulfur atom are in 1.5 moles of As2 S3. (ii) Determine the number of moles of arsenic in 0.005 moles of the same compound.

Solution : (i) 4.5 mole (ii) 9.033x1020 atom

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Molar mass defined as the mass (in grams) of 1 mole of units (such as atom or molecules) of a substance. For 1 mole H2SO4 = 98 gram and contains 6.022 x 1023 particles of H2SO4. For 1 mole gas H2 = 2 gram and contains 6.022 x 1023 molecules of H2. To determine how many moles in the substance, the following formula is used:

Mole of substance A = Mass of substance A(g) Mass of Formula A(g.mol-1)

EXAMPLE 5 (i) How many moles of (C2H6) are present in 0.334g of C2H6 ? (ii) Calculate the mass (gram) of 1.0 x 1012 Pb atoms. (Relative atomic mass C : 12, ; H : 1 ; Pb : 207 ) SOLUTION: (i) Mass of 1 mole ethane :

(2C + 6H ) = 2(12) + 6(1) gram = 30 gram (Formula). 30 gram ethane = 1 mol ∴0.334 gram ethane =

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(ii) 1 mole Pb = 207 g and contain 6.022 x 1023 Pb atoms

∴1.0 x 1012 atoms = g 10 x 3.44 = 10x 6.022 g 207x 10 x 1.0 10-

2312 Pb

EXERCISE 5 Calculate : Mass of 7.88 X 1020 molecules of ethylene glycol is 0.0681 g. Find the molar mass of the compound. (NA = 6.022 X 1023)

Solution : 0.0192 g

0.334g x 1mole = 0.011 mole of ethane 30g


1-7 NAMES OF CHEMICAL COMPOUND Compounds, can divided into six categories: binary compounds (non-metals), ionic compounds, hydrates, polyatomic ions, salts and acid.

Table 1.1: Names of Binary Compound

Chemical Name Symbol Dinitrogen pentoxide N2O5Carbon tetrachloride CCl4

Nitrous oxide N2O Ammonia NH3

Table 1.2: Names of Ionic Compound

Chemical Name Symbol Potassium sulfide K2S

Iron(II) sulfide FeS Calcium chloride CaCl2

Zinc nitrate Zn(NO3)2Sodium sulfate Na2SO4

Table 1.3: Names of Hydrates

Chemical Name Symbol

Sodium carbonate decahydrate Na2CO3.10H2O (washing soda)

Magnesium sulfate heptahydrate MgSO4.7H2O (epsom salt)

Calcium sulfate dihydrate (gypsum) CaSO4.2H2O Sodium tetraborate decahydrate

(borax) Na2BB4O7.10H2O

Table 1.4: Names of Polyatomic Ions

Chemical Name Symbol

Chromate Cr2O7Perchlorate ClO4

-

Hypochlorite ClO-

Bicarbonate HCO3-

Sulfate SO42-

Table 1.5: Names of Acid

Chemical Name Symbol Hydroiodic acid HI

Hydrosulfuric acid H2S Phosphoric acid H3PO4Phosphorus acid H3PO3

Chlorous acid HClO2 10


Table 1.6: Names of Salts

Chemical Name Symbol Sodium Fluoride NaF Sodium Sulfite Na2SO3

Sodium bicarbonate NaHCO3Sodium monohydrogen phosphite Na2HPO3

Potassium chlorate KClO3 Calculation of Oxidation Number An oxidation number signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely.

For any monatomic ion, the oxidation number is equal to the charge on the ion. thus:

Na+ ion, oxidation number = +1; and Cl- ion, oxidation number = – 1.

For any polyatomic, the sum of oxidation numbers of all the elements in the ion must

be equal to the net charge of the ion.

For example sulfate ion, (ion charge = – 2). The oxidation number of atom sulfur, S is calculated based on oxidation number of O = – 2

−24SO

(S) + 4 ( O) = – 2 X + 4(– 2) = – 2

∴ oxidation number of sulfur (X) = +6

In a neutral molecule, the sum of the oxidation numbers of all the atoms must be

zero. For example CO2 molecule Y + 2 (– 2) = 0,

∴Oxidation number C, Y = + 4.

Anion and cation charge that combine in ionic compound must be equal to zero. In naming compound cation must write first and then follow by anion.

For example: Sodium sulfate substance

Sodium ion, Na+ (+1), sulfate ion, SO4

2- (– 2);

2 mole of Na+ ion (total charge +2) need to balance 1 mole SO42- (– 2)

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Formula of substance:

2Na+ + SO42- → Na2SO4

And the name of ionic compound is sodium sulfate.

EXAMPLE 6 Write the chemical formula and name the compound that produce from the combination a pair of following ion: (i) Lead ion, Pb2+ with nitrate ion, −

3NO(ii) Potassium ion, K+ with carbonate ion, −2

3CO

(iii) Aluminium ion, Al3+ with sulfate ion, −24SO

SOLUTION: (i) Combination 1 Pb2+ with 2 to produce zero charge −

3NO Formula of substance : Pb2+ + 2 → Pb(NO−

3NO 3)2. Name of substance : lead nitrate.

(ii) Combination 2 K+ with 1 to produce zero charge −23CO

Formula of substance : 2 K+ + → K−23CO 2CO3.

Name of substance : potassium carbonate.

(iii) Combination 2 Al3+ with 3 −24SO to produce zero charge

Formula of substance: 2 Al3+ + 3 → Al−24SO 2(SO4)3

Name of substance : aluminium sulfate. EXERCISE 6 Write the formula and name the substance produce from the combination a pair of following ion :

(i) Barium ion, Ba2+ with iodide ion, I- (ii) Sodium ion, Na+ with nitrade ion, N3- (iii) Aluminium ion, Al3+ with oxide ion, O2- .

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1-8 CHEMICAL FORMULA A chemical formula provides information on the composition of molecules and ionic compounds in terms of chemical symbols. By composition we mean not only the elements present but also the ratios in which the atoms are combined. There are 3 types of formulas: molecular formula, empirical formula and structural formula. Empirical Formula – The empirical formula gives the simplest whole-number ratio of atoms of each element present in compound. For example: acetylene (C2H2) and benzene (C6H6), the simplest ratio C: H = 1:1 and the empirical formula is CH Molecular Formula – The molecular formula gives the actual number of atoms of each element present in a molecule. Structural Formula – Structural formula, which show how atoms are bonded to one another in a molecule.

For example: benzene molecule

Empirical Formula : CH, Molecule Formula : C6H6

Structural formula: or

CH2

CH2CH2

CH2

CH2

CH2

For each example, the molecular formula is given in parentheses: acetylene (C2H2), glucose (C6H12O6) and water (H2O). Experimental Determination of Empirical and Molecular Formulas Given the percent composition, the empirical and molecular formula is determined according to the following steps :

Write the mass or percentage of each element in the compound Calculate the number of moles of each element in the compound by dividing

the mass or percentage of the element by the relative atomic mass of the element.

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No of moles = Mass in gram Relative atomic mass

Next, divide each number of moles by the smallest value to obtain the simplest ratio.

Finally, write the empirical formula of the compound based on the ratio of elements.


EXAMPLE 7 11.66 g of Fe combines with 5.0 g of O to form a compound. Determine the empirical formula of this compound. SOLUTION:

Fe O

Mass

11.6 g

5.01 g

Mole g55.85

mol1xg11.6 = 0.209 mol g16

mol1xg5.01 = 0.313 mol

Simplest ratio 0 2090 209..

= 1 0 3130 209..

.= 1.5

Ratio (whole number)

=1 x 2 = 2 = 1.5 x 2 = 3

Empirical Formula : Fe2O3 EXERCISE 7 Determine the empirical and molecular formular for substance with molar mass equal to 32g with mass percentage as follows; 87.5% of N and 12.5% of H.

Solution : NH2 EXAMPLE 8 The empirical formula of a compound is CH. If the molar mass of this compound is about 78 g.mol-1. What is its molecular formula? SOLUTION:

(CH)n = 78 (12 + 1)n = 78 n = 6

∴Molecular Formula = (CH)6 atau C6H6. EXERCISE 8 Pure magnetite is composed of an iron-oxygen binary compound. It contains 72.41% of iron atom and 27.59 % of oxygen atom. Find the empirical formula of the compound.

Solution : FeO

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Percent Composition The percent composition is the percent by mass of each element in a compound.

Percent composition of = n x molar mass of element x 100 an element molar mass of compound

n = the number of moles of the element in 1 mole of the compound. EXAMPLE 9 Calculate the percent composition of N atom in Ca(NO3)2. Relative atomic mass: Ca = 40.08, N = 14, O =16. SOLUTION: Relative Molecular Mass = 40.08 + 2 (14 + 16 (3))

= 164.08

EXERCISE 9

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Calculate the percent composition by mass of atom C and O in following compound: (i) CH3COOH (ii) NaHCO3

Solution : (i) %C = 40 00; %O = 53.33 ;(ii) %C = 14.29 ; %O = 57.1

Percent composition N = mass of N (g) x 100 mass of Ca(NO3)2(g) = 2(14) g x 100 164.08 g = 17.08%

1-9 STOICHIOMETRY OF CHEMICAL REACTION The stoichiometry relation can be used to give conversion factors for relating quantities of reactants and products in a chemical reaction. Chemical Equations A chemical equation use chemical symbols to show what happens during a chemical reaction:

2H2(g) + O2(g) → 2H2O(ℓ) Reactant 1 Gas phase

React with Reactant 2 gas phase

To yield Product of liquid phase


The quantities 2 mole of H2, 1 mole of O2 and 2 mole of H2O which are given the coefficients in equation above are called stoichiometrically equivalent quantities. Balancing Chemical Equation

In general, we can balance a chemical equation by the following steps.

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

Begin balancing the equation by trying different coefficients to make the number of atoms of each element the same on both sides of the equation. We can change the coefficients (the number preceding the chemical formulas). Changing the subscripts would change the identity of the substance.

First, 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. Therefore there is no need to adjust the coefficient of these elements at this point.

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

Finally, balance elements that appear in two or more formula on the same side of the equation. Check your balanced equation to be sure that you have the same total number of each type of atoms on both sides of the equation arrow.

EXAMPLE 10 Write the balanced equation for the reaction:

P4O10 + H2O → H3PO4 SOLUTION: From this information, we write :

aP4O10 + bH2O → cH3PO4 Write the atom using algebra concept : Atom P : 4a = c ; atom O : 10a + b = 4c ; atom H : 2b = 3c The value of a = 1 Reactant aP4O10 become P4O10 ⇒ 4 atom P Product of CH3PO4, have 4 atom P (reactant) write as 4H3PO4 ⇒ c = 4

To find value of b : 2b = 3c ⇒ b = 62

4x3=

aP4O10 + bH2O → cH3PO4 ⇒ P4O10 + 6H2O → 4H3PO4 Check : P : reactant 4 P = product 4 P : O : reactant 16 O = product 16 O : H : reactant 12 H = 12 H Balanced Equation : P4O10 + 6H2O → 4H3PO4

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EXERCISE 10 Write the balanced equation for the reaction: (i) KClO3 → KCl + O2 (ii) NH4NO3 → N2O + H2O

Application in Balancing Chemical Equation Stoichiometry is the quantitative study of reactants and products in a chemical reaction. Whether the units given for reactant (or products) are moles, gram, liter (for gases) or some other units. We use moles to calculate the amount of product formed in a reaction. This approach is called mole method, which means simply that the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance.

No of moles = Mass in gram Relative atomic mass (g/mol)

The reaction equation:

2CH3OH + 3O2 ⎯→ 2CO2 + 4H2O The relationship between mole and mass in gram:

Reactant 1 Reactant 2 Product 1 Product 2 2CH3OH

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+ 3O2 + 4H2O 2CO2→

2 mole 3 mole 2 mole 4 mole 2 x 32 g = 44 g 3 x 32 g = 96 g 2 x 44 g = 88 g 4 x 18 g = 72 g

The conversion steps are:

Gram of reactant

Moles of reactant

Moles of product

Gram of product


EXAMPLE 11 The balanced equation for the combustion of butane is :

O(c)10H(g)8CO(g)13O(c)H2C 222104 +→+ Calculate the mass (g) of CO2 produced if 1.00 g of butane is completely combusted. SOLUTION: Change 1.00 gram of C4H10 to mole :

1042

104

104104 HCmol1.72x10

HCg58.0HCmol1

xHCg1.00 −=

Stoichiometry between reactant, C4H10 and product CO2 is 104

2

HCmol2COmol8

Mole of CO2 produced = 22

1043

104

2 CO6.88x10HCmol1.72x10xHCmol2

COmol8 −− =

Gram of CO2 = 22

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2 COg3.03COmol1COg44.0xCOmol6.88x10 =−

EXERCISE 11 Calculate the mass of H2 released when 6.082 g diborane (B2H6) completely react with water using following equation: BB2H6(g) + 6H2O(c) → 2H3BO3(p) + 6H2(g)

Solution : 2.644 g

Reaction Yield Theoretical yield of a reaction is the amount of product that would result if the entire limiting reagent reacted. The theoretical yield then is the maximum obtainable yield, predicted by the balanced equation. Actual yield is the amount of product actually obtained from a reaction which is always less than the theoretical yield. Percent yield which describe the proportion of the actual yield to the theoretical yield. It is calculated as follows: EXAMPLE 12

Urea is prepared through the reaction below : 2NH3 + CO2 → (NH2)2CO + H2O In one process, 637 g of NH3 are treated with 1142 g of CO2. Determine: (i) Limiting reagent. (ii) The mass of urea formed. (iii) How much excess reagent (in grams) is left at the end of the reaction? (iv) Calculate the percent yield if 622 g of mass is obtained.

% yield = Actual yield x 100 Theoretical yield

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SOLUTION:

(i) From balanced equation: 2 moles of NH3 react with1 mol of CO2 or 2[14 + 3(1)] g NH3 react with 12 + 2(16) g of CO2 34 g NH3 react with 44 g of CO2

∴637 g NH3 react with 2COg824.43NHg342COg44

x3NHg637 =

Actual Reactant of CO2 is 1142 g. Need only 824.4 g, so CO2 is excessive.

∴NH3 limiting reagent. (ii) The relation between mole, limiting reagents and products : 2 moles of ammonia produce 1 mole of urea or 2(17) g of NH3 produce 60 g (NH2)2CO

∴637 g of NH3 produce ureag11243NHg34

ureag60x3NHg637 =

(iii) Mass of excessive CO2 reactant= (1142 – 824.4) g of CO2 = 317.6 g of CO2.

(iv) Percent yield = = x100%urea g 1124urea g 622 = 55.34 %.

EXERCISE 12

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15.3 g of CH4 and 43.0 g of chlorine gas react to produce 80 % product. From the balanced equation, determine the reaction yield and calculate the mass of chloromethane, CH3Cl produced. The other product is hydrogen chloride. Note : Write the balanced equation first!

Solution : 24.5 g

Actual yield x 100% Theoretical yield

1-10 VOLUMETRIC ANALYSES To study solution stoichiometry and to know how much of the reactants are present in a solution and also how to control the amount of reactants used to bring about a reaction in aqueous solution. Concentration of Solutions (i) Gram per liter (g/L) or (g/mL) = (ii) Parts per million (ppm) = (iii) Molarity (M), Unit of Standard Concentration Molarity (M) of Compound A = =

Moles of solute (mol) Liters of solution (L) Concentration of solution (g/L) Molar mass (g/mol)

Mass of solute (g) Volume of solution (L)

Mass of component in solution x 106

Total mass of solution


(iv) Mole Fraction Mole fraction is a dimensionless quantity. The mole percentage or molar percentage, denoted “mol%” and equal to 100% times the mole fraction, is sometimes quoted instead of the mole fraction.

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Mole of component A (nA) = Mole of component B (nB) = B

Total number of mole nT = nA + nB ; mole fraction A, XB A = nA ; mole fraction B, XBB = nB nT nT Summary = mole fraction of = component A

Notes: (ii) is used in small concentration, for example: content of mercury in blood and content of Pb in water.

Mass of A (g) Relative atomic mass A (g/mol)

Mass of A (g) Relative atomic mass A (g/mol)

Moles of A Sum of moles of all components


EXAMPLE 13 1.2 g of KOH dissolved in 250 cm3. Calculate : (i) concentration of solution in g L-1 (ii) molarity of solution in mole L-1 SOLUTION:

(i) Mass of KOH = 1.2 g

Volume of solution = L 0.25cm1000

L 1xcm 250 33 =

Concentration KOH, g/L = L 0.25g 1.2

= 5.6 g.L-1

(ii) Molarity = L

mol

Mole KOH = mol 0.021KOH g 56KOH mol 1 x KOH g 1.2 =

Molarity, M KOH = M 0.084L 0.25

mol 0.021=

EXERCISE 13 The concentration of cholesterol (C27H46O) in normal blood is 0.005 M. Calculate mass of cholesterol (g) in 750 mL normal blood.

Solution : 1.448 g

EXERCISE 14 Given that the balanced equation for the reaction is as below 2 HCl(aq) + K2S(aq) H2S(g) + 2 KCl(aq)

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What is the volume of 0.350 M of HCl that will completely react with 475 mL of 0.125 M of K2S?

Solution : 2.947 L

Dilution of Solution Dilution is the procedure for preparing a less concentrated solution from a more concentrated one. The relationship between molarity of solution with volume:

M1V1 = M2V2

M1, V1 are moles of solute and volume before dilution, M2, V2 are moles of solute and volume after dilution,

MV = moles of solutes

The units of V1 and V2 must be the same (mL or L)


EXAMPLE 14 Find the volume of 5 M NaOH solution required to prepare 50 mL of 0.1 M NaOH. SOLUTION: Using formula M1V1= M2V2 M1, V1 are initial concentration and initial volume M2, V2 are final concentration and final volume

M1 = 5 M V1 =? M2 = 0.1 M V2 = 50 mL

V1 = 1

22M

VM

= M 5

mL) M)(50 (0.1= 1 mL

1 mL of NaOH 5 M (add with ≈ 49 mL air) required producing 50 mL, 0.1 M NaOH.

EXERCISE 14 Describe how to prepare 1.00 L of 0.646 M HCl solution, starting with a 2.00 M HCl solution.

Solution : Add 0.323 L of water to 2M of HCl Acid-Base Titration Acid-base titration is the determination of the concentration of an acid or base by exactly neutralizing the acid/base with an acid or base of known concentration. Neutralization of acid-base produce salt and water. The reaction is:

aHA + bMOH → cMA + dH2O acid base salts water

The relation of stoichiometry reaction and titration are:

Mole acid = a ⇒ (MV) acid = aMole base b (MV) base b M and V are molarity and volume acid-base

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EXAMPLE 15 In an acid-base titration, 15 00 cm3 of NaOH solution is needed to neutralize 18 00cm3 of 0.25 M H2SO4. What is the concentration (in molarity) of the NaOH solution? SOLUTION: First, write the balanced equation for the reaction :

H2SO4 + 2NaOH → Na2SO4 + 2H2O

a = 1, b = 2, Ma = 0.25 M, Mb = ? Va = 18.00 cm3, Vb = 15.00 cm3

Molarity of NaOH = 0.6 M EXERCISE 15 (i) Calculate the concentration of HNO3 if 68.50 mL is needed to neutralize 25

mL of 0.150 M KOH solution. (ii) In a neutralization reaction, 0.15 M sulfuric acid react with 0.20 M KOH to

produce K2SO4 (potassium sulfate). (a) Write the chemical equation for this reaction. (b) Calculate the volume that neutralize 25 mL of acid. (c) If 5 x 10-3 mol of base react with 3.0 x 10-3 mole of acid, which of the two

reactants is the limiting reagent? (d) Find the number of mole of salt produced.

Solution : (i) 0.055 M (ii) (b) 37.5 mL (c) KOH (d) 2.5 mmol potassium sulfate

(MV)acid = 1 ⇒ Mbase = 2 x (MV)acid = 2 x (0.25M) (18.00cm3) = 0.6M (MV)base 2 Vbase 15.00 cm3

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1-11 TUTORIAL QUESTIONS

1. Classify each of the following as an element, a compound, a homogeneous mixture or a heterogeneous mixture (a) seawater (b) helium gas, (c) sodium chloride, (d) a bottle of soft drinks, (e) milk-shake, (f) air in a bottle and (g) concrete.

2. Give the number of protons, neutrons and electrons in each of the following species (a) (b) and (c) O17

8 Hg19980 Hg200

80

3. Write the appropriate symbol for each of the following compounds: (a) Z = 74, A = 186 (b) Z = 80, A = 201

4. Give the number of proton and electrons in each of the following common ions: K+, Mg2+, Fe3+, Br-, Mn2+, and Cu2+

5. What are the empirical formulas of the following compounds? (a) Al2Br6 (b) Na2S2O4 (c) N2O3 (d) K2Cr2O7

6. How many amu are there in 8.4 g?

7. How many moles of cobalt (Co) atoms are there in 6.00 X 109 (6 billion) Co atoms?

8. How many grams of gold (Au) are there in 15.3 moles of Au?

9. How many atoms are present in 3.14 grams of copper (Cu)?

10. Which of the following has a greater mass: 2 atoms of lead or 5.1 X 10-23 mole of helium?

11. Calculate the molar mass of the following substance (a) Li2CO3 (b) CS2 (c) CHCl3 (d) C6H8O6 (e) KNO3 and (f) Mg3N2

12. How many molecules of ethane (C2H6) are present in 0.334g of C2H6

13. Urea [(NH2)2CO] is used for fertilizer and many other things. Calculate the number of N, C, O and H atoms in 1.68 X 104 g of urea.

14. Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N and O. Determine the percent composition of oxygen and the empirical formula from the following percent composition by mass: 19.8% C, 2.50% H, and 11.6% N. What is the molecular formula given that its molar mass is about 120g.

15. Tin (II) fluoride (SnF2) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in 24.6 g of the compound?

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16. What are the empirical formulas of the compounds with the following compositions? (a) 40.1% of C, 6.6% of H and 53.3% of O. (b) 18.4% C, 21.5% N and 60.1% K.

17. Balance the following equations

(a) N2O5 N2O4 + O2

(b) KNO3 KNO2 + O2

(c) NH4NO3 N2O + H2O

(d) NH4NO2 N2 + H2O

(e) NaHCO3 Na2CO3 + H2O + CO2

(f) P4O10 + H2O H3PO4

(g) HCl + CaCO3 CaCl2 + H2O + CO2

(h) Al + H2SO4 Al2(SO4)3 + H2

18. Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas

Si(s) + 2Cl2 (g) SiCl4(l)

In one reaction, 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction?

19. Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide

C6H12O6 2C2H5OH + 2CO2

Starting with 500.4 g of glucose, what is the maximum amount of ethanol in grams and liters that can be obtained by this process? (Density of ethanol = 0.789g/ml)

20. The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high altitude jet plane, the SST. The reaction is

O3 + NO O2 + NO2

If 0.740 g of O3 reacts with 0.670 g of NO, how many grams of NO2 will be produced? Which compound is the limiting reagent? Calculate the number of moles of the excess reagent remaining at the end of the reaction.

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21. Consider the reaction

MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

If 0.86 mole of MnO2 and 48.2 g of HCl react, which reagent will be used up first? How many grams of Cl2 will be produced?

22. Ethylene (C2H4), an important industrial organic chemical, can be prepared by heating hexane (C6H14) at 800 °C

C6H14 C2H4 + other products

If the yield of ethylene production is 42.5 percent, what mass of hexane must be reacted to produce 481 g of ethylene?

23. Disulfide dichloride (S2Cl2) is used in the vulcanization of rubber, a process that prevents the slippage of rubber molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine.

S8(l) + 4Cl2(g) 4S2Cl2(l)

What is the theoretical yield of S2Cl2 in grams when 4.06 g of S8 are heated with 6.24 g of Cl2? If the actual yield of S2Cl2 is 6.55 g, what is the percent yield?

24. How many grams of KOH are present in 35.0 mL of a 5.50 M solution?

25. Calculate the molarity of each of the following solutions. (a) 6.57 g of methanol (CH3OH) in 1.50 X 102 mL of solution, (b) 10.4 g of calcium chloride (CaCl2) in 2.20 X 102 mL of solution, and (c) 7.82 g of naphthalene (C10H8) in 85.2 mL of benzene solution.

26. Water is added to 25.0 mL of a 0.866 M KNO3 solution until the volume of the solution is exactly 500 mL. What is the concentration of the final solution?

27. You have 505 mL of a 0.125 M HCl solution and you want to dilute it exactly to 0.100 M. How much water should you add?

28. Calculate the concentration (in molarity) of NaOH solution if 25.0 mL of the solution are needed to neutralized 17.4 mL of 0.312 M HCl solution.

29. What volume of of a 0.5 M HCl solution is needed to neutralize each of the following; (a) 10 mL of a 0.300 M NaOH solution, (b) 10 mL of 0.200 M Ba(OH)2 solution.

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30. The SO2 present in air is mainly responsible for the acid rain phenomenon. Its concentration can be determined by titrating against a standard permanganate solution as follows:

5SO2 + 2MnO4- + 2H2O 5SO4

2- +2Mn2+ + 4H+

Calculate the number of grams of SO2 in a sample of air if 7.37 mL of 0.00800 M KMnO4 solution are required for the titration.

31. The concentration of hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in acidic medium according to the following equation;

2MnO4- + 5H2O2 + 6H+ 5O2 + 2Mn2+ + 8H2O

If 36.44 mL of a 0.01652 M KMnO4 solution are required to oxidize 25 mL of a H2O2 solution, calculate the molarity of H2O2 solution.

Answers

1. (a) Homogeneous mixture, (b) Element, (c) Compound, (d) Homogeneous mixture, (e) Heterogeneous mixture (f) Homogeneous mixture, (g) Heterogeneous mixture

2. (a) neutron = 9 electron = 8 proton = 8

(b) neutron = 119 electron = 80 proton = 80

(c) neutron = 120 electron = 80 proton = 80

3. (a) (b) W18676 Hg201

80

4. K+ (p= 19, e= 18), Mg2+ (p=12, e=10), Fe3+ (p=26, e=23), Br- (p=35, e=36), Mn2+ (p=25, e=23), C4- (p=6, e=10), Cu2+ (p=29, e=27)

5. (a) AlBr3, (b) NaSO2, (c) N2O5, (d) K2Cr2O7

6. 5.1 X 1024 amu

7. 9.96 X 10-15 mol

8. 3.01 X 103 g Au

9. 2.98 X 1022

10. Pb

11. (a) 73.89 g, (b) 76.14 g, (c) 119.35 g, (d) 176 g, (e) 101.11 g, (f) 100.93 g

12. 6.70 X1021 C2H6 molecules

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13. N: 3.37 X 1026 atoms, C: 1.69 X 1026 atoms, O: 1.69 X 1026 atoms,

H: 6.74 X 1026 atoms

14. C2H3NO5

15. 5.97 g F

16. (a) CH2O (b) KCN

17.

(a) 2N2O5 2N2O4 + O2

(b) 2KNO3 2KNO2 + O2

(c) NH4NO3 N2O + 2H2O

(d) NH4NO2 N2 + 2H2O

(e) 2NaHCO3 Na2CO3 + H2O + CO2

(f) P4O10 + 6H2O 4H3PO4

18. 1.01 mol

19. 255.76 g

20. O3, 0.709 g NO2, 6.913 X 10-3 mol

21. HCl, 23.4 g

22. 3.47 X 103 g

23. 8.55 g, 76.6%

24. 10.8 g

25. (a) 1.37 M (b) 0.43 M (c) 0.72 M

26. 0.0433 M

27. 126 mL

28. 0.217 M

29. (a) 6.00 mL (b) 8.00 mL

30. 9.45 X 10-3 g

31. 0.06020 M 28


Table 1.1: Mass of Atoms

Element Symbol No of Atom Atomic mass Element Symbol No of Atom Atomic massActinium Ac 89 227.0278 Mercury Hg 80 200.59 Aluminum Al 13 26.98154 Molybdenum Mo 42 95.94 Americium Am 95 (243) Neodymium Nd 60 144.24 Antimony Sb 51 121.75 Neon Ne 10 20.179 Argon Ar 18 39.948 Neptunium Np 93 237.0482 Arsenic As 33 74.9216 Nickel Ni 28 58.70 Astatine At 85 (210) Niobium Nb 41 92.9064 Barium Ba 56 137.33 Nitrogen N 7 14.0067 Berkelium Bk 97 (247) Nobelium No 102 (259) Beryllium Be 4 9.01218 Osmium Os 76 190.2 Bismuth Bi 83 208.9804 Oxygen O 8 15.9994 Boron B 5 10.81 Palladium Pd 46 106.4 Bromine Br 35 79.904 Phosphorus P 15 30.97376 Cadmium Cd 48 112.41 Platinum Pt 78 195.09 Calcium Ca 20 40.08 Plutonium Pu 94 (244) Californium Cf 98 (251) Polonium Po 84 (209) Carbon C 6 12.011 Potassium K 19 39.0983 Cerium Ce 58 140.12 Praseodymium Pr 59 140.9077 Cesium Cs 55 132.9054 Promethium Pm 61 (145) Chlorine Cl 17 35.453 Protactinium Pa 91 231.0359 Chromium Cr 24 51.996 Radium Ra 88 226.0254 Cobalt Co 27 58.9332 Radon Rn 86 (222) Copper Cu 29 63.546 Rhenium Re 75 186.207 Curium Cm 96 (247) Rhodium Rh 45 102.9055 Dysprosium Dy 66 162.50 Rubidium Rb 37 85.4678 Einsteinium Es 99 (254) Ruthenium Ru 44 101.07 Erbium Er 68 167.26 Samarium Sm 62 150.4 Europium Eu 63 151.96 Scandium Sc 21 44.9559 Fermium Fm 100 (257) Selenium Se 34 78.96 Fluorine F 9 18.998403 Silicon Si 14 28.0855 Francium Fr 87 (223) Silver Ag 47 107.868 Gadolinium Gd 64 157.25 Sodium Na 11 22.98977 Gallium Ga 31 69.72 Strontium Sr 38 87.62 Germanium Ge 32 72.59 Sulfur S 16 32.06 Gold Au 79 196.9665 Tantalum Ta 73 180.9479 Hafnium Hf 72 178.49 Technetium Tc 43 (97) Helium He 2 4.00260 Tellurium Te 52 127.60 Holmium Ho 67 164.9304 Terbium Tb 65 158.9254 Hydrogen H 1 1.0079 Thallium Tl 81 204.37 Indium In 49 114.82 Thorium Th 90 232.0381 Iodine I 53 126.9045 Thulium Tm 69 168.9342 Iridium Ir 77 192.22 Tin Sn 50 118.69 Iron Fe 26 55.847 Titanium Ti 22 47.90 Krypton Kr 36 83.80 Tungsten W 74 183.85 Lanthanum La 57 138.9055 Uranium U 92 238.029

Lawrencium Lr 103 (260) Vanadium V 23 50.9414

Lead Pb 82 207.2 Xenon Xe 54 131.30

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Lithium Li 3 6.941 Ytterbium Yb 70 173.04 Lutetium Lu 71 174.97 Yttrium Y 39 88.9059 Magnesium Mg 12 24.305 Zinc Zn 30 65.38 Manganese Mn 25 54.9380 Zirconium Zr 40 91.22 Mendelevium Md 101 (258)

Chapter 1 - Mole and Atomic Concepts

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