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Chapter 1: Matter
1.0 MATTER
Anything that occupies space and possesses mass is called matter. Example: air, earth, animals, trees, atoms,…
The three states of matter are solid, liquid and gas
Figure 1.1.1: Classification of matter
1.1 Atoms and Molecules
1.1.1 Atom
In an atom, there are three subatomic particles: proton (p), neutron (n) and electron (e). The number of protons = the number of electronsThe protons and neutrons of an atom are packed in an extremely small nucleus. Therefore, nucleus contains most of the atom’s mass. The relative mass of an atom is almost the same as its nucleon number (p + n). The nucleon number is sometimes used as the approximate relative mass in calculations.The number of proton it contains determines the positive charge of the nucleus.
The electrons move rapidly around the nucleus of an atom, and the space they occupy as they move defines the volume of the atom – electrons cloud.
Particle Charge Relative Mass
p +1 1.0 amun 0 1.0 amue -1 0.0
1.1.2 Atomic (Nuclide) Symbol or Notation for NuclidesX = element symbol
A = Nucleon Number of the nuclide X = Z + n
Z = Proton Number of the nuclide X
= p
2
charge
Pure Substanc
es
XAZ
Separation by
chemical method
Separation by physical methodMixtures
Heterogeneous Mixtures
Homogeneous Mixtures
Compoun Elements
Matte
nucleuselectron cloud
Chapter 1: Matter
Example 1.1.1:
the right superscript shows that the total charge on the ion is 2+ the right subscript shows that the ion is formed from two mercury atoms. the left superscript: nucleon number of mercury, A = 202 the left subscript: proton number of mercury, Z = 80
the number of neutrons in a nucleus of mercury = 202 – 80 = 122
Example 1.1.2: Give the number of protons, neutrons and electrons in each of the following species:
(a) (b) (c) (d)
Symbol
Number of
Chargeproton neutron electron
80 120 80 0
29 34 29 0
8 9 10 -2
27 32 24 +3
Example 1.1.3: Write the appropriate notation for each of the following nuclide:
SpeciesNumber of Notation for
nuclidesproton neutron electron
A 2 2 2
B 1 2 0
C 1 1 1
D 7 7 10
Exercise 1.1.1:
1) Atom 1939 K consists of ___ neutrons and ___ electrons. Ans: (20, 19)
2) An element Y contains 22 neutrons and 18 electrons. What is the proton number and nucleon number of the element? Write the symbol of Y. Ans: (18, 40)
3) An ion X2 has 34 neutrons and 30 protons. Determine the nucleon number and the number of electrons in the ion. Ans: (64, 32)
4) Determine the number of electrons and neutrons in the following species: ,
3
Chapter 1: Matter
Ans: Mn2+ : (21,32) F- : (10,10)
Note:
1) The proton number, Z, is the nuclear charge and also the number of electrons in a neutral atom of the element.
2) no. p = no. e atom (neutral)3) no. p > no. e positively charged – cation (atom lost electrons)4) no. p < no. e negatively charged – anion (atom gained electrons)
1.1.3 Isotope
Isotopes are termed as two or more atoms of the same element that have the same number of protons in their nucleus but different number of neutrons.
Protium Deuterium Tritium
Isotopes of hydrogen
Proton number 1 1 1
Number of neutrons 0 1 2
Example 1.1.4 : For the following nuclides: , and Z
a) write the nuclides that are isotopes. and (reason: same proton number but different nucleon number)
b) state the atoms that have the same number of neutrons. (no. neutrons = 14 6 = 8) and Z (no. neutrons =15 7 = 8)
Notes:Isotopes of an element have the same,(a) number of protons (proton number) (b) charge of nucleus of the atoms
(ionization energy; electron affinity; size of the atom; electronegativity are the same)(c) number of electrons in a neutral atom (d) electronic configuration (the number of valence electrons)(e) chemical properties
Isotopes of an element have different,(a) number of neutrons (nucleon number) in the nucleus of the atoms(b) relative isotopic mass
Uranium-235 Uranium-238
Isotopes of uranium
Proton number 92 92
Number of neutrons 143 146
4
Chapter 1: Matter
(c) physical properties (e.g boiling point / melting point, density, effusion rate,…)
1.1.4 Molecule
A molecule consists of a small number of atoms joined together by covalent bond.Diatomic molecule: contains two atoms (example: H2, Cl2, HCl, CO,)Polyatomic molecule: contains more than two atoms (example: H2O, NH3)
1.1.5 Ion
An ion is a charged species formed from a neutral atom or molecule when electrons are gained or lost as the result of a chemical reaction.Cation: a positively charged ion (number e < number p) (example: Mg2+, )
Anion: a negatively charged ion (number e > number p) (example: Cl, OH)Monatomic ion: ion contains only one nucleus (example: Fe3+, S2)Polyatomic ion: ion contains more than one nucleus (example: H3O+, CN)
1.1.6 Relative Mass
(a) Relative Atomic Mass, Ar of an element
Ar =
The average relative isotopic mass of the atoms must be used for calculation because most elements consist of a mixture of isotopes.
(b) Relative Molecular Mass, Mr of a molecular substance
Mr = or
= sum of the relative atomic masses of all the atoms shown in the molecular
formula.
Example 1.1.5: Mr of H2O = 2 (Ar of H) + Ar of O = 2(1.008) + 15.999 = 18.15
relative molecular mass of CaCl2 = 40 + 2(35.5) = 111 relative molecular mass of SO4
2 = 32 + 4(16) = 96
Notes: Ar and Mr – dimensionless Molecular mass, atomic mass – atomic mass unit (amu)
Exercise 1.1.2
1) Determine the relative atomic mass of an element X if the ratio of the atomic mass of X to carbon-12 atom is 0.75. Ans: (9.0)
5
Magnetic field
Acceleration chamber (Electric field)
Ionization chamber
Vaporisation chamber
Chapter 1: Matter
2) Relative atomic mass of chlorine atom is 35.5. Explain why this value is not a whole number.
3) Relative atomic mass of bromine atom is 80. What is the mass of bromine atom compared to carbon-12 atom? Ans: (6.67)
4) Calculate the relative molecular mass of C5H5N. Ans: (79)
1.1.7 Mass Spectrometer
Mass spectrometer is used to determine:i) Relative atomic mass of an element ii) Relative molecular mass of a compoundiii) Types of isotopes that are found in the naturally occurring element including the
abundance of the isotopes and its relative isotopic mass.iv) Recognize the structure of the compound in an unknown sample.
A pump maintains a vacuum inside the mass spectrometer as any air molecules inside would block the movement of the ions and to avoid the contamination of the sample.
There are five main stages:
A B 37Cl
Vacuum 35Cl
Figure 1.1.2: Mass Spectrometer
(a) Vaporisation Chamber
Sample of the element is vaporised into gaseous atom
(b) Ionisation ChamberA gaseous sample (atom or molecule) is bombarded by a stream of high-energy electrons that are emitted from a hot filament. Collisions between the electrons and the gaseous atom (or molecule) produce positive ions by dislodging an electron from each atom or molecule.
M + e M+ + e + e
M M+
(c) Acceleration Chamber (Electric field)
The positive ions are accelerated by an electric field towards the two oppositely charged plates. The electric field is produced by a high voltage between the two plates. The emerging ions are of high and constant velocity.
6
Ion detector
Chapter 1: Matter
(d) Magnetic field
The positive ions are separated and deflected into a circular path by a magnet according to its mass/charge (m/e) ratio.
Positive ions with small m/e ratio are deflected most and appear near A. Ions with large m/e ratio are deflected least and appear near B (Figure 1.1.2).
(e) Ion detector
The numbers of ions and types of isotopes are recorded as a mass spectrum. In practice, the ion detector is kept in a fixed position. The magnetic field is varied so
that the positive ions of different masses arrive at the detector at different times.
Mass spectrum: the horizontal axis shows the m/e ratio or nucleon number or isotopic mass or relative atomic mass of the ions entering the detector. The vertical axis shows the abundance or detector current or relative abundance or ion intensity or percentage abundance of the ions. The height is proportional to the amount of each isotope present.
Information from a mass spectrum of an elementi) the isotopes which are present in the elementii) the relative isotopic mass of each isotopeiii) the abundance of each isotope
Thus, the relative atomic mass of the element can be determined
Example 1.1.6 :
Relative abundance Mass Spectrum of Magnesium 63
8.1 9.1
0 24 25 26 m/e
The mass spectrum of magnesium shows that naturally occurring magnesium consists of three isotopes: 24Mg, 25Mg and 26Mg.The height of each line is proportional to the abundance of each isotope. In this example, 24Mg is the most abundance of the three isotopes.
of Mg = (24 a.m.u. x 63) + (25 a.m.u. x 8.1) + (26 a.m.u x 9.1) (63 + 8.1 + 9.1)
= 24.33
Ar =
7
Chapter 1: Matter
where Q = the abundance of an isotope of the element = the percentage of the isotope found in the naturally occurring elementm = the relative isotopic mass of the element
Notes :
1) The height of each peak measures the relative abundance of the ion which gives rise to that peak.
2) The total number of peaks in the mass spectrum of an element shows the types of naturally occurring isotopes.
3) The ratio of mass/charge for each species is found from the value of the accelerating voltage associated with a particular peak. Many ions have a charge of +1 elementary charge unit, and the ratio m/e is numerically equal to m, the mass of the ion.
(1 elementary charge unit = 1.60x1019 C)
4) The ion with the highest value of m/e is the molecular ion, and its mass gives the molecular mass of the compound
Exercise 1.1.3
1. Fig 1.3 shows the mass spectrum of the element rubidium, Rb.a) What isotopes are present in Rb?b) What is the percentage abundance of each isotope?c) Calculate the relative atomic mass of Rb.
2. The mass spectrum of neon, Ne consists of three lines corresponding to m/e ratio of 20, 21 and 22 with relative intensities of 0.910, 0.0026 and 0.088, respectively. a) Explain the significance of these data
[The mass/charge ratios refer to the singly-charged ions from the isotopes present in the sample:
the relative intensities show the relative abundance of each isotope in the sample.]
b) Calculate the relative atomic mass of Ne. Ans:(20.2)c) Sketch the mass spectrum that would be obtained from naturally occurring Ne.
3. The atomic masses of and are 6.0151 amu and 7.0160 amu respectively. What is the
relative abundance of each isotope if the relative atomic mass of lithium is 6.941 amu? Ans:(92.51%)
4. Naturally occurring iridium, Ir is composed of 2 isotopes and in the ratio of 5:8. The relative isotopic mass of and are 191.021 and 193.025 respectively. Calculate the relative atomic mass of iridium. Ans:(192.254)
5. Determine the relative atomic mass of copper from the mass spectrum shown below.
relative abundance r
8
Abundance
18
7
85 87
Fig 1.3
Chapter 1: Matter
7
3
m/e Ans: (63.6) 63 65
6. The ratio of relative abundance of naturally occurring of chlorine isotopes is as follow:
Based on the carbon-12 scale, the relative atomic mass of 35Cl=34.9689 and 37Cl=36.9659. Calculate the Ar of chlorine. Ans:(35.45)
7. A sample of carbon dioxide gas that composes of isotopes 12C, 13C, 16O and 18O is analysed using the mass spectrometer. How many peaks will be observed in the mass spectrum?
Ans: (6)
1.1.4 Chemical Formula
Empirical Formula indicates which elements are present and the simplest whole-number ratio of their atoms in a molecule.
Molecular Formula shows the exact number of atoms of each element in the smallest unit of a substance.
Note:The empirical formula and the molecular formula of a compound could be the same (NH3, CO2).Two molecules might have different molecular formulae but the same empirical formula.
A. Calculating the empirical formula from the masses of constituents
Example: 18.3 g sample of hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula.
Solution:
Molecules Empirical formula Molecular Formula
n
Water H2O H2O 1Hydrogen peroxide HO H2O2 2
BenzeneCH C6H6 6
Etyne CH C2H2 2
9
molecular formula = (empir ical formula) n
Example: Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. The molar mass of ascorbic acid is 176 g mol1. Determine its empirical formula and molecular mass.
Solution:
Element C H O
Mass/g 40.92 4.58 54.50
Amount/mol = 3.41 4.58 = 3.41
Simplest ratio = 1 = 1.33 = 1
1x3 1.33x3 1x33 4 3
Empirical formula = C3H4O3 Molecular formula = (C3H4O3)n
n = = (C3H4O3)2
= = 2 = C6H8O6
Chapter 1: Matter
Constituent Ca Cl H2OMass/g 4.0 7.1 7.2
Amount/mol = 0.10 = 0.20 = 0.40
Simplest ratio of relative amount = 1.0 = 2.0 = 4.0
1 2 4Empirical formula = CaCl2
.4H2O
Exercise 1.1.4:
1) A sample of a hydrated compound was analyzed and found to contain 2.10 g of cobalt, 1.14 g of sulphur, 2.28 g of oxygen and 4.50 g of water. Calculate its empirical formula.
2) 10.00 g sample of a compound contains 3.91 g of carbon; 0.87 g of hydrogen and the remainder is oxygen. Calculate the empirical formula of the compound.
3) 10.00 g of hydrated barium chloride is heated until all the water is driven off. The mass of anhydrous compound is 8.53 g. Determine the value of x in BaCl2
.xH2O.
n = ; molecular formula = (empirical formula)n
B. Calculating the empirical formula from percentage composition by mass
Note: You must never round off values close to whole number in order to get a simple ratio, but multiply the value by a factor until we get a whole number.However, if the value is very close to whole number (0.01), it is allowed to round off the value.
Exercise 1.1.5
10
Chapter 1: Matter
1) Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%; H: 6.21%; S: 39.5%; O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g mol1.
2) Determine the formula of a mineral with the following mass composition: Na=12.1%, Al=14.2%, Si=22.1%, O=42.1%, H2O=9.48%.
C Calculating the empirical formula from elemental analysis data
Example: 1.00 g sample of compound A was burnt in excess oxygen producing 2.52 g of CO 2
and 0.443 g of H2O. Determine the empirical formula of the compound.Solution: In 1 mol of CO2 there is 1 mol of C In 1 mol of H2O there are 2 moles of
H atoms
= 0.688 g C
= 0.0492 g H
the mass of oxygen = mass of sample – (mass of C + mass of H)= 1.00 g – (0.688 g + 0.0492 g)= 0.263 g O
Element C H O
Mass/g0.688 0.0492 0.263
Amount/mol = 0.0573 = 0.0492 = 0.0164
Simplest ratio of relative amount= 3.49 = 3.00 = 1.00
3.49x2 3.00x2 1.00x27 6 2
Empirical formula = C7H6O2
Exercise 1.1.6:
1) The combustion of 0.146 g of compound B gave 0.374 g of carbon dioxide and 0.154 g of water. Assuming that B contains carbon, hydrogen and oxygen only, determine its empirical formula.
2) An organic compound, X which contains only carbon, hydrogen and oxygen, has a molar mass of about 85 g mol1. When 0.43 g of X is burnt in excess oxygen, 1.10 g of carbon dioxide and 0.45 g of water are formed. a) What is the empirical formula of X?b) What is the molecular formula of X?
3) A compound Y with chemical formula as shown below: CH2=CHCOOCH3
a) Write the empirical formula and molecular formula of the compound.
b) Calculate the percentage composition of carbon in the compound X. Ans:(55.8%)
1.2 Mole Concept
11
Chapter 1: Matter
Mole: The amount of substance that contains as many elementary particles (atoms/molecules/ions) as there are atoms in exactly 12.000 g of carbon-12.
One mole of carbon-12 atoms has a mass of exactly 12.000 grams and contains 6.023x1023 atoms.
1 mole of atoms of an element = the relative atomic mass of the element in grams = Ar grams = 6.023x1023 elementary particles/entities
n (mole)=
Molar Mass: The mass (in grams) of 1 mole of units (atoms/molecules/electrons/ions) of a substance.
Avogadro Constant (symbol: L @ NA)
NA = 6.023x1023 mol1 elementary particles/entities/units
Example 1.2.1:
1.0 mol of chlorine atoms = 6.023x1023 chlorine atoms = 35.5 g Cl1.0 mol of chlorine molecules = 6.023x1023 chlorine molecules= 2(35.5) g = 71.0 g Cl2
= 6.023x1023 x 2 chlorine atoms1.0 mol of calcium bromide = 6.023x1023 units of CaBr2 = 200 g CaBr2
= 6.023x1023 calcium ions= 6.023x1023 x 2 bromide ions
1.0 mol of ammonia, NH3 contains 1.0 mol of nitrogen atoms and 3.0 mol of hydrogen atoms.1.0 mol of phosphorus, P4 contains 4.0 mol of phosphorus atoms.1.0 mol of Na2SO4
.10H2O contains 2.0 mol of natrium ions, 1.0 mol of sulphur atoms, 4.0 mol of oxygen atoms and 10.0 mol of water molecules.
Relationship between mole, mass, Ar @ Mr and amount of particlesIn general, if the relative molecular mass of SO2 = 32 + 2(16) = 64
the weight of SO2 = 64 a.m.u. the mass of 1.0 mol of SO2 = 64 g the mass of 6.023x1023 SO2 molecules = 64 g the mass of one SO2 molecule = ________ g the molar mass of SO2 = 64 g/mol
Notes :
12
The amount of A (mol) =Mass A (g)
Molar mass A (g/mol)
= Number of particles ANA
Chapter 1: Matter
1.0 mol substance = 6.023x1023 elementary entities/units/particles (atoms/ions/molecules/electrons)
= Ar @ Mr grams
= 22.4 litres at STP (i.e.: P= 1 atm & T= 273.15K)(molar
volume of gas)
= 24.0 dm3 at room temperature
Example 1.2.2: Calculate the number of atoms in 0.20 mol of magnesium.
The number of Mg atoms =
= 1.2 x 1023 atoms.
Example 1.2.3: Calculate the mass of (NH4)2CO3 that containsa) 0.300 mol NH4
+ b) 6.023 x 1023 hydrogen atoms
(a) The mass of (NH4)2CO3 = 0.300 mol NH4+ x = 14.4 g
(b) The mass of (NH4)2CO3 = 6.023x1023 H atoms x
= 12.0 g
Exercise 1.2.1:
1) How many atoms are there in 17 g of NH3 gas? Ans:(1.8x1024)
2) Calculate the amount of oxygen atoms in 80 g of CuSO4. Ans:(2 moles)
3) What is the mass of one HCl molecule? Ans:(6.06x1023)
4) How many moles of Fe2O3 are there in 1.00 kg of rust? Ans:(6.3)
5) What is the number of atoms in 2.5 g of phosphorus, P4? Ans:(4.85x1022)
6) What mass of sulphur dioxide, SO2 contains the same number of molecules as are in 1.00 g of ammonia? Ans:(3.77 g)
1.2.1 Mole Concept of Gases
Molar volume of any gas at STP, Vm = 22.4 dm3mol 1 STP = standard temperature and pressureWhere T = 273.15 K P = 1 atm @ 760 mmHg
13
Amount of gas at STP, n =V, Volume of gas (L)Vm (22.4 dm3 mol1)
Chapter 1: Matter
Example 1.2.4: How many moles are there in 6.5 L oxygen at STP?
= 6.5 L O2 x = 0.29 mol
Note: Do not use 22.4 L as the molar volume at temperatures and pressures other than 273 K and 1 atm.
Exercise 1.2.2:
1) A balloon is filled with hydrogen gas at STP. If the volume of the balloon is 2.24 dm3, calculate the amount of hydrogen gas. Ans:(0.1 mol)
2) Calculate the volume of 24x1023 molecules of gas at STP. Ans:(89.6 L)
3) A sample of carbon dioxide has a volume of 56 cm3 at STP. Calculate (a) the number of moles of gas molecules, Ans:(0.0025)(b) the number of molecules, and Ans:(1.5x1021)(c) the number of oxygen atoms in the sample. Ans:(3.0x1021)
1.2.2 Concentration Units
The concentration of solutions is the quantity of dissolved substance per unit quantity of solvent in a solution. Concentration is measured in various ways: concentration (formerly molarity), weight percent, weight/volume percent, molal concentration (or molality), mass concentration, mole fraction, or parts per million (ppm).
(a) Molarity, M The number of mole of solute dissolved per unit volume.(unit: mol L1 @ mol dm3 @ M)
Example 1.2.5: A matriculation student prepared a solution by dissolving 0.586 g of sodium carbonate, Na2CO3 in 250.0 cm3 of water. Calculate its molarity.
M (Na2CO3(aq)) = = 0.0221 mol
L1.
Exercise 1.2.3:
14
Molarity, M
= no. mole solute1L = 1 dm3
no. L solution 1mL = 1 cm3
1 dm3 = 1000 cm3
Chapter 1: Matter
1) What is the molarity of 85.0 mL ethanol solution that contains 1.77 g of ethanol, C 2H5OH?[Ar C = 12.0, H = 1.0, O = 16.0] Ans:(0.453)
2) Calculate the molarity of a solution of 1.71 g sucrose (C12H22O11) dissolved in ½ litre of water.[Ar H = 1.0, C = 12.0, O = 16.0] Ans:(0.0100)
3) How many grams of potassium dichromate, K2Cr2O7 required to prepare a solution of 250 mL with 2.16 M? Ans:(159)
4) Calculate the amount of moles of solute in 200 cm3 of ammonia solution, having a concentration of 0.125 mol dm3. Ans:(0.0250)
(b) Molal Concentration @ molality, m
The number of mole of solute per unit mass of solvent in kg.(unit: mol kg1 @ molal, m)
Example 1.2.6: Calculate the molal concentration of ethylene glycol (C2H6O2) solution containing 8.40 g of ethylene glycol in 200 g of water. The molar mass of ethylene glycol is 62 g/mol.
m (C2H6O2) = = 0.677 mol
kg1
Note:Mass of solution = mass of solute + mass of solventVolume of solution volume of solvent
Exercise 1.2.4:
1) A mixture is prepared from 45.0 g of benzene, C6H6 and 80.0 g of toluene, C7H8. Calculate the molarity of the solution. [Ar C=12 ; H=1] Ans: (7.21m)
2) A solution containing 121.8 g of Zn(NO3)2 per litre has a density of 1.107 g/mL. Calculate its molal concentration. [Ar Zn=65; N=14; O=16] Ans(0.654m)
3) What is the molal concentration of a solution prepared by dissolving 0.30 mol of CuCl 2
in 40.0 mol of water? Ans(0.417m)
(c) Weight Percent(% w/w) @ Percent by Mass
15
m = No. mole solute 1 kg = 1000 gMass of solvent in kg
% w/w = Weight solute × 100%Weight solution
Chapter 1: Matter
10% w/w NaOH 10 g of NaOH dissolved in 100 g of solution
10 g of NaOH dissolved in 90 g of solvent (water)
Note: mass of solution = mass of solute + mass of solvent
Example 1.2.7 : A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.3 g of water. What is the percent by mass of KCl in this solution?
Percent by mass of KCl = = 1.61%
Exercise 1.2.5:
1) Calculate the percent by mass of the solute in a aqueous solution containing 5.50 g of NaBr in 78.2 g of solution. Ans:(7.03)
2) Calculate the amount of water (in grams) that must be added to 5.00 g of urea, (NH 2)2CO in the preparation of a 16.2 percent by mass solution. Ans:(25.9)
3) How many grams of NaOH and water are needed to prepare 250.0 g of 1.00% NaOH solution? Ans:(2.50; 248)
4) Hydrochloric acid can be purchased as a solution of 37% HCl. What mass of this solution contains 7.5 g of HCl? Ans:(20.3)
(d) Weight/volume percent (%w/v)
5% w/v of KCl 5 g of KCl dissolved in 100 mL of solution
Note: mass of solution = volume of solution x density of solution
Example 1.2.8: What mass of NaCl is needed to prepare 250 mL of 0.9% w/v solution?
0.9% w/v NaCl =
16
% w/v = Weight of solute × 100%Volume of solution, mL
Chapter 1: Matter
mass of NaCl = 2.25 g
Exercise 1.2.6:
Describe how to prepare a 10% w/v KOH solution.
(e) Mole Fraction, X
The mole fraction of component A is given by XA = , where nA is the number of mole of
one component in a mixture, A (for a given entity) and ntotal is the total number of mole of all substances present in the mixture (for the same entity).
Exercise 1.2.7:
1) What is the mole fraction of CuCl2 in a solution prepared by dissolving 0.30 mol of CuCl2
in 40.0 mol of H2O? Ans:(0.0074)
2) A solution is prepared by mixing 55 g of toluene, C7H8 and 55 g of bromobenzene, C6H5Br. What is the mole fraction of each component? Ans:(0.63; 0.37)
3) A mixture containing benzene and toluene has 18.4 g of toluene and its percentage composition is 30%. Calculate the number of moles of benzene in this solution.
[Mr benzene = 78 and toluene = 92] Ans:(0.55)
(f) Parts per million, ppm
17
X =n
× 100%ntotal
ntotal = n + n + …
Cppm = Weight solute × 100%
weight solution
=Weight solute,mg
Volume solution, L
ppm = mg/L @ g /g @ g /mL @ mg / kg
Chapter 1: Matter
weight solute
Example 1.2.9: The concentration of calcium ions in blood is 100.0 ppm. Calculate the mass of calcium ions in 500.0 g of blood.
100.0 ppm =
mass of Ca2+ = 0.05 g
Exercise 1.2.8:
1) What is the concentration of K+ in a solution that contains 63.3 ppm of K3Fe(CN)6?[relative formula mass of K3Fe(CN)6 = 329.3] Ans:(5.77x104)
2) If the drinking water contains 1.5 ppm of NaF, how much water is needed to dissolve 454 g of NaF? Ans:(3.0x105L)
3) A sample of 500 L of air has a density of 1.20 g dm3 and it contains 2.40x103 g of SO2. Calculate the concentration of SO2 in the air in percent by mass and ppm. Ans:(4x104g,4ppm)
4) A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, a known carcinogen. The level of contamination was 12.4 ppm (by weight).a) Express this in percent by weight.b) What is the molar concentration of the CHCl3 in water?
1.2.3 Oxidation Number
i) For monatomic ions, the oxidation number is equal to the charge on the ion.ii) For covalently bonded atom, the oxidation number is the charge on an atom calculated
by assigning both electrons of a shared pair to the more electronegative atom.
Oxidation number signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely.
Rules:
1) In free elements (that is, an element not combined chemically with a different element), each atom has an oxidation number of zero.Thus each atom in H2, Br2, Na, Be, P4 and O2 has the same oxidation number: zero.
2) For ions composed of only one atom (that is, monatomic ions) the oxidation number is equal to the charge on the ion.
Oxidation number of Al3+ = +3Ca2+ = +2S2- = 2
18
Chapter 1: Matter
3) The oxidation number of oxygen in most compounds (for example: MgO and H2O) is –2, but
in hydrogen peroxide (H2O2) and peroxide ion (O2 ), it is –1.
4) The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compound. In these cases (for example: LiH, NaH, CaH2), its oxidation number is –1.
5) Fluorine has an oxidation number of –1 in all its compounds. Other halogens (Cl, Br, and I) have negative oxidation numbers when they occur as halide ions in their compounds. When combined with oxygen – for example in oxoacids and oxoanions – they have positive oxidation numbers.
6) In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero: CO 2, FeSO4, NaCl.
7) In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion: Cr2O7
2, NO3.
Example 1.2.10 : What are the oxidation number of
a) Fe atom in FeCl3 b) Cr atoms in Cr2O72- c) Cl atoms in Cl2O7
d) S atoms in Na2S4O6e) N atom in NH4
Oxidation number ( Ox.No.): a) +3 b) +6 c) (2 atoms)(Ox.No. of Cl) + 7(-2) = 0 d) +2.5
(2 atoms)(Ox.No. of Cl) = +14 (Ox.No. of Cl)= +7
e) (Ox.No. of N) + 4(+1) = 1 (Ox.No. of N) = 3
1.2.4 Oxidation Number & Redox Reaction
1) Oxidation – is an increase in oxidation number. 2) Reduction – is a decrease in oxidation number.
A substance is oxidized when it shows an increase in its oxidation number. This substance reacts as a reducing agent (reductant) in a chemical reaction.
A substance is reduced when it shows a decrease in its oxidation number. This substance reacts as an oxidation agent (oxidant) in a chemical reaction.
Example 1.2.11:
0 0 +1 -1 H2 + Cl2 2HCl
19
decrease, reduction
increase, oxidation
Chapter 1: Matter
The hydrogen is oxidized (increased in oxidation number) and the chlorine is reduced (decreased in oxidation number). Thus, the reducing agent is H2 and the oxidizing agent is Cl2.
Exercise 1.2.9:
1) Assign oxidation numbers to each of the following underlined elements: a) MnO4 MnO2 b) S2O3
2- SO42-
2) Identify the substance oxidized and the substance reduced as well as the oxidizing and reducing agents in the reaction:
2KCl + MnO2 + 2H2SO4 K2SO4 + MnSO4 + Cl2 + 2H2O
1.2.5 Chemical Equation and Stoichiometry
A chemical reaction is a process in which one set of substances called reactants is converted to a new set of substances called products.
Chemical equation is a way of denoting a chemical reaction using the symbols for the participating particles (atoms, molecules, ions, etc.); formulae of the reactants are written on the left side of the equation and formulae of the products, on the right.
xA + yB zC + wD( ) is used for an irreversible reaction; double arrows ( ) are used for reversible reactions.
When reactions involve different phases it is usual to put the phase in brackets after the symbol (s = solid, l = liquid; g = gas; aq = aqueous).
Because atoms can neither be created nor destroyed in a chemical reaction, the total number of atoms of each element is the same on both sides in a balanced equation.
The numbers x, y, z, and w, showing the relative numbers of molecules reacting, are called the stoichiometric coefficients.
Stoichiometry is the quantitative study of reactants and products in a chemical reaction. --- x moles of A react with y moles of B to yield z moles of C and w moles of D.
Example 1.2.12: CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)
The chemicals reacting together and the chemicals produced.Reactants: CaCO3 and HCl Products: CaCl2; CO2; and H2O
The physical states of reactants and products. The coefficients in a chemical equation provide the atom ratios or the mole ratios by which
moles of one substance react with or form moles of another. 1 mole of CaCO3 reacts with 2 moles of HCl to yield 1 mole of CaCl2, 1 mole of CO2 and
1 mole of H2O. or
20
Chapter 1: Matter
1 molecule CaCO3(s) + 2 molecules HCl(aq) 1 molecule CaCl2(aq) + 1 molecule CO2(g) + 1 molecule H2O(l)
Symbol ‘’: irreversible reaction.
1.2.6 Balancing a Chemical Equation
Method 1: By Inspection
1) Write correct formulae for all reactants and products.2) Assign the most complex formula a coefficient of 1,
Or if an element occurs in only one compound on each side of the equation, balance this element first,
Or balance the atoms of each element (except H and O)
The equation can be balanced only by adjusting the coefficients of the formulae, as necessary.Never introduce any species that are not involved in the reaction.Never change formulae for the purpose of balancing an equation. (consider polyatomic ions:
etc. as one unit)
When one of the reactants or products exists as the free element, balance this element last.
3) Then, work on the hydrogen and the oxygen.4) Check and make sure the total numbers of the atoms of each element is the same on both sides.5) Eliminate the fractional coefficient by multiplying a factor (an appropriate integer) on each
coefficient in order to get the smallest set of whole number coefficients.6) The states of the reacting substances and products can be included in small brackets after the
formulae.
Exercise 1.2.10:
Balance the following chemical equations by applying inspection method.
1) NH3 + CuO Cu + N2 + H2O2) C6H6 + O2 CO2 + H2O3) AgNO3 + Na2CrO4 Ag2CrO4 + NaNO3
4) Fe(OH)3 + H2SO4 Fe2(SO4)3 + H2O5) N2O5 + H2O HNO3
Method 2: Algebraic Method
(Mainly for the chemical equation that is difficultly balanced by inspection method)1) Write correct formulae for all reactants and products. 2) Assign a stoichiometric coefficient (example: a, b, c and d) in front of each formula.3) Form an algebraic equation, as there is the same number of atoms of each element indicated on
both sides of the arrow. Use the elimination method to solve the algebraic equations.Assume a=1, then substitute this value into the algebraic equations to solve b, c and d.
21
Chapter 1: Matter
4) Eliminate the fractional coefficient by multiplying a factor (an appropriate integer) on each coefficient in order to get the smallest set of whole number coefficients.
Example 1.2.13: a Al2(SO4)3 + b KOH c Al(OH)3 + d K2SO4
For element: Al : 2a = c if a = 1K : b = 2d then, c = 2OH : b = 3c b = 3(2) = 6SO4 : 3a = d d = 3(1) = 3
The balanced chemical equation is: Al2(SO4)3 + 6KOH 2Al(OH)3 + 3K2SO4
Exercise 1.2.11:Balance the following chemical equations by applying algebraic method.
1) N2H4 + H2O2 HNO3 + H2O2) ClO2 + H2O HClO3 + HCl
Method 3: The Ion – Electron Method
(Mainly for balancing the oxidation – reduction equations)Reactions that involve both REDuction and OXidation are called REDOX reactions
Oxidation is an Increase in the oxidation number and the substance Loses one or more electrons (OIL). Reduction, conversely, is the process in which some substance gains one or more electrons and the oxidation number of an atom in that substance decreases.
Oxidation Agent (Oxidant): A substance that gains electrons and brings about oxidation in other substances. Reducing Agent (Reductant): A substance that loses electrons and brings about reduction in other substances.
The sequence of steps used in balancing a redox equation in acidic solution is as follows:1) Identify the oxidizing agent and its reduced form as well as the reducing agent and its oxidized
form, and write two skeletal half-reaction. 2) In each half-reaction, balance the atoms of the elements undergoing changes in oxidation
number. 3) Balance oxygen atoms by using H2O.4) Balance hydrogen atoms by using H+(aq).5) Balance the charge in each half-reaction by adding electrons to equalise the ionic charges. The
number of electrons to be added is determined without reference to the oxidation number. Electrons are added simply to balance the charge.
6) Multiply the half-reactions by appropriate integers to ensure that the number of electrons lost in oxidation is equal to the number of electrons gained in reduction.
7) Sum the half-reaction. Simplify the overall equation algebraically so that any H 2O or H+(aq) appears on only one side of the equation. Check the charge balance as well as the balance of all atoms.
[Balancing the redox reactions in basic solution: Balance the equation first as if the solution were acidic, and then add enough OH to both sides of the equation balanced so that wherever H+(aq) appears, it can be combined with OH to form H2O. The number of hydroxide ions added is equal to the number of hydrogen ions in the equation. Be sure to combine H2O if it appears on both sides of the equation.]
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Chapter 1: Matter
Exercise 1.2.12:Balance the following chemical equations by applying the ion-electron method.
1. MnO + C2O 42 + H+ Mn2+ + CO2 + H2O (acidic solution)
2. Cr2O + Fe2+ + H+ Cr3+ + Fe3+ + H2O (acidic solution)
3. MnO 4 + SO2 + H+ SO 4
2 + Mn2+ + H2O (acidic media)
4. Cr(OH)3 + IO3 + OH CrO3
2 + I + H2O (basic media)5. ClO + Cl + (basic media)
6. Cl2 + Cl
7. NO2 + NO
8. P + Cu2+ Cu3P + H3PO3
1.2.7 Stoichiometric calculations and limiting reactant
When all the reactants are completely and simultaneously consumed in a chemical reaction, they are exactly in stoichiometric proportions/amounts (the proportions indicated by the balanced equation).
However, some reactions proceed better when one reactant is in stoichiometric excess. Therefore, in many situations a chemist deliberately mixes reactants in a mole ratio that does not agree with the coefficients of the equation. In this sense, the reactant that is completely consumed first (the limiting reagent) determines the amount of products that form.
Excess Reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent.
The concept of the limiting reagent is analogous to the relationship between men and women in a dance contest at a club. If there are fourteen men and only nine women, then only nine female/male pairs can compete. Five men will be left without partners. The number of women thus limits the number of men that can dance in the contest, and there is an excess of men.
In stoichiometric calculations involving limiting reagents, the first step is to decide / identify which reactant is the limiting reagent. Then, calculate the amount of product formed based on the amount of the limiting reagent available.
Notes:1) The first step in a stoichiometric calculation is to write a balanced equation. Either this
must be given or you must be able to supply your own.2) The coefficients in the balanced equation tell you only the molar ratios in which the
species combine or are formed.
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System(before the reaction)
System(after the reaction)
Limiting reactant
Excess Reactant Product
Balanced
Chapter 1: Matter
3) Always base the calculation of the maximum yield of product on the stoichiometric equivalency (molar ratio) between it and the limiting reagent.
4) To identify the limiting reagent:
Example 1.2.14:
Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide:2NH3(g) + CO2(g) (NH2)2CO(aq) + H2O(l)
In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.a) Which of the two reactants is the limiting reagent?b) Calculate the mass of (NH2)2CO formed.c) How much of the excess reagent (in grams) is left at the end of the reaction?
a) Since we cannot tell by inspection which of the two reactants is the limiting reagent, we
have to proceed by first converting their masses into numbers of moles.
Moles of NH3 = 637.2 g NH3 x = 37.42 mol NH3
Moles of CO2 = 1142 g CO2 x = 25.95 mol CO2
Method 1: The balanced equation shows that 2 mol NH3 1 mol CO2; therefore, the number of moles of NH3 needed to react with 25.95 mol CO2 is given by
25.95 mol CO2 x = 51.90 mol NH3
Since there are only 37.42 moles of NH3 present, not enough to react completely with the CO2, NH3 must be the limiting reagent and CO2 the excess reagent.
Method 2: ratio of
for NH3: = 18.71 for CO2: = 25.95
Since the NH3 has the smallest ratio, NH3 is the limiting reagent.
b) To calculate the maximum yield of (NH2)2CO formed, we must based on the stoichiometric proportions (the proportions indicated in the balanced equation) between the product and the limiting reagent, NH3.The balanced equation shows that 2 mol NH3 1 mol (NH2)2CO
Mass of (NH2)2CO = 37.42 mol NH3 x
= 1124 g (NH2)2CO
c) Since NH3 is the limiting reagent, CO2 must be the reactants left over after the reaction is finished. The difference between the amount of CO2 available and the amount consumed is the amount left over.
24
The limiting reagent is the reactant with the smallest ratio of
Chapter 1: Matter
The number of moles of CO2 (the excess reagent) left over is
25.95 mol CO2 (37.42 mol NH3 x ) = 7.24 mol CO2
mass of CO2 left over = 7.24 mol CO2 x = 319 g
CO2
Exercise 1.2.13:
1) The reaction between aluminium and iron (III) oxide can generate temperatures approaching 3000C and is used in welding metals:
2Al + Fe2O3 Al2O3 + 2Fe In one process 124 g of Al are reacted with 601 g of Fe2O3.a) Identify the limiting reagent.b) Calculate the mass (in grams) of Al2O3 formed.c) How much of the excess reagent (in grams) is left over at the end of the reaction?
2) Aluminium, Al reacts with sulphuric acid, H2SO4, which is the acid in automobile batteries, according to the equation: 2Al + 3H2SO4 Al2(SO4)3 + 3H2
If 20.0 g of Al is put into a solution containing 115 g of H2SO4,a) Which reactant will be used up first? (Ans:Al)b) Calculate the number of moles of Al2(SO4)3 will be produced? (Ans:1.11 mol)c) How many grams of Al2(SO4)3 can be formed? (Ans:127 g)d) Calculate the mass (in grams) of the excess reagent remaining at the end of the reaction.
(Ans: 6 g)
3) Hydrazine, N2H4 and hydrogen peroxide, H2O2 are used as rocket reagent in which the mixture of these two substances will produce nitric acid, HNO3 and water.a) Write a balanced chemical equation for this reaction.b) How many moles of HNO3 will be formed from 3.5 g of N2H4? (Ans:0.22 mol)c) Calculate the number of moles of H2O2 required to react with 22.0 g of N2H4?d) (Ans: 4.81 mol)e) How many moles of HNO3 are produced when 30.5 g H2O2 is allowed to react with 25.6 g of
N2H4? Which compound is the limiting reagent? (Ans:0.256 mol)f) How many grams of the reactant in excess remain after the reaction is completed?
(Ans:21.5 g)
1.2.8 Reaction yield and percentage yield
The amount of limiting reagent present at the start of a reaction determines the theoretical yield of the reaction, that 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.
In practice, the actual yield, or the amount of product actually obtained from a reaction.
There are several reasons for the difference between actual and theoretical yield: 1. Side reaction may occur, that is other reactions in addition to the principle one may
take place.
25
Chapter 1: Matter
2. Procedures may be necessary to separate the product from the reaction mixture and obtain it in a pure state. Product may be lost during the separation and purification processes.
3. The reaction may not go to completion. 4. There may have been impurities in one or more of the reactants.
To determine how efficient a given reaction is, chemists often figure the percentage yield, which describes the proportion of the actual yield to the theoretical yield.
% yield = x 100%
Exercise 1.2.14:
1) Chlorofluorocarbons, commonly known as Freons, have been implicated in the gradual destruction of the earth’s ozone shield. One of these, Freon-12 (CCl2F2), is a gas that is used as a refrigerant and is prepared by the reaction:
3CCl4 + 2SbF3 3CCl2F2 + 2SbCl3
In a certain experiment, 14.6 g of SbF3 was allowed to react with an excess of CCl4. After the reaction was finished, 8.62 g of CCl2F2 was isolated from the reaction mixture.
a) What was the theoretical yield of CCl2F2 in this experiment expressed in grams? (Ans:14.8g)
b) What was the actual yield of CCl2F2 in grams? (Ans:8.62g)
c) What was the percentage yield of CCl2F2 in this experiment? (Ans:58.2%)
2) In an experiment, a student allowed benzene, C6H6 to react with excess bromine, Br2 in an attempt to prepare bromobenzene, C6H5Br. This reaction also produced, as a by-product, dibromobenzene, C6H4Br2. On the basis of the equation:
C6H6 + Br2 C6H5Br + HBr
a) What is the maximum number of grams of C6H5Br that the student could have hoped to obtainfrom 15.0 g of benzene? (Ans:30.1 g)
b) In this experiment, the student obtained 2.50 g of C6H4Br2. How many grams of C6H6 were not converted to C6H5Br? (Ans:0.828 g)
c) What was the student’s actual yield of C6H5Br in grams? (Ans:28.5 g)d) Calculate the percentage yield for this reaction. (Ans:94.7%)
3) A chemist wishes to synthesize a certain compound that has a molecular mass of 100. The synthesis requires six consecutive steps, each giving a 50% yield (computed on a mole basis). If the chemist begins with 30.0 g of starting material having a molecular mass of 80.0, how many grams of final product will be obtained? How many grams of starting material will be required to produce 10.0 g of final product. (Ans:0.590g; 508g)
4) Industrially, vanadium metal, which is used in steel alloys, can be obtained by reacting vanadium(V) oxide with calcium at high temperatures:
5Ca + V2O5 5CaO + 2VIn one process 1.54 x 103 g of V2O5 react with 1.96 x 103 g of Ca.
26
Chapter 1: Matter
a) Calculate the theoretical yield of V. (Ans:863 g)b) Calculate the percent yield if 803 g of V are obtained. (Ans:93.0 %)
5) Ethylene, C2H4 an important industrial organic chemical, can be prepared by heating hexane, C6H14 at 800C:
C6H14 C2H4 + other products
If the yield of ethylene production is 42.5%, what mass of hexane must be reacted to product 481g of ethylene? (Ans:3.47x103 g)
1.2.9 Stoichiometry of reactions in solution
(a) Dilution of solution
Dilution is a procedure for preparing a less concentrated solution from a more concentrated one.All the solute in the initial more concentrated solution, appears in the final diluted solution.
moles of solute before dilution = moles of solute after dilution
Notes:
Adding more solvent to a given amount of the stock solution changes (decreases) the concentration of the solution without changing the number of moles (or mass) of solute present in the solution.
Example 1.2.15:
A particular analytical chemistry procedure requires 0.0500 M K2CrO4. What volume of 0.250 M K2CrO4 must be diluted with water to prepare 100 mL of 0.0500 M K2CrO4.
= 0.0500 M = 0.250 M
= 100.0 mL = ?
(0.0500 M)(100 mL) = (0.250 M)
= 20.0 mL
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Volume of water (solvent) added
Dilution
ci
Vi
Vf
cf
= Molarity, M = ; V in litre/dm3
= i = initial – concentrated solution f = final – diluted solution
Chapter 1: Matter
The laboratory procedure for preparing a solution by solution is as follow:A pipette is used to withdraw a 20.0 mL sample of 0.250 M K2CrO4(aq).The pipetteful of 0.250 M K2CrO4 is discharged into a 100.0 mL volumetric flask.Following this, water is added to bring the level of the solution to the calibration mark etched on the neck of the flask. At this point the solution is 0.0500 M K2CrO4.
Exercise 1.2.15:
1) When left in an open beaker for a period of time, the volume of 275 mL of 0.105 M NaCl is found to decrease to 237 mL. What is the new concentration of the solution?
2) Distilled water is added to a 25.0 mL of 0.866 M KNO3 so that the final volume is brought up to 500 mL. What is the new concentration of this solution? (Ans:0.0433M)
3) You have a solution of 505 mL 0.125 M HCl and you would like to dilute it to 0.100 M. What volume of water should you add? (Ans:126mL)
4) A solution of 46.2 mL 0.568 M calcium nitrate, Ca(NO3)2 is added to a solution of 80.5 mL 1.396 M calcium nitrate. What is the final concentration of the mixture?
(Ans:1.09M)
(b) Titration / Titrimetric analysis / titrimetry
The titration is a method of volumetric analysis in which a volume of one reagent (the titrant) is added slowly (from burette) through the stopcock into another vessel (conical flask) that contains a known volume of another reagent (analyte).
The trick is to stop the titration at the point where both reactants (titrant and analyte) have been consumed simultaneously, a condition called the equivalence point of the titration.
In a titration we need some means of signalling when the equivalence point is reached. A small quantity of indicators added to the reaction mixture and it changes colour at or very near the equivalence point.
The point at which an excess one-drop of titrant changes the colour of the indicator is called the end point of the titration - the delivery of the titrant is stopped and the volume of the titrant used in the reaction recorded.
Given the concentration of one (standard solution), the aim is to find the concentration of the other.
analyte + titrant product
28
Titrant:: concentration of titrant: volume of added titrant or titre value
Analyte:: concentration of analyte: volume of pippeted analyte
Calculations:
= stoichiometric proportions
Chapter 1: Matter
Types of Titration:i. Acid-base titrationsii. Redox titrations
iii. Back titrations
(i) Acid-Base Titrations
Reaction between acid and base
Example 1.2.16: How many millilitres of 0.112 M HCl will react exactly with the sodium carbonate in 21.2 mL of 0.150 M Na2CO3 according to the following equation?
2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + CO2(g) + H2O(l)
= 0.112 M HCl = 0.150 M Na2CO3
= ? = 21.2 mL Na2CO3
The balanced equation shows the stoichiometric ratio is 2:1.
= 56.8 mL HCl soln
Exercise 1.2.16:
1) In a titration between 20.0 cm3 standard solution of sodium carbonate and HCl, it needs 19.15 cm3 of HCl for neutralization. If the concentration of Na2CO3 is 0.052 M, calculate the concentration of HCl. (Ans:0.109M)
2) 0.321 g sample of sodium carbonate, contaminated by sodium chloride, was dissolved in water. The resulting solution required 35.4 mL of 0.144 M HCl to react completely with the sodium carbonate. (The impurity does not interfere.) The products of the reaction are sodium chloride, carbon dioxide and water.a) Write the balanced equation for this reaction.b) How many grams of sodium carbonate were in the sample? (Ans:0.270g)c) Calculate the percentage of sodium carbonate in the sample. (Ans:84.1% )
3) Calculate the volume of 0.12 mol dm3 KOH required to react with 25.0 cm3 of H3PO4 solution according to the equation:
2KOH + H3PO4 K2HPO4 + 2H2OThe H3PO4 solution contains 4.90 g of H3PO4 per dm3. [H=1.00; O=16.0; P=31.0]
(Ans:20.8cm-3)
4) A sample of analgesic drug was analyzed for aspirin, a monoprotic acid, HC9H7O4, by titration with a base. In a titration, a 0.500 g sample of the drug required 21.50 mL of 0.100 M NaOH for complete neutralization. What percentage by mass of the drug was aspirin? (Ans:77.4%)
5) 5.125 g of washing soda crystal are dissolved and made up to 250 cm3 of solution. A 25.0-cm3
portion of the solution requires 35.8 cm3 of 0.0500-mol dm3 sulphuric acid for neutralization. Calculate percentage of sodium carbonate in the crystal. (Ans:37.1%)
29
Chapter 1: Matter
6) Sodium carbonate crystal (27.8230 g) were dissolved in water and made up to 1.00 dm3. 25.0 cm3 of the solution were neutralized by 48.8 cm3 of hydrochloric acid of concentration 0.100 mol dm3. Find n in the formula Na2CO3
.nH2O. (Ans: 10)
7) 0.960 g of zinc is dissolved in 50.00 cm3 of 1.04 M HCl solution. The final volume is brought up to 250 cm3 by adding water. In a titration, 20.0-cm3 portion of the solution is completely neutralized with 17.20 cm3 of 0.105 M NaOH. What is the atomic mass of zinc? (Ans:65.3)
(ii) Redox titration
Reactions between oxidants and reductants.
In a redox titration that involves permanganate ion, or dichromate ion, , no indicator
is needed in the reaction as these ions have an obvious different colour between the ions and their reduced form.
Mn2+
(purple) (colourless)
MnO2 (p)
(purple) (brown)
Cr3+
(orange) (green)
For the redox reaction between I2 and , starch solution is added near the end point as an
indicator for the redox titration.
Example 1.2.17: A 16.42-mL volume of 0.1327 M KMnO4 solution is needed to oxidize 20.00 mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO 4
solution? The net ionic equation is
5Fe2+ + + 8H+ Mn2+ + 5Fe3+ + 4H2O
The balanced equation shows the stoichiometric proportion is 5:1 (5 mol Fe2+ 1 mol ).
(Fe2+) = ? ( ) = 0.1327 M
= 20.00 mL = 16.42 mL
= 0.5450 M
Exercise 1.2.17:
1) How many millilitres of a 0.206 M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution according to the following equation:
30
Chapter 1: Matter
10HI + 2KMnO4 + 3H2SO4 5I2 + 2MnSO4 + K2SO4 + 8H2O (Ans:204 mL)
2) In a titration, 25.00 cm3 of H2O2 solution is needed to react completely with 35.00 cm3 of 0.02 M KMnO4 solution. Calculate the concentration of H2O2 solution. (Ans:0.07M)
3) Ammonium iron(II) sulphate crystals have the following formula: (NH4)2SO4.FeSO4
.nH2O. In an experiment to determine n, 8.492 g of the salt were dissolved and made up to 250 cm 3 of solution with distilled water and dilute sulphuric acid. A 25.0 cm3 portion of the solution was further acidified and titrated against potassium manganate(VII) solution of concentration 0.0150 mol dm3. A volume of 22.5 cm3 was required. (Ans:12)
4) Vanadium(II) ions can be oxidised quantitatively by acidified manganate(VII) ions, to a higher oxidation number. Two half-equation for the reaction can be written:V2+ Vz+ + (z – 2)e
MnO4 + 8H+ + 5e Mn2+ + 4H2O
In an experiment, 25.0 cm3 of 0.02 moldm3 V2+ was found to react with 15.0 cm3 of 0.02 mol dm3 MnO4
, acidified with sulphuric acid.a) How many moles of V2+ were used in the titration? (Ans:0.0005 mol)b) How many moles of MnO4
were used in the titration? (Ans:0.0003 mol)c) How many moles of V2+ reacted with one mole of MnO4
? (Ans:5/3 mol)d) Hence deduce the number of electrons given up by one V2+ ion when it is oxidised. (Ans:3)e) What is the new oxidation number of vanadium after oxidation (ie what is z)? Ans:(+5)
5) An amount of a solution containing 0.092 M of iron(II) requires 21.7 cm 3 of M2Cr2O7 solution with a concentration of 4.63 g/dm3 for complete oxidation. Determine the relative atomic mass of M. (Ans:23)
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