chapter one chemistry

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Unit 1-1 Unit 1 Atoms, Molecules and Stoichiometry Section 1.1 The Atomic Structure ( 1 ) Protons, neutrons and electrons as constituents of the atom Atoms and molecules The first chemist to use the name ‘atom’ was John Dalton (1766-1844). Dalton used the word ‘atom’ to mean the smallest particle of an element. He then went on to explain how atoms could react together to form molecules : An atom is the smallest part of an element which can ever exist. A molecule is the smallest part of an element or a compound which can exist alone under ordinary conditions. Electrons In 1897, J. J. Thomson investigated the conductivity of electricity by gases at very low pressure. At ordinary pressures gases are electrical insulators, but when they are subjected to very high voltages at very low pressures (below 0.01 atm) they break down and conduct electricity. When Thomson applied 15000 volts across the electrodes of a tube containing a trace of gas, a bright green glow appeared on the glass. The green glow results from the bombardment of the glass by rays travelling in straight lines from the cathode. Thomson called these rays cathode rays. He also showed that when the rays were deflected by an electric field across a pair of charged plates, the rays moved away from the negative plate towards the positive plate. This suggested that the rays were negative. Thomson studied the blending of a thin beam of cathode rays by magnetic and electric fields and concluded that they consisted of electrons – tiny negatively charged particles. The same results were obtained with the cathode rays using different gases in the tube and with tubes and electrodes of different materials. This suggested that electrons were present in the atoms of all substances.

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Page 1: Chapter One Chemistry

Unit 1-1

Unit 1 Atoms, Molecules and Stoichiometry Section 1.1 The Atomic Structure ( 1 ) Protons, neutrons and electrons as constituents of the atom Atoms and molecules The first chemist to use the name ‘atom’ was John Dalton (1766-1844). Dalton used the word ‘atom’ to mean the smallest particle of an element. He then went on to explain how atoms could react together to form molecules : An atom is the smallest part of an element which can ever exist. A molecule is the smallest part of an element or a compound which can exist alone under ordinary conditions. Electrons In 1897, J. J. Thomson investigated the conductivity of electricity by gases at very low pressure. At ordinary pressures gases are electrical insulators, but when they are subjected to very high voltages at very low pressures (below 0.01 atm) they break down and conduct electricity.

When Thomson applied 15000 volts across the electrodes of a tube containing a trace of gas, a bright green glow appeared on the glass. The green glow results from the bombardment of the glass by rays travelling in straight lines from the cathode. Thomson called these rays cathode rays. He also showed that when the rays were deflected by an electric field across a pair of charged plates, the rays moved away from the negative plate towards the positive plate. This suggested that the rays were negative.

Thomson studied the blending of a thin beam of cathode rays by magnetic and electric fields and

concluded that they consisted of electrons – tiny negatively charged particles. The same results were obtained with the cathode rays using different gases in the tube and with tubes and electrodes of different materials. This suggested that electrons were present in the atoms of all substances.

Page 2: Chapter One Chemistry

Unit 1-2 Rutherford’s model of atomic structure In 1911, Ernest Rutherford had the idea of probing inside the atom using alpha-particles. He used alpha-particles from radioactive substances as ‘nuclear bullets’. Alpha-particles are helium ions, He2+, with positive charge. Rutherford expected that most of the very fast-moving alpha-particles would pass straight through the thin metal foil or be deviated a little.

Most of the alpha-particles did pass straight through the foil, but when the detecting screen and microscope were rotated from the straight-on position flashes could still be seen. Clearly, some of the alpha-particles were deflected by the foil, but to every one’s surprise, one particle in every 10000 appeared to rebound from the foil. Rutherford suggested that deflections and reflections could only be caused by the particles coming close to a concentrated region of positive charge.

Rutherford concluded that atoms in the metal foil consisted of a central positive nucleus composed of protons, where the mass of the atom was concentrated. This nucleus was surrounded by a much larger volume in which the electrons move. From the angles through which alpha-particles are deflected, Rutherford calculated that the nucleus of an atom would have a radius of about 10-14 m. This is about one ten-thousandth of the size of the whole atom which has a radius of about 10-10 m.

Page 3: Chapter One Chemistry

Unit 1-3 Neutrons In spite of the success of Rutherford in explaining atomic structure, one major problem remained unsolved. If the hydrogen atom contains one proton and the helium atom contains two protons, then the relative atomic mass of helium should be twice that of hydrogen. Unfortunately, the relative atomic mass of helium is four and not two. In 1932, Chadwick, one of Rutherford’s collaborators, was able to show where the extra mass in helium atoms came from. Chadwick bombarded a thin sheet of beryllium with alpha-particles. The alpha-particles can be traced by electric counter which detects charged particles. When the beryllium is in place, the counter registers nothing, showing that the alpha-particles are being stopped by the beryllium. However, if a piece of paraffin wax is placed between the beryllium and the counter, charged particles are detected again. Chadwick provided an explanation. He suggested that the alpha-particles striking the beryllium foil displaced uncharged particles called neutrons from the nuclei of beryllium atoms. These uncharged neutrons could not affect the charged-particle counter, but they could displace positively charged protons from the paraffin wax which would affect the counter. Further experiments showed that neutrons had almost the same mass as protons and Chadwick was able to explain the difficulty concerning the relative atomic masses of hydrogen and helium. Hydrogen atoms have one proton, no neutrons and one electron. Since the mass of the electron is negligible compared to the masses of the proton and neutron, a hydrogen atom has a relative mass of one unit. Helium atoms have two protons, two neutrons and two electrons, so the relative mass of a helium atom is four units. This means that a helium atom is four times as heavy as a hydrogen atom.

Page 4: Chapter One Chemistry

Unit 1-4 ( 2 ) The relative masses and charges of a proton, neutron and electron Nowadays scientists believe that all atoms are composed of three important sub-atomic particles : protons, neutrons and electrons.

Sub-atomic particle

Mass / Kg

Relative mass (to that of a proton)

Charge / C

Relative charge (to that on a proton)

Proton 1.6726 x 10-27 1 + 1.6022 x 10-19 + 1 Neutron 1.6750 x 10-27 1 0 0 Electron 9.1095 x 10-31

18361

- 1.6022 x 10-19 - 1

( 3 ) The atomic nucleus The nucleus of an atom is composed of protons and neutrons. Because protons and neutrons occupy the nucleus, they are sometimes collectively called nucleons. Virtually all the mass of the atom is concentrated in the nucleus, which occupies only a small fraction of the total volume of the atom. The neutron has no charge, whereas the proton carries one positive charge. Electrons with one negative charge occupy the space outside the nucleus. The mass of an electron is 1836 times less than that of a proton. The atomic number or the proton number of an element is the most important feature of an element’s individuality because it represents (i) the number of protons in the nucleus, (ii) the number of electrons in the neutral atom, (iii) the position in which the element appears in the periodic table. The number of protons + the number of neutrons in an atom is called the mass number or the nucleon number. The word nuclide is used to describe any atomic species of which the atomic number and the mass number are specified. The symbol A

Z X is used to represent the nuclide X with atomic number Z and mass number A. Example : Particle/Atom Proton Neutron Electron Hydrogen Helium

Symbol 11H or 11p 1

0n 0-1e 1

1H 42He

Atoms of the same element with different masses are called isotopes. All the isotopes of one particular element have the same atomic number because they have the same number of protons, but they have different mass numbers because they have different numbers of neutrons. Isotopes have the same number of electrons and hence the same chemical properties, because chemical properties depend upon the transfer and redistribution of electrons. As isotopes have different number of neutrons, they have different masses and hence different physical properties. For example, pure 37

17Cl2 has a higher density, higher melting point and higher boiling point than pure 35

17Cl2. Example : Hydrogen have three isotopes : hydrogen-1, hydrogen-2 and hydrogen-3. Write their symbols. Example : The nucleus of a fluorine atom has a diameter of about 1.0 x 10-12 cm and a mass of 3.1 x 10-23 g, calculate the density of the fluorine nucleus.

Page 5: Chapter One Chemistry

Unit 1-5

Section 1.2 Relative isotopic, atomic and molecular masses ( 1 ) Relative isotopic and atomic masses Carbon-12 scale

Chemists use a relative atomic mass scale to compare the masses of different atoms. In 1961 carbon-12 (12C) was chosen as standard against which the masses of other atoms were compared

Carbon-12, an isotope of carbon, has been assigned a relative atomic mass of exactly 12. This scale is called the carbon-12 scale. The isotope of carbon was chosen because carbon is a very common element. Being a solid, it is easy to store and transport. Relative isotopic mass

Relative isotopic mass of a particular isotope of an element is the relative mass of one atom of that isotope on the carbon-12 scale.

Relative isotopic mass = 12-carbon of atom one of Mass

121

elementan of isotopean of atom one of Mass

Atomic mass unit

Relative isotopic mass is a ratio, it has no unit. An atomic mass unit (a.m.u.) is taken as 121 of

the mass of the carbon-12 atom. Thus one atom of C-12 weighs 12.000 a.m.u. and its relative isotopic mass is 12.000. Example :

The mass of a carbon-12 atom is 1.9926 x 10-26 kg. What is the mass of 1 a.m.u. in kg ?

On the carbon-12 scale, the mass of the proton (1.0074 a.m.u.) is almost the same as that of the neutron (1.0089 a.m.u.), and the mass of the electron is very small in comparison (0.0005 a.m.u.). Now since the relative masses of the proton and neutron are very close to one and the electron has a negligible mass, it follows that all relative isotopic masses will be very close to whole numbers. In fact, the relative isotopic mass of an isotope will be very close to its mass number and the two are assumed to be almost identical in all but the most accurate work. Example : Symbol 12C 16O 17O 18O 35Cl 37Cl Relative isotopic mass 12.000 15.995 16.999 17.999 34.969 36.966 Mass / a.m.u. Relative atomic mass

Naturally occurring elements often consist of a mixture of isotopes. Relative atomic mass ( Ar ) of an element is the weighted average of the relative isotopic masses of

the natural isotopes of an element on the carbon-12 scale.

Relative atomic mass = 12-carbon of atom one of Mass

121

elementan of atom one of mass Average

Example : Element Isotopic mass Relative abundance %

16O 15.995 99.76 17O 16.999 0.04

Oxygen

18O 17.999 0.20

Relative atomic mass of oxygen :

12C 12.0000 98.89 13C 13.0034 1.11

Carbon

14C 14.0041 negligible

Average mass of one carbon atom :

Page 6: Chapter One Chemistry

Unit 1-6 Example : The relative atomic mass of chlorine is 35.45. It has two isotopes with mass numbers 35 and 37. Calculate the relative abundances of each isotope. ( 2 ) Relative molecular mass Relative molecular mass ( Mr ) is the mass of one molecule relative to

121 of the mass of a 12C

atom on the carbon-12 scale.

Relative molecular mass = 12-carbon of atom one of Mass

121

molecule one of mass Average

The relative mass of a molecule can be found by adding up the relative atomic masses of all the

atoms in it. Example :

Find the relative molecular mass of ethanol, CH3CH2OH.

A vast number of compounds consist of ions, not molecules. The compound sodium chloride, for example consists of sodium ions and chloride ions. For ionic compounds, the term formula unit is used to describe the ions which make up the compound. A formula unit of sodium chloride is NaCl.

Relative molecular mass of an ionic compound = 12-carbon of atom one of Mass

121

unit formula one of mass Average

Example :

Calculate the relative molecular mass of copper(II) sulphate-5-water.

Page 7: Chapter One Chemistry

Unit 1-7 ( 3 ) Mass spectrometer

Mass spectrometer consists of five parts : the vaporization chamber, ionization chamber, electric field, magnetic field, ion detector and the recorder.

Vaporization chamber The material to be analyzed may be an element or a compound. The sample of the material under investigation is injected and heated to vaporize in the vaporization chamber. Ionization chamber The vaporized sample then passes into the ionization chamber. Here, atoms or molecules of the sample are bombarded with a stream of high-energy electrons emitted from the electron gun. This causes ionization of the atoms or molecules to form positive ions which are mainly singly charged. X(g) + e- → X+

(g) + 2 e- Electric field and magnetic field

The ions are then accelerated by an electric field and deflected along a circular path by a magnetic field. The lighter the positive ions, the greater is the deflection. At a given electric and magnetic field strength, only ions of a particular mass/charge (m/e) ratio can hit the ion detector.

Ion detector and recorder

By varying the strength of accelerating electric field or deflecting magnetic field, ions of any m/e ratio can be brought to the ion detector. A mass spectrum showing the m/e ratio of the ions and the corresponding intensity (i.e. the relative amount of that ion) can then be traced out by a recorder.

Page 8: Chapter One Chemistry

Unit 1-8 In the mass spectrum of an element, the peaks can give information about various isotopes of the element. Whereas in the mass spectrum of a compound, the peak with highest m/e ratio will most likely corresponding to the ‘molecular ion’, i.e. the molecule which has lost only a single electron. In most mass spectra, the values of the m/e ratio can be converted to the relative masses of the particles if the charges on the ions are taken to be one. Example 1

When 13C and 12C are analyzed in a mass spectrometer, the ratio of their masses is found to be

=CC

12

13

Mass Mass 1.0836129

Calculate the relative isotopic mass of 13C. Example 2

The following figure shows a mass spectrometer trace for the isotopes of neon. Calculate the relative atomic mass of neon.

Example 3 The following figure shows the mass spectrum of HCl. The peak at mass 36 corresponds to the

molecular ion (1H35Cl)+.

(i) What ion is responsible for the peak at mass 38 ? (ii) What ions are responsible for the two lower peaks ? (iii) How do you explain the relative heights of the peaks at mass 36 and 38 ?

Page 9: Chapter One Chemistry

Unit 1-9

Section 1.3 The mole concept ( 1 ) The mole and the Avogadro constant The mole

The counting unit for atoms, molecules and ions is the mole ( mol ). It is defined as the amount of substance that contains as many elementary particles as there are atoms in exactly 12 g of carbon-12. The Avogadro constant

The Avogadro constant ( L ) relates the number of particles to the amount. It is represented as : L = 6.02 x 1023 mol-1

The molar mass

The molar mass ( M ) of a substance is the mass per mole of that substance. Molar mass has the unit : g mol-1. It follows from the definition of the mole that the molar mass of carbon-12 is exactly 12 g mol-1. The term molar mass applies not only to elements in the atomic state but also to all chemical species – atoms, molecules, ions, etc. The phrase ‘1 mol of chlorine’ has two possible meanings because it does not specify whether it refers to atoms or molecules. To avoid confusion, we must always specify the entity, either by formula or in words :

1.0 mol of Cl or one mole of chlorine atoms 1.0 mol of Cl2 or one mole of chlorine molecules

Example 1

Using the Periodic Table, calculate the molar masses of (a) ammonia, NH3,

(b) calcium bromide, CaBr2. Amount, mass and number

In chemistry, the amount ( n ) of substance can be calculated from the mass and molar mass of that substance :

Amount of substance = massmolar

mass or n = Mm

Example 2

A sample of ammonia, NH3, weighs 1.00 g. (a) What amount of ammonia is contained in this sample ?

(b) What mass of sulphur dioxide, SO2, contains the same number of molecules as are in 1.00 g of ammonia ?

If the number of particles in a given amount of substance is required, it can be found using the expression : N = nL

where N = the number of particles, n = the amount, L = the Avogadro constant. Example 3

Calculate the number of atoms in 7.20 g of sulphur, S8 .

Page 10: Chapter One Chemistry

Unit 1-10 ( 2 ) Molar volume of gases at R.T.P. and S.T.P.

The following table shows the results of an experiment to determine the volume of 1 mole of four common gases at R.T.P. (25℃ and 1 atm) :

Gas Mass of 1 dm3 of gas / g dm-3

Relative molecular mass of gas

Molar mass of gas / g mol-1

Volume of 1 mole of gas / dm3

O2 1.31 32.00 N2 1.15 28.02 CO 1.15 28.01 CO2 1.81 44.01

In fact, the results of a large number of experiments show that one mole of any gas at R.T.P. occupies

24.0 ± 0.1 dm3 or 22.4 ± 0.1 dm3 at S.T.P. (0℃ and 1 atm). The volume of 1 mole of gas is known as the molar volume.

Avogadro’s principle states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. Therefore, the number of moles of a given volume of a gas can be found by the following formula :

Number of moles of a gas = meMolar volu

gas theof Volume

Example 1

Find the number of molecules in 4.48 cm3 of carbon dioxide gas at standard temperature and pressure.

(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1 ; Avogadro constant = 6.02 x 1023 mol-1) Example 2

1.6 g of a gas occupies 1.2 dm3 at room temperature and pressure. What is the relative molecular mass of the gas ? (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)

Page 11: Chapter One Chemistry

Unit 1-11 ( 3 ) Ideal gas equation Boyle’s law

In 1662, Robert Boyle had discovered that the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature remains constant Charles’s law

In 1787, Charles showed that the volume of a fixed mass of gas is directly proportional to its absolute temperature, provided the pressure remains constant. Avogadro’s law

In 1811, the Italian chemist Amadeo Avogadro deduced that the volume of a gas is directly proportional to its amount, provided the pressure and temperature remain constant. Ideal gas equation

By combining Boyle’s law, Charles’s law and Avogadro’s law,

The constant is given the symbol R and is called the gas constant. For one mole of gas at S.T.P.,

Operating in SI units, the volume of one mole of gas is 0.0224 m3 at a pressure of 101325 Nm-2 (Pa) and 273 K.

For n moles of gas, PV = nRT, which is known as the ideal gas equation. A gas which obeys this equation is called an ideal gas or a perfect gas. In practice, real gases obey

the equation very closely at low pressure and high temperature.

Page 12: Chapter One Chemistry

Unit 1-12 Example 1

Assuming ideal behaviour, calculate the volume occupied by 2.00 g of carbon monoxide at 20℃ under a pressure of 6.25 kPa. Example 2 (a) What is the volume of 1.50 g of hydrogen, H2, at 15℃ and a pressure of 750 mm Hg ? ( 1 atm = 760 mm Hg)

(b) At what temperature will 4.71 g of nitrogen occupy 12.0 dm3 at 1 atm ? Example 3

A widely used explosive is TNT which has a formula C7H5N3O6 . This is mixed with a solid oxidant so that oxygen required for combustion can be supplied rapidly.

(a) Write an equation for the combustion of TNT. Assume that C and H atoms are completely oxidized and that N atoms emerge as nitrogen gas.

(b) What amount of gas is produced from 1.00 mol of TNT, and what volume would it occupy at 1.00 atm and 400 ℃ ?

(c) Assume that the reaction occurs so fast that the gaseous products occupy only 2.00 dm3 at 600 ℃. What would be the resulting pressure ?

Page 13: Chapter One Chemistry

Unit 1-13 ( 4 ) Relative molecular mass determination 1. Finding the relative molecular mass of a gas by direct weighing

Using the Ideal gas equation, PV = nRT, and n = Mm where m is the mass of gas whose molar mass

is M.

By measuring m, p, V and T for a sample of gas and assuming ideal gas behaviour, it is possible to calculate M. Regnault’s method

A container of known volume (V) is weighed full of gas at a pressure P and temperature T. The same container is then weighed after evacuation in order to obtain the mass of gas inside (m). This method is accurate provided the following precautions are taken : 1. The container must be large, so as to give an appreciable weight of gas. Gases are so light that any

slight error in weighing produces a large percentage error if the mass of the substance weighed is very small.

2. The gas used must be perfectly pure and perfectly dry. 3. The container should be filled and emptied several times to make sure that all the air or the previous gas

has been removed. 4. The container must be evacuated as completely as possible. 5. All weighings must be carried out at the same temperature and pressure. Example :

A volume of 1.00 dm3 is occupied by 1.798 g of a gas at 298 K and 101 kPa. Calculate the molar mass of the gas.

Page 14: Chapter One Chemistry

Unit 1-14 2. Finding the relative molecular mass of a volatile liquid This method is a development of an earlier technique used by Victor Meyer. Figure below shows a suitable apparatus to use :

Procedures : 1. Draw a few cm3 of air into the graduated syringe and fit the self-sealing rubber cab over the nozzle.

Pass steam through the outer jacket until the temperature reading and the volume of air in the syringe become steady. Continue to pass steam through the jacket and record the temperature and the volume of air in the syringe.

2. Fill the hypodermic syringe with about 1 cm3 of the liquid under investigation. Weigh the hypodermic syringe and its contents and then push the needle through the self-sealing cap of the graduated syringe.

3. Inject about 0.2 cm3 of liquid into the graduated syringe and withdraw the hypodermic syringe. Immediately, re-weigh the hypodermic syringe and its contents.

4. The liquid injected into the graduated syringe will evaporate. The final volume of air plus vapour in the graduated syringe should be recorded when the volume becomes steadily. Finally, record the atmospheric pressure.

Questions : (a) Why is this method unsuitable for liquids which boil above 80℃ ? (b) Why should the hypodermic syringe be handled as little as possible between weighings ? (c) What are the main sources of error in this experiment ? Example : Find the relative molecular mass of a liquid X, given that : Mass of the liquid vaporized = 0.160 g Initial volume of air in graduated syringe = 10.0 cm3 Initial volume of air + vapour in graduated syringe = 56.0 cm3 Temperature = 100℃, Pressure = 1 atm, R = 8.31 JK-1mol-1

Page 15: Chapter One Chemistry

Unit 1-15 ( 5 ) Partial pressure of gas and mole fraction Partial pressure of gas In a mixture of gases, the partial pressure of a gas is the pressure that gas would exert if it alone occupied the container. Dalton’s law of partial pressure

In a mixture of ideal gases which do not react chemically, the total pressure of the mixture is the sum of the partial pressures of the constituent gases.

P = PA + PB + PC + ……… where PA, PB, PC ….. are partial pressures of the constituent gases A, B, C, ……

Dalton’s law of partial pressures is strictly true only for ideal gas mixtures. Consequently the law

is likely to de disobeyed at high pressures and low temperatures.

For example, in a mixture of three gases, A, B and C, with partial pressures PA, PB and PC, Mole fraction

In any mixture of substances, the mole fraction of a constituent A, XA , is given by the expressions :

Mole fraction of A = amounttotal

A ofamount or XA = nnA

There is a simple expression relating partial pressure of a gas to its mole fraction :

Partial pressure = total pressure x mole fraction Example 1 4.00 dm3 of oxygen at a pressure of 400 kPa and 1.00 dm3 of nitrogen at a pressure of 200 kPa are introduced into a 2.00 dm3 vessel. What is the total pressure in the vessel ? Example 2

A 500 cm3 globe contains oxygen at 1.00 atm. 300 cm3 of nitrogen, measured at the same temperature and 1 atm, is added under pressure, and then carbon dioxide is added till the total pressure is 3.10 atm. Calculate the partial pressure and mole fraction of each gas in the mixture.

Page 16: Chapter One Chemistry

Unit 1-16

Section 1.4 Empirical and Molecular Formulae ( 1 ) Empirical formulae The empirical formula shows the composition of a substance as the simplest ratio of the amount of the constituent elements. 1. Derivation of empirical formula from composition by mass Example 1

When 1.27 g of copper combine with oxygen, 143 g of an oxide are formed. What is the empirical formula of the oxide ?

Elements Cu O Masses / g 127 16 Molar mass / g mol-1 63.55 16.00 Amount / mol

Relative amount

Simplest ratio of relative amounts

Empirical formula is Example 2 10.00 g of hydrated barium chloride are 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.

Compounds present BaCl2 H2O Masses / g 8.53 1.47 Molar mass / g mol-1 208.2 18.0 Amount / mol

Relative amount

Simplest ratio of relative amounts

Empirical formula is and x =

Example 3

An organic compound was analyzed and was found to have the following percentage composition by mass : 48.8% carbon, 13.5% hydrogen and 37.7% nitrogen. Calculate the empirical formula of the compound.

Assume the mass of the sample is 100.0 g. Elements C H N Masses / g Molar mass / g mol-1 Amount / mol

Relative amount

Simplest ratio of relative amounts

The empirical formula of the organic compound is

Page 17: Chapter One Chemistry

Unit 1-17 2. Derivation of empirical formula using combustion data Example 1

An organic compound, X, contains only carbon, hydrogen and oxygen. 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. What is the empirical formula of X ?

Mass of carbon =

Mass of hydrogen =

Mass of oxygen =

Elements C H O Masses / g Molar mass / g mol-1 Amount / mol

Relative amount

Simplest ratio of relative amounts

The empirical formula of X is Example 2

Vitamin C is an organic compound known to contain the elements carbon, hydrogen and oxygen only. Complete combustion of a 0.200 g sample of this compound yields 0.2998 g CO2 and 0.0819 g H2O. What is the empirical formula of vitamin C ?

Mass of carbon =

Mass of hydrogen =

Mass of oxygen =

Elements C H O Masses / g Molar mass / g mol-1 Amount / mol

Relative amount

Simplest ratio of relative amounts

The empirical formula of vitamin C is

Page 18: Chapter One Chemistry

Unit 1-18 ( 2 ) Molecular formulae

A molecular formula shows the number of atoms of each element in a molecule or molecule unit of a substance. Molecular formula is an integral multiple of the empirical formula and so its relative molecular mass is also an integral multiple of the relative empirical formula mass. molecular formula = n (empirical formula) where n = 1, 2, 3, 4 ……….. etc. Questions (a) What are the molecular formula and empirical formula of benzene ? (b) What are the molecular formula and empirical formula of poly(ethene) ? Derivation of molecular formula from empirical formula and relative molecular mass Example 1

A liquid, Y, of relative molecular mass 88 contains 54.5% carbon, 36.4% oxygen and 9.1% hydrogen. Calculate the empirical formula of Y and deduce its molecular formula.

Assume the mass of Y is 100.0 g. Elements C H O Masses / g Molar mass / g mol-1 Amount / mol

Relative amount

Simplest ratio of relative amounts

The empirical formula of Y is

The molecular formula of Y is Example 2 Vitamin C was found to have a relative molecular mass lying somewhere between 170 – 180. Determine the molecular formula of vitamin C from its empirical formula, C3H4O3 and the range of relative molecular mass. Example 3

2.00 g of a hydrocarbon with the general formula CnH2n+2 gave 400 cm3 of vapour at 523 K and 1 atm. Given that the gas constant R is 0.082 atm dm3 K-1 mol-1, determine the molecular formula of this compound.

Page 19: Chapter One Chemistry

Unit 1-19

Section 1.5 Chemical Equation and Stoichiometry ( 1 ) The stoichiometry relationship between reactants and products Stoichiometry is the study of quantitative compositions of chemical substances and the quantitative changes that take place during chemical reactions.

A Chemical equation such as

N2(g) + 3 H2(g) → 2 NH3(g) is a kind of chemical balance sheet; it states that one mole of nitrogen reacts with three moles of hydrogen to yield two moles of ammonia. It does not tell us about the rate of the reaction or the conditions necessary to bring it about. The numbers 1,3 and 2 are called the stoichiometric coefficients, they tell us the mole ratio in which the substances react and in which the products are formed. Example

What mass of iodine will react completely with 10.0 g of aluminium ? Step 1: Write the balanced chemical equation for the reaction. Step 2: Find the mole ratio between aluminium and iodine from the stoichiometric coefficients. Step 3: Calculate the amount of aluminium. Step 4: Calculate the amount of iodine which reacts completely with this amount of Al. Step 5: Calculate the mass of iodine.

Page 20: Chapter One Chemistry

Unit 1-20 ( 2 ) Calculations involving reacting masses Example 1

What mass of oxygen would be produced by completely decomposing 4.25 g of sodium nitrate ? Example 2

Iron burns in chlorine to form iron chloride. An experiment showed that 5.60 g of iron combined with 10.65 g of chlorine. Deduce the equation for the reaction. Example 3

A mixture of 5.00 g of sodium carbonate and sodium hydrogencarbonate is heated. The loss in mass is 0.31 g. Calculate the percentage by mass of sodium hydrogencarbonate in the mixture.

Page 21: Chapter One Chemistry

Unit 1-21 Limiting reactant

In a chemical reaction, the amount of product is determined by the amount of the reactant that is not in excess and is used up completely in the reaction. This is called the limiting reactant. Example 4 5.00 g of iron and 5.00 g of sulphur are heated together to form iron(II) sulphide. Which reactant is present in excess ? What mass of product is formed ? Percentage yield

There are many reactions which do not go to completion. The percentage yield is found by comparing the actual mass of product obtained from the experiment and the theoretical mass of product calculated from the equation.

Percentage yield = product of mass calculated

product of mass actual x 100%

Example 5

24.0 g of ethyl ethanoate are obtained by 23.0 g of ethanol by esterification with ethanoic acid in the presence of concentrated sulphuric acid. What is the percentage yield of the reaction ? Example 6

29.5 g of ethanoic acid are obtained from the oxidation of 25.0 g of ethanol. What percentage yield does this represent ?

Page 22: Chapter One Chemistry

Unit 1-22 ( 3 ) Calculations involving volumes of gases Gay-Lussac’s law

In 1809, the French chemist Gay-Lussac stated that volumes of gaseous reactants and products of a reaction, measured at the same temperature and pressure, will be in a simple ratio to each other. Using the idea that the molar volume of any gas is approximately constant, the Italian chemist Amadeo Avogadro stated his principle in 1811 and provided an explanation for Gay-Lussac’s law. Example 1

What is the volume of oxygen needed for the complete combustion of 2 dm3 of propane ? Example 2

15 cm3 of a mixture of carbon monoxide and methane was mixed with excess oxygen and exploded. There was a contraction in volume of 21 cm3 at the same room temperature and 1.0 atm. Calculate the mole fraction of each gas in the mixture and their partial pressures. Example 3

20 cm3 of ammonia are burnt in an excess oxygen at 110℃. 10 cm3 of nitrogen and 30 cm3 of steam are formed. Deduce the formula for ammonia, given that the formula of nitrogen is N2, and the formula of steam is H2O.

Let the formula of ammonia be NaHb .

Page 23: Chapter One Chemistry

Unit 1-23 Finding molecular formula of a hydrocarbon by combustion

A hydrocarbon in the vapour phase is burned in an excess of oxygen to form carbon dioxide and water vapour. When the mixture of gases is cooled to room temperature, water vapour condenses to occupy a very small volume. The gaseous mixture consists of carbon dioxide and unused oxygen. The volume of carbon dioxide can be found by absorbing it in an alkali. From the volumes of gases, the equation for the reaction and the formula of the hydrocarbon can be found. Example 4

10 cm3 of the gaseous hydrocarbon were mixed with 33 cm3 of oxygen which was in excess. The mixture was exploded and, after cooling to room temperature, the residual volume of gas occupied 28 cm3. On adding concentrated potassium hydroxide the volume decreased to 8 cm3. Find the molecular formula of the hydrocarbon. Example 5 10 cm3 of a hydrocarbon, B, are exploded with an excess of oxygen. A contraction of 35 cm3 occurs, all volumes being measured at room temperature and pressure. On treatment of the products with sodium hydroxide, a contraction of 40 cm3 occurs. Deduce the molecular formula of B.

Page 24: Chapter One Chemistry

Unit 1-24 ( 4 ) Calculations involving concentrations and volumes of solutions Molar concentration

Molar concentration is the amount in moles of a substance present per dm3 of solution.

Concentration of solution ( mol dm-3 ) = )(dmsolution of Volume

(mol) solute ofAmount 3

Example 1

What is the concentration of a solution of 10.0 g of sodium hydroxide in 500 cm3 of solution ? Example 2

Calculate the amount of solute in 250 cm3 of a solution of ethanoic acid which has a concentration of 0.100 mol dm-3. Titration

A solution of known concentration is called a standard solution. A primary standard is a pure compound from which a standard solution of accurately known concentration can be prepared. To qualify as a primary standard the compound :

(i) must be pure; (ii) should dissolve in water easily; (iii) should not be hygroscopic or deliquescent. In volumetric analysis, the concentration of a solution is found by measuring the volume of solution

that will react with a known volume of a standard solution.

The procedure of adding one solution to another in a measured way until the reaction is complete is called titration.

The end-point is the point in a titration revealed by a significant change in the property of the liquid system being monitored. This is usually shown by a sharp change in the colour of an indicator present in the liquid. The equivalent point is the point in a titration at which the stoichiometric amounts of reactants have been added together. If 25.0 cm3 of 0.100 mol dm-3 NaOH(aq) are titrated with 0.100 mol dm-3 HCl(aq) from a burette, the equivalent point will be reached when exactly 25.0 cm3 of the acid have been added to the alkali. If a suitable indicator is used, the equivalent point will match the end-point of the titration.

Page 25: Chapter One Chemistry

Unit 1-25 1. Calculations on titrations involving acid-base reactions Example 1 : Standardizing hydrochloric acid

Anhydrous sodium carbonate is used as a primary standard in volumetric analysis. A solution of sodium carbonate of concentration of 0.100 mol dm-3 is used to standardize a solution of hydrochloric acid. 25.0 cm3 of the standard solution of sodium carbonate require 35.0 cm3 of the acid for neutralization. Calculate the concentration of the acid. Example 2 : Standardizing sodium hydroxide solution What is the concentration of a solution of sodium hydroxide, 25.0 cm3 of which requires 20.0 cm3 of ethanedioic acid of concentration 0.100 mol dm-3 for neutralization ?

Example 3 : Calculating the percentage of sodium carbonate in washing soda 5.125 g of washing soda crystals, Na2CO3(s), 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 dm-3 sulphuric acid for neutralization. Calculate the percentage of sodium carbonate in the crystals. (Mr of sodium carbonate = 106)

Page 26: Chapter One Chemistry

Unit 1-26 2. Calculations on titrations involving redox reactions Potassium manganate(VII) titrations

When potassium manganate(VII) acts as an oxidizing agent in acid solution, it is reduced to a manganese(II) salt :

Once it has been standardized, a solution of potassium manganate(VII) can be used to estimate reducing agents such as iron(II) salts. No indicator is needed as the oxidant changes from purple to colourless during titration. At the end point, permanent purple colour appears due to the addition of a drop of excess KMnO4(aq). Example 1 : Standardizing potassium manganate(VII) against primary standard, sodium ethanedioate

A 25.0 cm3 portion of sodium ethanedioate solution of concentration 0.200 mol dm-3 is titrated against a solution of acidified potassium manganate(VII). If 17.2 cm3 of potassium manganate(VII) are required, what is its concentration ? Example 2 : Oxidizing iron(II) compounds 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 make up to 250 cm3 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 dm-3. A volume of 22.5 cm3 was required. Find n. (Mr of (NH4)2SO4.FeSO4 = 284, H2O = 18)

Page 27: Chapter One Chemistry

Unit 1-27 Sodium thiosulphate titrations

Sodium thiosulphate reduces iodine to iodide ions, and forms solution tetrathionate, Na2S4O6 :

Iodine has a limited solubility in water. It dissolves in a solution of potassium iodide because it forms the very soluble complex ion, I3

- : An equilibrium is set up between iodine and triiodide ions, and if iodine molecules are removed from

solution by the reaction with thiosulphate, triiodide ions dissociate to form more iodine molecules. When sufficient of a solution of thiosulphate is added to a solution of iodine, the colour of iodine

fades to a pale yellow. Then 1 cm3 of starch solution are added to give a dark blue colour with iodine. Addition of thiosulphate is continued drop by drop, until the blue colour disappears.

Example 1 : Standardization of thiosulphate against potassium iodate(V)

Potassium iodate(V) is a primary standard. It reacts with iodide ions in the presence of acid to form iodine :

A standard solution of iodine can be prepared by weighing out the necessary quantity of potassium

iodate(V) and making up to a known volume of solution. When a portion of this solution is added to an excess of potassium iodide in acid solution, a calculated amount of iodine is liberated.

1.015 g of potassium iodate(V) are dissolved and made up to 250 cm3. To a 25.0 cm3 portion are

added an excess of potassium iodide and dilute sulphuric acid. The solution is titrated with a solution of sodium thiosulphate, starch solution being added near the end-point. 29.8 cm3 of thiosulphate solution are required. Calculate the concentration of the thiosulphate solution. (Mr of KIO3 = 214)

Page 28: Chapter One Chemistry

Unit 1-28 Example 2 : Estimation of chlorine

Chlorate(I) solution are often used as a source of chlorine as they liberate chlorine readily on reaction with acid :

The amount of chlorine available in a domestic bleach which contains sodium chlorate(I) can be

found by allowing the bleach to react with iodide solution to form iodine, and then titrating with standard thiosulphate solution.

A domestic bleach in solution is diluted by pipetting 10.0 cm3 and making this volume up to 250 cm3.

A 25.0 cm3 portion of the solution is added to an excess of potassium iodide and ethanoic acid and titrated against 0.0950 mol dm-3 sodium thiosulphate solution, using starch as an indicator. The volume required is 21.3 cm3. Calculate the mass of available chlorine in 10.0 cm3 of the bleach. Example 3 : Estimation of copper(II) salts

Copper(II) ions oxidizes iodide ions to iodine : The iodine produced can be titrated with standard thiosulphate solution, and from the amount of

iodine produced, the concentration of copper(II) ions in the solution can be found. A sample of 4.256 g of copper(II) sulphate-5-water is dissolved and made up to 250 cm3. A 25.0

cm3 portion is added to an excess of potassium iodide. The iodine formed required 18.0 cm3 of a 0.0950 mol dm-3 solution of sodium thiosulphate for reaction. Calculate the percentage of copper in the crystals. (Ar of Cu = 63.5)