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  • #3804

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    Atoms, Molecules and Stoichiometry

    Practice Question (15 min)

    The alums are a series of double salts formed when a cation, with a charge of +1, having a large radius and a cation, with a charge of +3, having a small radius combined with sulfate ions.

    Ammonium chromium alum has the formula of (NH4)aCr(SO4)b.xH2O. 1.28 g of the salt was dissolved in 100 cm3 of 0.0500 mol dm-3 ammonium chloride solution and the solution was divided into two equal portions.

    To one portion was added an excess of sodium hydroxide and the mixture was boiled. The ammonia that was evolved neutralized 25.60 cm3 of 0.150 mol dm-3 nitric acid.

    (a) Write a balanced ionic equation for the reaction between sodium hydroxide and the ammonium ions in solution. [1]

    (b) Calculate the number of moles of NH3 from the alum. [2]

    To the other portion, an excess of zinc was added which reduced the Cr3+(aq) to Cr2+(aq). The mixture was then filtered and the filtrate was titrated with acidified potassium dichromate(VI). It was found that 22.35 cm3 of 0.0100 mol dm-3 acidified potassium dichromate(VI) was required for the titration.

    (c) Write a balanced ionic equation for the reaction between Cr2+ ions and acidified potassium dichromate(VI). [1]

    (d) Calculate the number of moles of chromium ions from the alum. [1]

    (e) Use your answer to (b) and (d), calculate the values of a and b. [1]

    (f) Hence, find the relative formula mass of ammonium chromium alum and the value of x. [2]

    (g) Calculate the percentage of chromium in the sample. [2]

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    #3862

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    1. Concept: Redox Half Equations

    (a) Define the term disproportionation. [1]

    2. (Time allocation: 12 min)
    (a) Mixtures of CaCl2 and NaCl are used for salting roads to prevent ice formation. 1.95 g sample of such mixture was dissolved in water and excess aqueous Na2C2O4 was added to completely precipitate the Ca2+ ions as CaC2O4.

    The CaC2O4 formed was separated from the solution and dissolved in sulphuric acid. The resulting H2C2O4 solution was titrated with 37.8 cm3 of 0.102 mol dm-3 KMnO4 solution.

    (i) Given that C2O42- is oxidized to form carbon dioxide, write a balanced equation for the reaction of C2O42- and MnO4-. [1]

    (ii) Calculate the number of moles of H2C2O4 reacted with KMnO4. [1]

    (iii) Calculate the mass of CaCl2 in the original sample. [2]

    (iv) Hence, calculate the percentage mass of NaCl in the original sample.
    [2]

    (b) 20.0 cm3 of 0.100 mol dm-3 vanadium (II) solution, V2+(aq), reacted with 40.0 cm3 of a 0.100 mol dm-3 iron (III) solution. Iron (III) ions was reduced to form iron (II) ions in the reaction. Calculate the final oxidation state of vanadium in the above reaction. [2]

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    #3998

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    (Time allocation: 9 min)
    When sodium is burned in air, a mixture of sodium oxide, Na2O, and sodium peroxide, Na2O2, is formed. The mixture reacts with water according to the following equations.
    Na2O + H2O —> 2NaOH
    Na2O2 + 2H2O —> 2NaOH + H2O2
    The following information will allow you to calculate the relative amounts of the two oxides produced when sodium is burned.
    • The mixture obtained by burning a sample of sodium was dissolved in distilled water and made to 100 cm3 to give solution H.
    • A 25.0 cm3 of solution H was titrated with 0.100 mol dm-3 HCl. 22.50 cm3 of acid was required to reach the end-point. (A)
    • The H2O2 content of solution H was found by titration of another 25.0 cm3 portion with 0.0200 mol dm-3 KMnO4. 10.0 cm3 of KMnO4 solution was required to reach the end-point. (B)

    (a) Using the results of HCl titration, calculate the total number in moles of NaOH in 100 cm3 of solution H.

    (b) (i) Write a balanced equation for the oxidation of H2O2 with KMnO4 in an acidic medium.

    (ii) Using the results of the KMnO4 titration, calculate the amount in moles of H2O2 in 100 cm3 of solution H.

    (c) Hence, calculate the amount in moles of Na2O and Na2O2 formed during the burning of the sodium sample.
    Na2O + H2O —> 2NaOH …..(1)
    Na2O2 + 2H2O —> 2NaOH + H2O2 ….(2)

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    #4176

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    Redox – Exam Practice Question
    3 (Time allocation: 12 min)
    (a) Mixtures of CaCl2 and NaCl are used for salting roads to prevent ice formation. 1.95 g sample of such mixture was dissolved in water and excess aqueous Na2C2O4 was added to completely precipitate the Ca2+ ions as CaC2O4.

    The CaC2O4 formed was separated from the solution and dissolved in sulphuric acid. The resulting H2C2O4 solution was titrated with 37.8 cm3 of 0.102 mol dm-3 KMnO4 solution.

    (i) Given that C2O42- is oxidized to form carbon dioxide, write a balanced equation for the reaction of C2O42- and MnO4-. [1]

    (ii) Calculate the number of moles of H2C2O4 reacted with KMnO4. [1]

    (iii) Calculate the mass of CaCl2 in the original sample. [2]

    (iv) Hence, calculate the percentage mass of NaCl in the original sample.[2]

    (b) 20.0 cm3 of 0.100 mol dm-3 vanadium (II) solution, V2+(aq), reacted with 40.0 cm3 of a 0.100 mol dm-3 iron (III) solution. Iron (III) ions was reduced to form iron (II) ions in the reaction. Calculate the final oxidation state of vanadium in the above reaction. [2]

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    #4254

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    One.Tution Place students are encouraged to continue learning beyond the classroom. Online Chemistry FAQ resources with fully worked-out solutions are available to the students to equip them with handy strategies to handle higher order thinking questions

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    #4336

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    Atoms, Molecules and Stoichiometry – Notes

    Lecture Notes Outline

    1 Atoms & sub-atomic particles
    2 Relative masses
    3 The mole and related concepts
    4 Empirical and molecular formulae
    5 Concentration of a solution
    6 Stoichiometry
    7 Volumetric analysis
    8 Redox reactions
    9 Balancing redox equations
    10 Redox titrations
    11 Precipitation titrations
    12 Determination of oxidation number

    1. ATOMS & SUB-ATOMIC PARTICLES
    • All atoms are composed of three fundamental particles ¾ protons, neutrons and electrons. The
    relative charges and relative masses of sub-atomic particles are shown in the table below.

    • The atomic number of an element is the number of protons in the nucleus of an atom of that element. Each element has its own atomic number. The atomic number of a neutral atom is also equal to the number of electrons.

    • The nucleon number, also known as mass number, is the number of protons and neutrons in the nucleus of an atom. Protons and neutrons are both nucleons.

    • The word nuclide is used to describe any atomic species of which the proton number and the mass number are specified.

    • Isotopes of an element are atoms with the same atomic number but different mass numbers (i.e. they have the same number of protons but different number of neutrons in the nucleus).
    – Isotopes have the same number of electrons  the same chemical properties.
    – Isotopes have different numbers of neutrons (i.e. different masses)  different physical properties.

    • Most elements consist of mixtures of isotopes. The abundance of each in the mixture is called its isotopic abundance (in terms of percentages or fractions).

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    #4372

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    Atoms, Molecules and Stoichiometry – Notes

    3 THE MOLE & RELATED CONCEPTES

    • The word ‘mole’ is derived from the Latin word meaning a collection, mass or pile.
    Just as 1 dozen is a collective term for 12 items, 1 mole is a collective term for 6.02 x 10^23 of elementary entities or particles which may be electrons, ions, atoms or molecules.

    • The Avogadro constant (L) is the constant of proportionality between the number of specified entities of a substance and amount of that substance, experimentally found to be 6.02 x 10^23 mol−1.

    3.2 Molar Mass (g mol−1)

    • Molar mass refers to the mass of one mole of a substance (element or compound).
    (Units = g mol−1)
    • The molar mass of a substance is numerically equal to the Ar or Mr of that substance except that it has units of g mol−1 while both Ar and Mr have no units.

    E.g. Molar mass of NaCl = 23.0 + 35.5 = 58.5 g mol−1
    while Relative formula mass of NaCl = 58.5

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    #4437

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    Atoms, Molecules and Stoichiometry – Notes

    4. Empirical and Molecular Formulae

    Molecular Formula:
    The actual number of atoms of the different elements in the compound.

    Empirical Formula:
    The simplest whole number ratio for the atoms of different elements in the compound.

    The empirical formula of a compound may be calculated from experimental data (from combustion analysis or elemental analysis). The molecular formula can be determined from the empirical formula provided the molar mass or the relative molecular mass of the compound is known.

    Note:
    In the calculation of empirical formula from experimental data, it is a common procedure to round off figures to the corresponding nearest whole numbers in order to get the simplest ratio. Great care should be exercised when the following figures are obtained and these figures are usually multiplied by a factor in order to get the correct simplest ratio.

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    #4477

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    Atoms, Molecules and Stoichiometry – Notes

    5. Concentration of a Solution

    • A solution is a homogeneous mixture of 2 or more substances.
    • The substance which is in greater quantity is the solvent and the other substance is called the
    solute.
    • The term concentration is to designate the amount of solute dissolved in a given quantity of solvent or solution.

    Concentration can be expressed as:
    (a) mol dm−3
    (b) g dm−3

    • The above two units are related by the following expression:
    concentration in g dm−3 = concentration in mol dm-3 x molar mass (g mol−1)

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    #4535

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    Atoms, Molecules and Stoichiometry – Notes

    6 Stoichiometry

    6.1 Avogadro’s Hypothesis
    • Avogadro’s hypothesis states that:
    Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.
    • This means that 1 mole of a gas will contain 6.02 × 1023 molecules/atoms and occupy the same molar volume under the same conditions of temperature and pressure.

    6.2 Limiting Reagent
    • In carrying out chemical reactions, the reagents are not always present in stoichiometric amounts.
    • One or more reagents may be in excess of that theoretically needed for complete reaction. The deficient reactant is called the limiting reactant and is used up completely in the reaction.
    • The maximum or theoretical yield of a product is thus determined based on the limiting reactant.

    6.3 Percentage Yield
    • The yield of a product calculated from the chemical equation is called the theoretical yield.
    • The actual yield refers to the amount of product that is actually obtained in the reaction during an experiment. The actual yield obtained is usually less than the theoretical yield.
    • The percentage yield relates the actual yield to the theoretical yield and is expressed as follows:

    Percentage Yield = (Actual yield (g))/ (Theoretical yield (g)) x 100%

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    #4598

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    Atoms, Molecules and Stoichiometry – Notes

    7 Volumetric (or titrimetric) Analysis

    • Volumetric analysis (or titrimetric analysis) is a method of quantitative analysis which depends essentially on the accurate measurements of volumes of two solutions which react together completely.

    • A standard solution is defined as one which is of known concentration. This standard solution is used to determine the concentration of another solution through a titration process.

    • The equivalence point is then said to have been reached and it marks the completion of the titration.

    • In practice, the completion of a titration is detected by a distinct colour change brought about by the use of a suitable indicator. The point at which this distinct colour change occurs is called the end-point of the titration.

    • The use of a suitable indicator will ensure that the end-point is very close to the equivalence point.

    7.1Acid – Base Titrations

    • Acid-base titrations are carried out in order to establish the stoichiometric amounts of acid and base which are required to neutralise each other. There are four main types of acid-base titrations:
    (a) strong acid  strong base titration
    (b) strong acid  weak base titration
    (c) weak acid  strong base titration
    (d) weak acid  weak base titration

    7.2 Titration Curves (to be further discussed in Ionic Equilibrium)

    Below are titration curves showing the pH changes as a 0.1 mol dm-3 solution of a strong or weak alkali is added to a 0.1 mol dm-3 solution of a strong or weak acid. The precise shape of the curves depends on the actual strengths of the alkalis and acids concerned. Slightly different curves are obtained too, if solutions other than 0.1 mol dm-3 are used.

    7.3 Indicators

    • The indicators used in acid-base titrations are generally organic dyestuffs which are weak acids or weak bases.

    • In an acid-base titration, the indicator used is usually added in a small quantity to help to determine the end point of the titration. Each indicator has a pH range over which it changes colour. The choice of an indicator for an acid-base titration depends on the pH range of the indicator and the type of titration involved. It is important to pick an indicator which changes colour over the range in which there is a marked pH change during the titration.

    • Essential characteristics of a good indicator:

    1. The colour change of the indicator must be clear and sharp i.e. it must be sensitive. Thus it would be useless if 2 or 3 cm3 of the reagent were necessary to bring about the colour change.

    2. Its pH range for colour change must coincide with the region of rapid pH change in the titration curve. In other words, the pH range of the indicator must fall on the vertical portion of the titration curve.

    7.4 Back Titration

    • Back titrations are usually employed when a direct titration is not possible, e.g. solid substances (CaCO3 in colgate) where the end-point is difficult to detect and volatile substances (ammonia, iodine) where inaccuracy arises due to loss of substance during titration.

    • In back titration, a known excess of one reagent A is allowed to react with an unknown amount of a reagent B. At the end of the reaction, the amount of A that remains unreacted is found by titration with a standard reagent C. A simple calculation gives the amount of A that has reacted with B and also the amount of B that has reacted.

    Steps in back titration calculations:

    1. Write balanced equations for the reactions.

    2. Calculate the initial no. of mole of reactant A added.

    3. Calculate the no. of mole of reactant A remaining (that reacted with standard reagent C).

    4. Calculate the no. of mole of reactant A reacted with reactant B.

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    #4639

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    Atoms, Molecules and Stoichiometry – Notes

    8 Redox Reactions

    The term “redox” is used as an abbreviation for the processes of reduction and oxidation which occur simultaneously. A redox reaction is an oxidation-reduction reaction.

    (a) Loss/gain of Oxygen

    Oxidation is gain of oxygen.
    Reduction is loss of oxygen

    (b) Loss/gain of Hydrogen.

    Oxidation is loss of hydrogen.
    Reduction is gain of hydrogen.

    (c) Loss/gain of electrons

    Oxidation is loss of electrons.
    Reduction is gain of electrons.

    (d) Increase/decrease in oxidation number(oxidation state)

    Oxidation is increase in oxidation number.
    Reduction is decrease in oxidation number.

    8.1 Redox Processes in Terms of Electron Transfer

    • Consider the reaction between zinc and copper(II) sulfate solution.

    Zn(s) + Cu2+(aq) –> Zn2+(aq) + Cu(s) Ionic equation

    • The overall reaction can be separated into two simpler processes involving electron transfer.
    Oxidation: Zn(s) –> Zn2+(aq) + 2e
    Reduction: Cu2+(aq) + 2e –> Cu(s)

    • The two separate equations can be termed ion-electron equations but they are more commonly known as half-equations. One half-equation represents the oxidation process while the other represents the reduction process. Addition of the two half-equations gives the overall redox equation.

    • In the shown reaction,

    Zn acts as a reducing agent. It loses two electrons and is itself oxidised to Zn2+. Cu2+ acts as an oxidising agent. It gains two electrons and is itself reduced to Cu.

    8.2 Oxidation Numbers

    • An oxidation number is a number that is assigned to an element in a substance to show its state of oxidation

    8.3 Rules for Assigning Oxidation Number (O.N)

    1. The oxidation number of an atom in the elemental state = 0
    E.g. O.N. of Ca, I2 S8 or P4 = 0

    2. The oxidation number. of monoatomic ion is simply the charge on the ion.
    E.g. O.N. of Cl- = -1; O.N. of Ca2+ = +2

    3. The sum of the oxidation number of all the atoms or ions in a neutral compound = 0.
    E.g.
    CaBr2 +2 + 2(-1) = 0
    Al2O3 2(+3) + 3(-2) = 0
    CO2 +4+2(-2)=0

    4. In polyatomic ions, the sum of the oxidation numbers = the charge on the ion.

    E.g.

    SO4 2-

    : The sum of the oxidation numbers (O.N of S = +6, O.N. of O = -2)
    = +6 + 4(-2) = -2 = charge on the ion

    CrCl6 3-

    : The sum of the oxidation numbers (O.N of Cr = +3, O.N. of Cl = -1)
    = +3 + 6(-1) = -3 = charge on the ion

    5. In any compound, the more electronegative atom has the negative oxidation number while the less electronegative atom has the positive oxidation number.

    • Electronegativity is the ability of an atom in a molecule to attract shared electrons in a bond.

    • Electronegativity values : F > O > N > C > H

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    #4693

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    Atoms, Molecules and Stoichiometry – Notes

    10 Redox Titrations

    • In redox titrations, the chemical reaction involves the transfer of electrons from a reducing agent (electron donor) to an oxidising agent (electron acceptor).

    • Important half-equations to take note:

    • The following types of redox titrations will be discussed:
    1. Manganate(VII) titrations
    2. Dichromate(VI) titrations
    3. Iodine-thiosulfate titrations

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    #4731

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    Atoms, Molecules and Stoichiometry – Notes

    Determination of Oxidation Number

    Steps to determine value of oxidation number in a species:
    1. Let the unknown oxidation number be n. Write the two half equations.
    2. Calculate the no of moles of each ion
    3. Determine the mole ratio
    4. Write the overall equation wrt mole ratio in Step 3
    5. Balance wrt electrons and solve for the unknown oxidation no, n

    Note: In a redox reaction,
    the no. of moles of e− lost by one species = no. of moles of e− gained by other (i.e. e− on the LHS and RHS of overall equation must cancel off )

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    #4806

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    Atomic Structure – Notes

    1. THE SUB-ATOMIC PARTICLES OF MATTER

    (a) The 3 fundamental sub-atomic particles in matters are protons, neutrons and electrons.

    (b) Protons and neutrons are collectively known as nucleons.

    (c) An atom is electrically neutral and contains equal numbers of electrons and protons.

    DO YOU KNOW?

    • The nucleons reside in the small nucleus of the atom whereas the electrons revolve around it in the “vast” empty space.

    • The size of an atom is easily more than 10 000 times that of the nucleus.

    • The nucleus accounts for most of the mass of an atom since the mass of the electron is negligible as compared to the mass of the protons and neutrons.

    Try it out! 1

    Which of the following statements is incorrect?

    A The nucleus is positively charged.
    B An atom is electrically neutral.
    C The region where electron moves is mostly empty space.
    D An electron is 1840 times heavier than a proton.

    Ans : D

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