Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of.

Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element. 3. Divide each value in point 2 by the smallest of them. 4. If necessary, multiply all the numbers by the smallest integer possible to obtain approximate whole number values for each element. 5. Write these whole numbers as subscripts of the element symbols to give an empirical formula. 6. If the molecular weight is known, determine how many empirical formula weights are required to obtain the molecular weight. Use this factor to multiply the number of atoms of each element in the empirical formula to give the molecular formula.

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0)

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass.

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = 3.48

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = 3.48 4. Multiply by 2 to obtain approx. whole nos.: N2, H8, Cr2, O7

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = 3.48 4. Multiply by 2 to obtain approx. whole nos.: N2, H8, Cr2, O7 5. Write as subscripts: N 2 H 8 Cr 2 O 7

Q 14, 2001. Germanium has five naturally occurring isotopes: 70 Ge, 20.5%, 69.924 a.m.u.; 72 Ge, 27.4%, 71.922 a.m.u.; 73 Ge, 7.8%, 72.923 a.m.u.; 74 Ge, 36.5%, 73.921 a.m.u., 76 Ge, 7.8%, 75.921 a.m.u. What is the atomic weight of germanium? (a)73.42 (b) 72.63 (c) 73.63 (d) 72.32 (e) 72.92

Q 14, 2001. Germanium has five naturally occurring isotopes: 70 Ge, 20.5%, 69.924 a.m.u.; 72 Ge, 27.4%, 71.922 a.m.u.; 73 Ge, 7.8%, 72.923 a.m.u.; 74 Ge, 36.5%, 73.921 a.m.u., 76 Ge, 7.8%, 75.921 a.m.u. What is the atomic weight of germanium? (a)73.42 (b) 72.63 (c) 73.63 (d) 72.32 (e) 72.92 = (69.924 x 20.5 + 71.922 x 27.4 + 72.923 x 7.8 + 73.921 x 36.5 + 75.921 x 7.8) / 100

Q 14, 2001. Germanium has five naturally occurring isotopes: 70 Ge, 20.5%, 69.924 a.m.u.; 72 Ge, 27.4%, 71.922 a.m.u.; 73 Ge, 7.8%, 72.923 a.m.u.; 74 Ge, 36.5%, 73.921 a.m.u., 76 Ge, 7.8%, 75.921 a.m.u. What is the atomic weight of germanium? (a)73.42 (b) 72.63 (c) 73.63 (d) 72.32 (e) 72.92 = (69.924 x 20.5 + 71.922 x 27.4 + 72.923 x 7.8 + 73.921 x 36.5 + 75.921 x 7.8) / 100 = 72.63 a.m.u.

Atoms and Molecules- Summary: 1. Atomic structure, ions, isotopes. 2. Atomic number, mass no., atomic mass, chemical atomic weight, molecular weight. 3. Mass- no. of particles for practical use, the mole, Avogadro’s number. 4. Dalton atomic theory and early ideas on atoms and molecules. 5. Volume relationships in gas phase reactions (Gay-Lussac). 6. Volume - no. of particle - mass relationships in gases (Avogadro’s hypothesis). 7. Determination of empirical and molecular formulae.

Oxidation and Reduction

Oxidation: any process by which an entity loses electrons.

e.g.2Mg o + O 2 o 2Mg 2+ + O 2- H 2 o + F 2 o 2H + F - Mg and H are oxidised in these examples.

Oxidation: any process by which an entity loses electrons. e.g.2Mg o + O 2 o 2Mg 2+ + O 2- H 2 o + F 2 o 2H + F - Mg and H are oxidised in these examples. Reduction: any process by which an entity gains electrons. Oxygen and fluorine are reduced in the above examples.

Oxidation: any process by which an entity loses electrons. e.g.2Mg o + O 2 o 2Mg 2+ + O 2- H 2 o + F 2 o 2H + F - Mg and H are oxidised in these examples. Reduction: any process by which an entity gains electrons. Oxygen and fluorine are reduced in the above examples. Nomenclature:positive ions are called “ cations” negative ions are called “ anions”

Oxidation: any process by which an entity loses electrons. e.g.2Mg o + O 2 o 2Mg 2+ + O 2- H 2 o + F 2 o 2H + F - Mg and H are oxidised in these examples. Reduction: any process by which an entity gains electrons. Oxygen and fluorine are reduced in the above examples. Nomenclature:positive ions are called “ cations” negative ions are called “ anions” Oxidation state:the sign and magnitude of the charge on an ion (also called oxidation number).

Note: 1. Oxidation state of an element in the uncombined state always equal to zero.

Note: 1. Oxidation state of an element in the uncombined state always equal to zero. 2. The algebraic sum of the oxidation states of all the atoms in the molecular formula of a compound is zero.

Note: 1. Oxidation state of an element in the uncombined state always equal to zero. 2. The algebraic sum of the oxidation states of all the atoms in the molecular formula of a compound is zero. 3. In complex ions (contain more than one atom) the sum of the oxidation states of all the constituents must equal overall charge on the ion.

Note: 1. Oxidation state of an element in the uncombined state always equal to zero. 2. The algebraic sum of the oxidation states of all the atoms in the molecular formula of a compound is zero. 3. In complex ions (contain more than one atom) the sum of the oxidation states of all the constituents must equal overall charge on the ion. Common Oxidation States of Elements 1.Hydrogen usually +1. 2.Oxygen usually -2. 3.Halogens (i.e. F, Cl, Br, I) usually -1. 4.Metal ions are invariably positive.

Example: Calculate the oxidation state of S in the S 2 O 3 2- ion.

Solution: Oxidation state of oxygen is -2.

Example: Calculate the oxidation state of S in the S 2 O 3 2- ion. Solution: Oxidation state of oxygen is -2. Sum of oxidation states of all the elements in the ion must equal the charge on the ion.

Example: Calculate the oxidation state of S in the S 2 O 3 2- ion. Solution: Oxidation state of oxygen is -2. Sum of oxidation states of all the elements in the ion must equal the charge on the ion.  2 (ox. st. of S) + 3 (-2) = -2

Example: Calculate the oxidation state of S in the S 2 O 3 2- ion. Solution: Oxidation state of oxygen is -2. Sum of oxidation states of all the elements in the ion must equal the charge on the ion.  2 (ox. st. of S) + 3 (-2) = -2 2 (ox. st. of S) - 6 = -2

Example: Calculate the oxidation state of S in the S 2 O 3 2- ion. Solution: Oxidation state of oxygen is -2. Sum of oxidation states of all the elements in the ion must equal the charge on the ion.  2 (ox. st. of S) + 3 (-2) = -2 2 (ox. st. of S) - 6 = -2 2 (ox. st. of S) = +4

Example: Calculate the oxidation state of S in the S 2 O 3 2- ion. Solution: Oxidation state of oxygen is -2. Sum of oxidation states of all the elements in the ion must equal the charge on the ion.  2 (ox. st. of S) + 3 (-2) = -2 2 (ox. st. of S) - 6 = -2 2 (ox. st. of S) = +4  oxidation state of S = +2

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion.

Solution: Oxidation state of oxygen is -2. 

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion. Solution: Oxidation state of oxygen is -2.  2 (ox. st. of S) + 8 (-2) = - 2 2 (ox. st. of S) - 16 = - 2 2 (ox. st. of S) = + 14

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion. Solution: Oxidation state of oxygen is -2.  2 (ox. st. of S) + 8 (-2) = - 2 2 (ox. st. of S) - 16 = - 2 2 (ox. st. of S) = + 14  oxidation state of S = +7

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion. Solution: Oxidation state of oxygen is -2.  2 (ox. st. of S) + 8 (-2) = - 2 2 (ox. st. of S) - 16 = - 2 2 (ox. st. of S) = + 14  oxidation state of S = +7 Example: Calculate the oxidation state of Mn in KMnO 4.

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion. Solution: Oxidation state of oxygen is -2.  2 (ox. st. of S) + 8 (-2) = - 2 2 (ox. st. of S) - 16 = - 2 2 (ox. st. of S) = + 14  oxidation state of S = +7 Example: Calculate the oxidation state of Mn in KMnO 4. Solution: Oxidation state of oxygen is -2. 

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion. Solution: Oxidation state of oxygen is -2.  2 (ox. st. of S) + 8 (-2) = - 2 2 (ox. st. of S) - 16 = - 2 2 (ox. st. of S) = + 14  oxidation state of S = +7 Example: Calculate the oxidation state of Mn in KMnO 4. Solution: Oxidation state of oxygen is -2.  1 +  ox. st. of Mn + 4 (-2) = 0 1+ ox. st. of Mn - 8 = 0 ox. st. of Mn = + 7

Example: Calculate the oxidation state of S in the S 2 O 8 2- ion. Solution: Oxidation state of oxygen is -2.  2 (ox. st. of S) + 8 (-2) = - 2 2 (ox. st. of S) - 16 = - 2 2 (ox. st. of S) = + 14  oxidation state of S = +7 Example: Calculate the oxidation state of Mn in KMnO 4. Solution: Oxidation state of oxygen is -2.  1 +  ox. st. of Mn + 4 (-2) = 0 1+ ox. st. of Mn - 8 = 0 ox. st. of Mn = + 7  oxidation state of Mn = +7

Nomenclature of Cations Cation Name Al 3+ Aluminium Ba 2+ Barium Na + Sodium Cu + Cuprous or copper (I) Cu 2+ Cupric or copper (II) Fe 2+ Ferrous Fe 3+ Ferric Sn 2+ Stannous Sn 4+ Stannic NH 4 + Ammonium

Nomenclature of Anions AnionName Br - Bromide Cl - Chloride O 2- Oxide S 2- Sulphide OH - Hydroxide NO 3 - Nitrate CO 3 2- Carbonate SO 4 2- Sulphate PO 4 3- Phosphate CN - Cyanide

Oxidation States of Some Common Ions CationsAnions +1+2+3+4-1-2-3 Li + Be 2+ Al 3+ Si 4+ F - O 2- N 3- Na + Mg 2+ Sc 3+ Mn 4+ Cl - S 2- P 3- K + Ca 2+ Y 3+ U 4+ Br - Se 2- Rb + Sr 2+ Ga 3+ Th 4+ I - CO 3 2- Cs + Ba 2+ Sb 3+ Ce 4+ OH - SO 4 2- Cu + Mn 2+ Bi 3+ NO 3 - Ag + Fe 2+ V 3+ NO 2 - Tl + Co 2+ Cr 3+ CN - H + Ni 2+ Fe 3+ MnO 4 - NH 4 + Cu 2+ Co 3+ Zn 2+ Cd 2+

Oxidation and Reduction - Summary l Oxidation = loss of electrons

Oxidation and Reduction - Summary l Oxidation = loss of electrons l Reduction = gain of electrons

Oxidation and Reduction - Summary l Oxidation = loss of electrons l Reduction = gain of electrons l Oxidation state or number: sign and magnitude of charge on ion

Oxidation and Reduction - Summary l Oxidation = loss of electrons l Reduction = gain of electrons l Oxidation state or number: sign and magnitude of charge on ion l Hydrogen +1; oxygen -2; halides usually -1; metals usually +ve.

Oxidation and Reduction - Summary l Oxidation = loss of electrons l Reduction = gain of electrons l Oxidation state or number: sign and magnitude of charge on ion l Hydrogen +1; oxygen -2; halides usually -1; metals usually +ve. l Calculation of oxidation states in ions and molecules

Photography light 2 AgBr2Ag + Br 2

Photography light 2 AgBr2Ag + Br 2 light, h

Modern Theories of Atomic Structure

l Chemical reactions: rearrangements in electronic structure of atoms/molecules

l Chemical reactions: rearrangements in electronic structure of atoms/molecules l e - rearrangements use up or generate energy, often in form of light (electromagnetic radiation)

l Chemical reactions: rearrangements in electronic structure of atoms/molecules l e - rearrangements use up or generate energy, often in form of light (electromagnetic radiation) l Energy (light) used or generated can give info about electronic structure of atoms/molecules

l Chemical reactions: rearrangements in electronic structure of atoms/molecules l e - rearrangements use up or generate energy, often in form of light (electromagnetic radiation) l Energy (light) used or generated can give info about electronic structure of atoms/molecules First step: find out about nature of light!

Electromagnetic Radiation

Energy (work): force x distance

Electromagnetic Radiation Energy (work): force x distance Electric Magnetic

Electromagnetic Radiation Energy (work): force x distance Electric Magnetic through Space Material

Electromagnetic Radiation Energy (work): force x distance Electric Magnetic through Space Material Various types of e.m.r.: radio waves, X-rays, light, etc.

Electromagnetic Radiation Energy (work): force x distance Electric Magnetic through Space Material Various types of e.m.r.: radio waves, X-rays, light, etc. Wave Nature of E.m.r. What is a wave?

Electromagnetic Radiation Energy (work): force x distance Electric Magnetic through Space Material Various types of e.m.r.: radio waves, X-rays, light, etc. Wave Nature of E.m.r. What is a wave? A way in which energy travels through a medium

Wave characteristics: l Velocity (c): in vacuum, same for all types of e.m.r. (3 x 10 8 m s-1 )

Wave characteristics: l Velocity (c): in vacuum, same for all types of e.m.r. (3 x 10 8 m s-1 ) Wavelength ( ): distance between 2 consecutive peaks

Wave characteristics: l Velocity (c): in vacuum, same for all types of e.m.r. (3 x 10 8 m s-1 ) Wavelength ( ): distance between 2 consecutive peaks Frequency ( ): number of cycles per second passing a point in space

Wave characteristics: l Velocity (c): in vacuum, same for all types of e.m.r. (3 x 10 8 m s-1 ) Wavelength ( ): distance between 2 consecutive peaks Frequency ( ): number of cycles per second passing a point in space l Amplitude (A): height (strength) of wave at peak

Wave characteristics: l Velocity (c): in vacuum, same for all types of e.m.r. (3 x 10 8 m s-1 ) Wavelength ( ): distance between 2 consecutive peaks Frequency ( ): number of cycles per second passing a point in space l Amplitude (A): height (strength) of wave at peak c =  or = c/

Particulate (Quantised) Nature of E.M.R. Max Planck (1900): Energy gain/loss can occur only in whole number multiples of h (h = Planck’s constant = 6.626 x 10 -34 J s).

Particulate (Quantised) Nature of E.M.R. Max Planck (1900): Energy gain/loss can occur only in whole number multiples of h (h = Planck’s constant = 6.626 x 10 -34 J s).  E = nh where n = 1, 2, 3, etc.

Particulate (Quantised) Nature of E.M.R. Max Planck (1900): Energy gain/loss can occur only in whole number multiples of h (h = Planck’s constant = 6.626 x 10 -34 J s).  E = nh where n = 1, 2, 3, etc. Energy is transferred in “ packets “ or quanta

Energy Continuous energy levelsQuantised energy levels

Only when energy is greater than threshold value are electrons ejected Increasing intensity increases # of electrons ejected only when threshold energy has been attained

Einstein (1905): Energy of e.m.r. is quantised E photon = h = hc/ (1) E.m.r.stream of ” particles “ called photons

Einstein (1905): Energy of e.m.r. is quantised E photon = h = hc/ (1) E.m.r.stream of ” particles “ called photons Also:Energy has mass! i.e. E = mc 2 or m = E/c 2 (2)

Einstein (1905): Energy of e.m.r. is quantised E photon = h = hc/ (1) E.m.r.stream of ” particles “ called photons Also:Energy has mass! i.e. E = mc 2 or m = E/c 2 (2) Combining (1) and (2) : “ Mass “ of e.m.r. photon = m = h/ c

Einstein (1905): Energy of e.m.r. is quantised E photon = h = hc/ (1) E.m.r.stream of ” particles “ called photons Also:Energy has mass! i.e. E = mc 2 or m = E/c 2 (2) Combining (1) and (2) : “ Mass “ of e.m.r. photon = m = h/ c Overall conclusion: Light (e.m.r.) has both wave and particulate properties

Einstein (1905): Energy of e.m.r. is quantised E photon = h = hc/ (1) E.m.r.stream of ” particles “ called photons Also:Energy has mass! i.e. E = mc 2 or m = E/c 2 (2) Combining (1) and (2) : “ Mass “ of e.m.r. photon = m = h/ c Overall conclusion: Light (e.m.r.) has both wave and particulate properties Question: Does matter have wave-like properties?

Does matter have wave-like properties?

Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses

Does matter have wave-like properties? Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses Similar equation applies as for light: m = h/.v or = h/mv (v = velocity of particle)

Does matter have wave-like properties? Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses Similar equation applies as for light: m = h/.v or = h/mv (v = velocity of particle) For an electron (m = 9.1 x 10 -28 g, v = 1.2 x 10 5 m/s): = 6.1 x 10 -9 m

Does matter have wave-like properties? Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses Similar equation applies as for light: m = h/.v or = h/mv (v = velocity of particle) For an electron (m = 9.1 x 10 -28 g, v = 1.2 x 10 5 m/s): = 6.1 x 10 -9 m For a cyclist (m = 1 x 10 5 g, v = 3m/s): = 3.5 x 10 -40 m

Energy of photon E = h = hc/

Energy of photon E = h = hc/ = 500 nm

Energy of photon E = h = hc/ = 500 nm E = 6.626 x 10 -34 x 3.00 x 10 9 / 500 x 10 -9 = 3.97 x 10 -18 J s m s -1 m -1 = 3.97 x 10 -18 J

Energy of photon E = h = hc/ = 500 nm E = 6.626 x 10 -34 x 3.00 x 10 9 / 500 x 10 -9 = 3.97 x 10 -18 J s m s -1 m -1 = 3.97 x 10 -18 J 100 W bulb emits 100 J s -1  100/ 3.97 x 10 -18 = 2.5 x 10 19 photons s -1.

Summary so far: l On the atomic scale energy is transferred in discrete quantities (quanta) l Light (e.m.r.) exhibits both wave and particulate behaviour

Summary so far: l On the atomic scale energy is transferred in discrete quantities (quanta) l Light (e.m.r.) exhibits both wave and particulate behaviour l Matter, if small enough in mass, exhibits wave behaviour which is measurable

Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of.

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Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of.

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Presentation on theme: "Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of."— Presentation transcript:

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