SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 2 by hmv21438

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									SURVEY OF CHEMISTRY I

         CHEM 1151

           CHAPTER 2

 DR. AUGUSTINE OFORI AGYEMAN
   Assistant professor of chemistry
   Department of natural sciences
       Clayton state university
     CHAPTER 2

ATOMS AND MOLECULES
           THE ATOMIC THEORY OF MATTER


                      Law of Constant Composition
   - The relative numbers and kinds of atoms are constant in a given compound
- All samples of a given chemical compound have the same elemental composition

                                   Example
            - Water (H2O) always contains 1 g of H for every 8 g of O


                Law of Conservation of Mass (Matter)
 - The total mass of materials after a chemical reaction is equal to the total mass
                           before the chemical reaction
           THE ATOMIC THEORY OF MATTER


                     Law of Multiple Proportions
- When two or more elements combine to form a compound, their masses in that
                 compound are in a fixed and definite ratio

             - Elements combine in a ratio of small whole numbers

- If two elements form more than one compound, the ratios of the masses of the
 second element combined with a fixed mass of the first element will be in ratios
                            of small whole numbers
THE ATOMIC THEORY OF MATTER


     Law of Multiple Proportions

  - C and O can combine to form CO and CO2

                    CO
       1.33 g O combine with 1.00 g C

                   CO2
       2.66 g O combine with 1.00 g C

     - Ratio of O is 2.66 g : 1.33 g = 2 : 1
       THE ATOMIC THEORY OF MATTER

                   Dalton’s Atomic Theory

1. All matter (every element) is made up of very small particles
                           called atoms
             - Atoms are indivisible and indestructible


   2. All atoms of a given element are identical in mass and
                            properties
   - Atoms of a given element are different from atoms of all
                          other elements
      THE ATOMIC THEORY OF MATTER

                 Dalton’s Atomic Theory

3. Compounds are formed from a combination of two or more
                   different kinds of atoms
 - A given compound always has the same relative number and
                         kind of atoms

     4. A chemical reaction is a rearrangement of atoms
- Atoms are neither created nor destroyed in a chemical reaction
    THE ATOMIC THEORY OF MATTER

 Modern atomic theory is more involved but based on
                  Dalton’s theory

- Atoms can be destroyed by nuclear reactions but not by
                   chemical reactions

 - There are different kinds of atoms within an element
      (isotopes - different masses, same properties)
         THE ATOMIC STRUCTURE

                                Atom
- Is the smallest particle of an element that retains the chemical
                        identity of the element
          - Is the basic building block of ordinary matter

- Made up of smaller particles (the building blocks of an atom)
                  called subatomic particles

             Three Types of Subatomic Particles
      Electron: possesses a negative (-) electrical charge
       Proton: possesses a positive (+) electrical charge
             Neutron: has no charge (it is neutral)
             THE ATOMIC STRUCTURE

            Electronic Charge equals 1.602177 x 10-19 C
                          (C = coulombs)

- Charges are usually expressed as multiples of the electronic charge

         Charge of an electron = -1.602177 x 10-19 C = -1

          Charge of a proton = +1.602177 x 10-19 C = +1

   Atoms have no net electrical charge since they have equal
              number of electrons and protons
              THE ATOMIC STRUCTURE
   - Protons and neutrons have very large masses (about 2000 x) as
                        compared to electrons
            - Atoms generally have extremely small masses
      - Atomic Mass Unit (u) is used to express such small masses

           1 u = 1.66054 x 10-24 g or 1 g = 6.02212 x 1023 u


Particle   Charge        Mass (g)       Mass (u)       Relative Mass

Electron Negative (-1) 9.109 x 10-28    5.486 x 10-4   1
Proton   Positive (+1) 1.673 x 10-24    1.0073         1837
Neutron Neutral (0)    1.675 x 10-24    1.0087         1839
          THE ATOMIC STRUCTURE

- The center of an atom is small, dense, and positively charged
                      called the nucleus


- The nucleus contains all protons and neutrons and are referred
                         to as necleons


 - The nucleus is, therefore, positively charged and contributes
              about 99.9% of the mass of an atom
   THE ATOMIC STRUCTURE

- The electrons move rapidly around the nucleus

 - Outer region called the extranuclear region

  - Account for most of the volume of an atom

                Electron Cloud
        - Volume occupied by electrons
             - Negatively charged
       ATOMIC NUMBER (Z)

- The number of protons in the nucleus of an atom
     - determines the identity of the element

   - Since atoms have no net electrical charge
    number of protons = number of electrons


  Z = number of protons = number of electrons
                 MASS NUMBER (A)

             - The sum of the number of protons and
        the number of neutrons in the nucleus of an atom

-The total number of subatomic particles in the nucleus of an atom

              - The number of nucleons of an atom

          A = number of protons + number of neutrons

   number of neutrons = mass number - atomic number = A - Z
ATOMIC AND MASS NUMBERS
      MASS NUMBER
                    CHEMICAL SYMBOL
 ATOMIC NUMBER

                A
                    SYMBOL
                Z


 12                 16                 40
      C                 O                Ca
                    8                  20
  6


 Mass number is the superscript to the left
 Atomic number is the subscript to the left
        ATOMIC AND MASS NUMBERS

 An atom has an atomic number of 56 and a mass number of 138.
 What are the numbers of protons, electrons, and neutrons present
 in the atom? What is the number of subatomic particles present
                  in the nucleus of the atom?

           Number of protons = atomic number = 56
           Number of electrons = atomic number = 56
Number of neutrons = mass number – atomic number = 138-56 = 82
Number of subatomic particles in the nucleus = mass number = 138
       CHEMICAL PROPERTIES OF ATOMS


- The number of protons (the atomic number) characterizes an atom

     - Electrons determine the chemical properties of an atom

  - Atoms with the same atomic number have the same chemical
                           properties

   - Atoms with the same atomic number are atoms of the same
                            element
       CHEMICAL PROPERTIES OF ATOMS


               Chapter 1 definition of An Element
- Is a pure substance that cannot be reduced to a simpler substance
                    by normal chemical means


               Chapter 2 definition of An Element
  - Is a pure substance in which all atoms present have the same
                           atomic number
                        ISOTOPES
- Atoms of an element with the same atomic number but different
                         mass numbers

- Atoms of an element with the same number of protons and the
   same number of electrons but different numbers of neutrons

  - Isotopes of an element have the same chemical properties
            but slightly different physical properties

 - The atomic number is usually omitted since it is the same for
                 isotopes of a given element
                       ISOTOPES

11       12       13       14
    C         C        C       C    Most abundant is carbon-12
6         6        6       6



         1         2       3
             H         H   1H
                                    Most abundant is hydrogen-1
         1         1


        28        29       30
          Si        Si         Si   Most abundant is silicon-28
        14        14       14
            AVERAGE ATOMIC MASS

- Determined by using the masses of an element’s various
     isotopes and their respective natural abundances

                          Units
     1 u = 1.66054 x 10-24 g or 1 g = 6.02214 x 1023 u
                u (amu): atomic mass unit

u is defined by assigning a mass of exactly 12 u to an atom
               of carbin-12 (reference point)
             AVERAGE ATOMIC MASS

         For an element with n isotopes which have
        atomic masses in u (m1, m2, m3,….., mn) and
natural abundances expressed as fractions (x1, x2, x3,……,xn)

  Average Atomic Mass = m1x1 + m2x2 + m3x3 +….+ mnxn


    The natural abundance is usually expressed as a percentage

Divide by 100 to convert to the decimal form (fractional abundance)
           AVERAGE ATOMIC MASS


The mass spectrometer is an instrument used to measure
  the masses and relative (natural) abundances of the
       isotopes present in a sample of an element


                       Homework
Describe the operation and uses of the mass spectrometer
        One page maximum and double spaced
              AVERAGE ATOMIC MASS
Naturally occurring copper is 69.09% 63Cu, which has a relative
mass of 62.93 u, and 30.91% 65Cu, which has a relative mass of
     64.93 u. Calculate the average atomic mass of copper.
                             63Cu

                  natural abundance = 69.09%
          fractional abundance = 69.09/100 = 0.6909
                             65Cu

                  natural abundance = 30.91%
          fractional abundance = 30.91/100 = 0.3091
  Average Atomic Mass = (62.93)(0.6909) + (64.93)(0.3091)
                      = 63.5478
                      = 63.55 u
                FORMULA MASS

- The sum of atomic masses of all the atoms present in the
             chemical formula of a substance


- Relative mass based on the carbon-12 relative-mass scale


-It is advisable to use two decimal places for atomic masses
          FORMULA MASS

      Calculate the formula mass of H2O
            H: 2 x 1.01 u = 2.02 u
           O: 1 x 16.00 u = 16.00 u
   Formula mass = (2.02 + 16.00) u = 18.02 u


      Calculate the formula mass of H2SO4
             H: 2 x 1.01 u = 2.02 u
            S: 1 x 32.06 u = 32.06 u
            O: 4 x 16.00 u = 64.00 u
Formula mass = (2.02 + 32.06 + 64.00) u = 98.08 u
             FORMULA MASS

       Calculate the formula mass of CaCO3
            Ca: 1 x 40.08 u = 40.08 u
             C: 1 x 12.01 u = 12.01 u
             O: 3 x 16.00 u = 48.00 u
Formula mass = (40.08 + 12.01 + 48.00) u = 100.09 u


      Calculate the formula mass of Fe2(SO4)3
             Fe: 2 x 55.85 u = 111.70 u
              S: 3 x 32.07 u = 96.21 u
             O: 12 x 16.00 u = 192.00 u
Formula mass = (111.70 + 96.21 + 192.00) u = 399.91 u
                      THE MOLE

 The amount of substance of a system, which contains as many
elementary entities as there are atoms in 12 grams of carbon-12

                      - abbreviated mol

          1 mole (mol) = 6.02214179 x 1023 entities

 - known as the Avogadro’s number (after Amedeo Avogadro)

              - usually rounded to 6.022 x 1023
                     THE MOLE

The number of entities (or objects) can be atoms or molecules

               1 mol C = 6.022 x 1023 atoms C

           1 mol CO2 = 6.022 x 1023 molecules CO2

        2 conversion factors can be derived from each
               THE MOLE
How many atoms are there in 0.40 mole nitrogen?
                    6.022 x 1023 nitrogen atoms
0.40 mol nitrogen x
                          1 mol nitrogen
           = 2.4 x 1023 nitrogen atoms



How many molecules are there in 1.2 moles water?
                6.022x 1023 water molecules
1.2 mol water x
                        1 mol water
           = 7.2 x 1023 water molecules
                        THE MOLE

     How many H atoms are there in 1.2 moles water?

                (6.022x 1023 water molecules)      (2 H atoms)
1.2 mol water x                               x
                        (1mol water)            (1water molecule)

                    = 1.4 x 1024 H atoms
                    MOLAR MASS

- The mass of a substance in grams that is numerically equal to
              the formula mass of that substance


     - Add atomic masses to get the formula mass (in u)
                  = molar mass (in g/mol)


       - The mass, in grams, of 1 mole of the substance
               MOLAR MASS
              Consider the following

    Sodium (Na) has an atomic mass of 22.99 u
This implies that the mass of 1 mole of Na = 22.99 g
          Molar mass of Na = 22.99 g/mol

         Formula mass of NaCl = 58.44 u
       The mass of 1 mole of NaCl = 58.44 g
        Molar mass of NaCl = 58.88 g/mol

       Formula mass of CaCO3 = 100.09 u
     The mass of 1 mole of CaCO3 = 100.09 g
      Molar mass of CaCO3 = 100.09 g/mol
           MOLAR MASS

 Calculate the mass of 2.4 moles of NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)

           = 85.00 g /mol NaNO3

                            85.00 g NaNO 3
g NaNO 3  2.4 mol NaNO 3 x
                             1 mol NaNO 3

               = 204 g NaNO3

            = 2.0 x 102 g NaNO3
           MOLAR MASS

How many moles are present in 2.4 g NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)

           = 85.00 g /mol NaNO3
                              1 mol NaNO 3
 mol NaNO 3  2.4 g NaNO 3 x
                             85.00 g NaNO 3

             = 0.028 mol NaNO3

           = 2.8 x 10-2 mol NaNO3
             CHEMICAL FORMULA

   Subscripts represent both atomic and molar amounts

                     Consider Na2S2O3:


- Two atoms of sodium, two atoms of sulfur, and three atoms of
        oxygen are present in one molecule of Na2S2O3


- Two moles of sodium, two moles of sulfur, and three moles of
         oxygen are present in one mole of Na2S2O3
            CHEMICAL FORMULA
How many moles of sodium atoms, sulfur atoms, and oxygen
  atoms are present in 1.8 moles of a sample of Na2S2O3?

  I mol Na2S2O3 contains 2 mol Na, 2 mol S, and 3 mol O

                                        2 mol Na
      mol Na 1.8 mol Na 2S2 O 3 x                     3.6 mol Na
                                     1 mol Na 2S2 O 3

                                         2 mol S
     mol S 1.8 mol Na 2S2 O 3 x                       3.6 mol S
                                     1 mol Na 2S2 O 3

                                        3 mol O
     mol O 1.8 mol Na 2S2 O 3 x                       5.4 mol O
                                     1 mol Na 2S2 O 3
            CHEMICAL CALCULATIONS

  Calculate the number of molecules present in 0.075 g of urea,
                          (NH2)2CO

                     Given mass of urea:
         - Convert to moles of urea using molar mass
    - Convert to molecules of urea using Avogadro’s number

                     1mole (NH2 ) 2 CO 6.022x 1023 molecules ( NH2 ) 2 CO
0.075g (NH2 ) 2 CO x                    x
                     60.07g (NH2 ) 2 CO       1mole (NH2 ) 2 CO


                  = 7.5 x 1020 molecules (NH2)2CO
          CHEMICAL CALCULATIONS
How many grams of carbon are present in a 0.125 g of vitamin C,
                         C6H8O6?

                   Given mass of vitamin C:
       - Convert to moles of vitamin C using molar mass
  - Convert to moles of C (1 mole C6H8O6 contains 6 moles C)
     - Convert moles carbon to g carbon using molar mass

                      1 mol C 6 H 8O 6      6 mol C        12.01 g C
0.125 g C 6 H 8O 6 x                    x                x
                     176.14 g C 6 H 8O 6 1 mol C 6 H 8O 6 1 mol C

                      = 0.0511 g carbon
       PERCENTAGE COMPOSITION

       - Percentage by mass contributed by individual
                  elements in a compound


                                 mass of element
                  % element                     x 100%
                                mass of compound


              (atomic mass of element)(n umber of atoms of element)
% element                                                          x 100%
                           formula mass of compound
       PERCENTAGE COMPOSITION

Calculate the percentage of carbon, hydrogen, and oxygen, in
                     ethanol (C2H5OH)

                    (12.01u)(2)
             %C                x 100%  52.14 %
                      46.07 u

                  (1.01u)(6)
             %H             x 100%  13.13 %
                   46.07 u

                 (16.00 u)(1)
            %O               x 100%  34.73 %
                   46.07 u
      PERCENTAGE COMPOSITION

Calculate the percent composition by mass of each element
                in the following compounds

                        C9H8O4
                      (NH4)2PtCl4
                        C2H2F4
                       C8H10N4O2
                      Pt(NH3)2Cl2
         EMPIRICAL FORMULA
                Given mass % elements:

   - Convert to g elements assuming 100.0 g sample

     - Convert to mole elements using molar mass

                 - Calculate mole ratio
     (divide each by the smallest number of moles)

           - Round each to the nearest integer

   - Multiply through by a suitable factor if necessary
( __.5 x 2      or    __.33 x 3       or       __ .25 x 4)
            EMPIRICAL FORMULA
Determine the empirical formula for a compound that gives the
   following percentages upon analysis (in mass percents):
    71.65 % Cl            24.27 % C             4.07 % H

   - Assume 100.0 g of sample and convert grams to moles
                                        1 mol Cl
          71.65 g Cl    71.65 g Cl x               2.021 mol Cl
                                       35.45 g Cl

                                        1 mol C
          24.27 g C     24.27 g C x               2.021 mol C
                                       12.01 g C

                                       1 mol H
           4.07 g H     4.07 g H x               4.04 mol H
                                       1.01 g H
         EMPIRICAL FORMULA
              - Calculate mol ratios
                          2.021
                   Cl :          1.000
                          2.021

                          2.021
                   C:            1.000
                          2.021

                          4.04
                    H:           1.99
                          2.021

- Round to nearest integers and write empirical formula

          Cl: 1, C: 1, H: 2 giving CH2Cl

								
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