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CH 9 10

VIEWS: 17 PAGES: 41

  • pg 1
									   Chemistry

 Chapters 9 & 10:
Chemical Formulas
     Writing Formulas of Ionic Compounds
chemical formula: has neutral charge;
                shows types of atoms
                and how many of each
 To write an ionic compound’s formula, we need:
   1. the two types of ions
   2. the charge on each ion
         Na1+   and    F1–        NaF
         Ba2+   and   O2–         BaO
         Na1+   and   O2–         Na2O
         Ba2+   and   F1–         BaF2
criss-cross rule: charge on cation / anion
       “becomes” subscript of anion / cation

     ** Warning: Reduce to lowest terms.

Al3+ and O2–    Ba2+ and S2–     In3+ and Br1–



   Al 2 O 3         Ba 2S 2          In 1 Br 3

    Al2O3            BaS              InBr3
     Writing Formulas w/Polyatomic Ions
Parentheses are required only when you need more
than one “bunch” of a particular polyatomic ion.

      Ba2+    and   SO42–       BaSO4
      Mg2+ and      NO21–       Mg(NO2)2
      NH41+ and     ClO31–      NH4ClO3
       Sn4+   and   SO42–       Sn(SO4)2
      Fe3+    and   Cr2O72–     Fe2(Cr2O7)3
      NH41+ and     N3–         (NH4)3N
  Ionic Compounds (cation/anion combos)
 Single-Charge Cations with Elemental Anions

                                i.e., “pulled off the
The single-charge cations are:         Table” anions
     groups 1, 2, 13, and Ag1+, Cd2+, and Zn2+
           Na
A. To name, given
   the formula:
           Ba
     1. Use name of cation.
    2. Use name of anion (it has the ending “ide”).

             NaF         sodium fluoride
             BaO         barium oxide
             Na2O        sodium oxide
              BaF2       barium fluoride
                                      Zn
            Ca
                                 Ag
B. To write formula,
   given the name:


      1. Write symbols for the two types of ions.
      2. Balance charges to write formula.

      silver sulfide     Ag1+   S2–     Ag2S
      zinc phosphide     Zn2+   P3–     Zn3P2
      calcium iodide     Ca2+ I1–       CaI2
                        Common Polyatomic Ions
                                              Names of Common Polyatomic Ions

             Ion                  Name                            Ion       Name

             NH4 +               ammonium                        CO3 2-     carbonate
             NO2 -               nitrite                         HCO3 -     bicarbonate
             NO3 -               nitrate                         IO3-       iodate
             SO3 2-              sulfite                         IO4-       periodate
             SO4 2-              sulfate                         ClO -      hypochlorite
             HSO4-               hydrogen sulfate                ClO2 -     chlorite
             OH -                hydroxide                       ClO3 -     chlorate
             CN -                cyanide                         ClO4 -     perchlorate
             C2H3O2-             acetate                         BrO3-      bromate
             MnO4 -              permanganate                    C2O42-     oxalate
             PO4 3-              phosphate                       Cr2O7 2-   dichromate
             PO3 3-              phosphite                       CrO4 2-    chromate
             H2PO4 -             dihydrogen phosphate            O2 2-      peroxide




Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 100
               Polyatomic Ion Trends
 +1 oxygen        “Normal”     -1 oxygen      -2 oxygens
“Per____ate”     “–ate” form   “-ite” form   “hypo___ite”
   ClO4-             ClO3-       ClO2-           ClO-
 perchlorate       chlorate      chlorite     hypochlorite
   BrO4-             BrO3-       BrO2-           BrO-
perbromate         bromate      bromite      hypobromite
    IO4-              IO3-        IO2-            IO-
 periodate          iodate        iodite      hypoiodite
                     NO3-         NO2-
                    nitrate      nitrite
                     CO32-       CO22-
                  carbonate    carbonite
                     SO42-       SO32-
                    sulfate      sulfite
                     PO43-       PO33-
                 phosphate     phosphite
  Multiple-Charge Cations with Elemental Anions

                                 i.e., “pulled off the
                                        Table” anions
The multiple-charge cations are:
        Pb, Sn, and the transition elements
       (but – of course! – not Ag, Cd, or Zn)
A. To name, given the formula:               Cu
                                        Fe
1. Figure out charge on
  cation.
 2. Write name of cation.
3. Write Roman numerals
   in ( ) to show cation’s charge.      Stock System
4. Write name of anion.                of nomenclature

     FeO         2+
              Fe? oxide
              iron O2–               iron (II) oxide
     Fe2O3    iron oxide O2– O2–O2– iron (III) oxide
              Fe? Fe?
                 3+   3+


     CuBr        ?
              Cu1+ Br1–
              copper bromide         copper (I) bromide
     CuBr2    Cu? Br1– Br1–
                2+
              copper bromide         copper (II) bromide
B. To find the formula, given the name:
    1. Write symbols for the two types of ions.
    2. Balance charges to write formula.
                       Co



                                          Sn

     cobalt (III) chloride Co3+ Cl1–      CoCl3
     tin (IV) oxide         Sn4+ O2–      SnO2
     tin (II) oxide         Sn2+ O2–      SnO
    Traditional System of Nomenclature
…used historically (and still some today) to name
 compounds w/multiple-charge cations




To use: 1. Use Latin root of cation.
         2. Use -ic ending for higher charge;
               -ous ending for lower charge.
         3. Then say name of anion, as usual.
Element       Latin root       -ic           -ous
gold, Au      aur-             Au3+          Au1+
lead, Pb      plumb-           Pb4+          Pb2+
tin, Sn       stann-           Sn4+          Sn2+
copper, Cu    cupr-            Cu2+          Cu1+
iron, Fe      ferr-            Fe3+          Fe2+
Write formulas:        Write names:
                                             3– 3–
                                 ? Pb? Pb4+ P P
                                 4+  4+  ?
cuprous sulfide        Pb3P4 Pb              P3– P3–
 Cu1+ S2–     Cu2S         plumbic phosphide
auric nitrite          Pb3P2 Pb? Pb? Pb? P3– P3–
                                2+   2+  2+


 Au3+ NO21– Au(NO2)3     plumbous phosphide
                                           OH1–
ferrous fluoride       Sn(OH)4 Sn? OH1– OH1–OH1–
                                 4+


 Fe2+ F1–      FeF2        stannic hydroxide
    Compounds Containing Polyatomic Ions
 Insert name of ion
 where it should go
 in the compound’s
 name.
Write formulas:
 iron (III) nitrite   Fe3+ NO21–   Fe(NO2)3
 ammonium phosphide   NH41+ P3–    (NH4)3P
 ammonium chlorate    NH41+ ClO31– NH4ClO3
 zinc phosphate       Zn2+ PO43–   Zn3(PO4)2
 lead (II) permanganate Pb2+ MnO41– Pb(MnO4)2
Write names:


  (NH4)2S2O3                ammonium thiosulfate

  AgBrO3                    silver bromate

  (NH4)3N                   ammonium nitride
             CrO42–
              ?
  U(CrO4)3 U6+ CrO42–      uranium (VI) chromate
             CrO42–
  Cr2(SO3)3 Cr? SO32–
                3+         chromium (III) sulfite
             3+
           Cr? SO32– SO32–
                   Acid Nomenclature
• Identifying the compound as an acid.                  H2SO4
   – Acids begin with “H”
• Binary Acids have only 2 elements.                    HNO3
   – Identify the other element. (Not Hydrogen)
                                                        HCl
   – Change the ending of the element’s name to “-ic”
   – Put a “hydro-” in front                            HF
       • HCl
           – Chlorine is the second element
           – Chlorine becomes “chloric”
           – Hydrochloric acid
       • HF
           – fluorine is the second element
           – fluorine becomes “fluoric”
           – Hydrofluoric acid
                         Acid Nomenclature
•   Oxyacids have a polyatomic ion.                        H2SO4
     – Identify the polyatomic ion.
     – If the polyatomic ion ends with “-ate” it will      HNO3
       be an “-ic acid”
          • H2SO4                                          HCl
               – The polyatomic ion is sulfate
               – Sulfuric acid                             HF
          • HNO3
               – The polyatomic ion is nitrate
               – Nitric acid
     – If the polyatomic ion ends with “-ite” it will be
       an “-ous acid”
          • H2SO3
               – The polyatomic ion is sulfite
               – Sulfurous acid
          • HNO2
               – The polyatomic ion is nitrite
               – Nitrous acid
       Acid Nomenclature
• Examples
  –HBr          –Hydrobromic acid
  –H3PO4        –Phosphoric acid
  –HClO4        –Perchloric acid
  –H2S          –Hydrosulfuric acid
  –HClO2        –Chlorous acid
             Covalent Compounds
-- contain two types of
      nonmetals
       ** Key:
 FORGET CHARGES!
                  Use Greek prefixes to indicate how
   What to do:     many atoms of each element, but
                  don’t use “mono” on first element.
           1 – mono          6 – hexa
           2 – di            7 – hepta
           3 – tri           8 – octa
           4 – tetra         9 – nona
           5 – penta        10 – dec
EXAMPLES:



 carbon dioxide         CO2
 CO                     carbon monoxide
 dinitrogen trioxide    N2O3
 N2O5                   dinitrogen pentoxide
 carbon tetrachloride   CCl4
 NI3                    nitrogen triiodide
molar mass: the mass of one mole of a substance
  PbO2      Pb:   1 (207.20 g) = 207.20 g
                                            239.20g
             O:   2 (16.00 g) = 32.00 g

  HNO3       H:   1 (1.01 g) = 1.01 g
             N:   1 (14.01 g) = 14.01 g      63.02 g
             O:   3 (16.00 g) = 48.00 g
ammonium phosphate      NH41+ PO43–
  (NH4)3PO4 N: 3 (14.01 g) = 42.03 g
             H: 12 (1.01 g) = 12.12 g
                                            149.12 g
             P:   1 (30.97 g) = 30.97 g
             O:   4 (16.00 g) = 64.00 g
 percentage composition: the mass % of each
                         element in a compound
                       g element
 % of element =                        x 100
                molar mass of compound

Find % composition.    (see calcs above)

  PbO2 207.20 g Pb : 239.20 g= 86.6% Pb
        32.00 g O : 239.20 g= 13.4% O

(NH4)3PO4 42.03 g N   : 149.12 g=   28.2% N
          12.12 g H   : 149.12 g=    8.13% H
          30.97 g P   : 149.12 g=   20.8% P
          64.00 g O   : 149.12 g=   42.9% O
              zinc acetate

           Zn2+     CH3COO1–
             Zn(CH3COO)2


Zn: 1 (65.39 g)= 65.39 g                  = 35.6% Zn
C: 4 (12.01 g)= 48.04 g                   = 26.2% C
                             : 183.49 g
H: 6 (1.01 g) =     6.06 g                = 3.30% H
O: 4 (16.00 g) = 64.00 g                  = 34.9% O
                  183.49 g
           Dihydrogen Monoxide (DHMO):
                A Tale of Danger and
                   Irresponsibility
-- major component of acid rain
-- found in all cancer cells
-- inhalation can be deadly
-- excessive ingestion results in
       acute physical symptoms:
     e.g., frequent urination,
           bloated sensation,
           profuse sweating
-- often an industrial byproduct of chemical
   reactions; dumped wholesale into rivers and lakes
Empirical Formula and Molecular Formula

    lowest-terms           shows the true number
      formula           and type of atoms in a m’cule

                   Molecular      Empirical
 Compound
                   Formula        Formula
   glucose         C6H12O6          CH2O

   propane           C3H8            C3H8

   butane            C4H10           C2H5

 naphthalene         C10H8          C5H4

   sucrose         C12H22O11       C12H22O11

   octane            C8H18           C4H9
Finding an Empirical Formula from Experimental Data
1. Find # of g of each element. (If %, assume 100g)
2. Convert each g to mol.
3. Divide each “# of mol” by the smallest “# of mol.”
4. Use ratio to find formula.
 A compound is 45.5% yttrium and 54.5% chlorine.
 Find its empirical formula.
            1 mol Y 
  45.5 g Y             0.512 mol Y  0.512  1
            88.91 g Y 
            1 mol Cl 
  54.5 g Cl              1.537 mol Cl  0.512  3
            35.45 g Cl 
                              YCl3
     A ruthenium/sulfur
     compound is 67.7% Ru.
     Find its empirical formula.


           1 mol Ru 
67.7 g Ru               0.670 mol Ru  0.670  1
           101.07 g Ru 
           1 mol S 
 32.3 g S              1.007 mol S  0.670  1.5
           32.07 g S 

               RuS1.5          Ru2S3
      A 17.40 g sample of a
      technetium/oxygen compound
      contains 11.07 g of Tc. Find
      the empirical formula.

            1 mol Tc 
11.07 g Tc             0.113 mol Tc  0.113  1
            98 g Tc 
           1 mol O 
 6.33 g O              0.396 mol O  0.113  3.5
           16.00 g O 

                TcO3.5        Tc2O7
 A compound contains 4.63 g lead, 1.25 g nitrogen,
 and 2.87 g oxygen. Name the compound.
           1 mol Pb 
4.63 g Pb               0.0223 mol Pb  0.0223  1
           207.2 g Pb 
           1 mol N 
 1.25 g N              0.0892 mol N  0.0223  4
           14.01 g N 
                    
2.87 g O  1 mol O  0.1794 mol O  0.0223  8
          16.00 g O 
                 ?    PbN4O8    ?
                                     lead (IV) nitrite
                     Pb(NO2)4
                                    (plumbic nitrite)
                     Pb? 4 NO21–
To find molecular formula…
  A. Find empirical formula.
  B. Find molar mass of
     empirical formula.
  C. Find n = mm molecular
               mm empirical
  D. Multiply all parts of
     empirical formula by n.

  (How many empiricals “fit into” the molecular?)
A carbon/hydrogen compound is 7.7% H and has a
molar mass of 78 g. Find its molecular formula.
             1 mol H 
    7.7 g H             7.62 mol H  7.62  1
             1.01 g H 
             1 mol C 
   92.3 g C              7.69 mol C    7.62    1
             12.01 g C 

                                 emp. form.  CH


                           78 g
  mmemp = 13.02 g                 =6         C6H6
                          13.02 g
A compound has 26.33 g nitrogen, 60.20 g oxygen,
and molar mass 92 g. Find molecular formula.
           1 mol N 
26.33 g N              1.879 mol N  1.879  1
           14.01 g N 
           1 mol O 
60.20 g O              3.763 mol O  1.879  2
           16.00 g O 


                                         NO2


                        92 g
 mmemp = 46.01 g              =2        N2O4
                        46.01 g
          Hydrates and Anhydrous Salts
anhydrous salt: an ionic compound (i.e., a salt) that
                attracts water molecules and forms
                loose chemical bonds with them;
                symbolized by MN
                    “anhydrous” = “without water”
                 Uses: “desiccants” in leather
                       goods, electronics, vitamins
hydrate: an anhydrous salt with the water attached
     -- symbolized by MN . ? H2O
     Examples: CuSO4 . 5 H2O          BaCl2 . 2 H2O
               Na2CO3 . 10 H2O        FeCl3 . 6 H2O
  H 2O H 2O                                  H 2O
                                       H 2O
H2O MN H2O     HEAT                         HO
                       MN   +       H 2O H O 2
  H 2O H 2O                               2    H 2O

 hydrate          anhydrous salt        water


              ENERGY            +




              ENERGY                +
     Finding the Formula
        of a Hydrate


1. Find the # of g of MN and # of g of H2O.
2. Convert g to mol.
3. Divide each “# of mol” by the smallest “# of mol.”

4. Use the ratio to find the hydrate’s formula.
 Find formula of hydrate for each problem.
    sample’s mass before        MN . ? H2O
         heating = 4.38 g        (hydrate)
    sample’s mass after        MN
         heating = 1.93 g (anhydrous salt)
  molar mass of anhydrous salt = 85 g
           1 mol MN 
1.93 g MN             0.0227 mol MN  0.0227  1
           85 g MN 
            1 mol H 2O 
2.45 g H2O                 0.1360 mol H 2O  0.0227  6
            18.02 g H 2O 

                        MN . 6 H2O
     A. beaker = 46.82 g                    beaker
     B. beaker + sample before             beaker +
          heating = 54.35 g               salt + water
     C. beaker + sample after
                                          beaker + salt
          heating = 50.39 g
 molar mass of anhydrous salt = 129.9 g
           1 mol MN 
3.57 g MN               0.0275 mol MN  0.0275  1
           129.9 g MN 

3.96 g H2O  1 mol H 2O   0.220 mol H O  0.0275  8
                                     2
           18.02 g H 2O 
                         MN . 8 H2O
    A. beaker = 47.28 g                     beaker
    B. beaker + sample before              beaker +
         heating = 53.84 g                salt + water
    C. beaker + sample after
                                         beaker + salt
         heating = 51.48 g
molar mass of anhydrous salt = 128 g
           1mol MN 
4.20 g MN             0.0328 mol MN  0.0328  1
           128 g MN 
            1 mol H 2O 
2.36 g H2O                0.1310 mol H 2O  0.0328  4
            18.02 g H 2O 
                       MN . 4 H2O
 For previous problem, find % water and
 % anhydrous salt (by mass).

                     g H2O in formula
           % H2O 
                   molar mass of hydrate

            4 (18.02 g H 2O)
% H 2O                        36.0% H2O
         4 (18.02 g )  128 g   64.0% MN
                          or…
                2.36 g H2O
      % H2O                   36.0% H2O
              2.36 g  4.20 g
                                64.0% MN
Strontium chloride is an anhydrous salt on which the
following data were collected. Find formula of hydrate.
    A. beaker = 65.2 g                     beaker
    B. beaker + sample before              beaker +
         heating = 187.9 g                salt + water
    C. beaker + sample after
                                         beaker + salt
         heating = 138.2 g
   Sr2+   Cl1–    SrCl2
           1 mol MN 
73.0 g MN                0.461 mol MN  0.461  1
           158.52 g MN 
            1 mol H 2O 
49.7 g H2O                 2.76 mol H2O  0.461  6
            18.02 g H 2O 
                                    SrCl2 . 6 H2O

								
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