Determination of the Empirical Formula of Zinc Oxide

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Unit 4 – Conservation
     of Mass and
    Stoichiometry
  Ions
 Cation: A positive ion
               Mg2+, NH4+

 Anion: A negative ion
               Cl-, SO42-

 Ionic Bonding: Force of attraction
  between oppositely charged ions.
       Predicting Ionic Charges
Group 1: Lose 1 electron to form 1+ ions
     H+   Li+ Na+   K+
     Predicting Ionic Charges
Group 2: Loses 2 electrons to form 2+ ions
   Be2+   Mg2+   Ca2+   Sr2+   Ba2+
      Predicting Ionic Charges
                    Group 13: Loses 3
B3+   Al3+   Ga3+   electrons to form
                    3+ ions
       Predicting Ionic Charges
Neither! Group 13      Group 14: Lose 4
elements rarely form   electrons or gain
ions.                  4 electrons?
 Predicting Ionic Charges
 N3- Nitride     Group 15: Gains 3
 P3- Phosphide   electrons to form
As3- Arsenide    3- ions
 Predicting Ionic Charges
 O2- Oxide      Group 16: Gains 2
 S2- Sulfide    electrons to form
Se2- Selenide   2- ions
          Predicting Ionic Charges
F1- Fluoride   Br1- Bromide   Group 17: Gains 1
                              electron to form
Cl1-Chloride   I1- Iodide
                              1- ions
Predicting Ionic Charges
              Group 18: Stable
              Noble gases do not
              form ions!
        Predicting Ionic Charges
Groups 3 - 12: Many transition elements
have more than one possible oxidation state.
Iron(II) = Fe2+   Iron(III) = Fe3+
        Predicting Ionic Charges
Groups 3 - 12: Some transition elements
have only one possible oxidation state.
Zinc = Zn2+    Silver = Ag+
   Writing Ionic Compound Formulas
Example: Barium nitrate
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges are
                                 Ba2+ ( NO - )
balanced.                                 3 2
3. Balance charges , if necessary,     Not balanced!
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
   Writing Ionic Compound Formulas
Example: Ammonium sulfate
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges    ( NH4+) SO42-
are balanced.
                                        2
3. Balance charges , if necessary,
using subscripts. Use parentheses      Not balanced!
if you need more than one of a
polyatomic ion.
   Writing Ionic Compound Formulas
Example: Iron(III) chloride
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges
                                     Fe3+ Cl-
are balanced.
                                                 3
3. Balance charges , if necessary,
                                       Not balanced!
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
   Writing Ionic Compound Formulas
Example: Aluminum sulfide
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges
are balanced.                          Al 3+
                                           2
                                               S2-
                                                   3
3. Balance charges , if necessary,
using subscripts. Use parentheses       Not balanced!
if you need more than one of a
polyatomic ion.
   Writing Ionic Compound Formulas
Example: Magnesium carbonate
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges
are balanced.                     Mg2+ CO32-
                                       They are balanced!
   Writing Ionic Compound Formulas
Example: Zinc hydroxide
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges are
balanced.                        Zn2+     ( OH- )2
3. Balance charges , if necessary,
using subscripts. Use parentheses      Not balanced!
if you need more than one of a
polyatomic ion.
   Writing Ionic Compound Formulas
Example: Aluminum phosphate
1. Write the formulas for the cation
and anion, including CHARGES!

2. Check to see if charges are
balanced.                        Al 3+   PO4   3-


                                 They ARE balanced!
       Naming Ionic Compounds
• 1. Cation first, then anion

• 2. Monatomic cation = name of the
  element
          • Ca2+ = calcium ion

• 3. Monatomic anion = root + -ide
           • Cl- = chloride
      • CaCl2 = calcium chloride
         Naming Ionic Compounds
                     (continued)

      Metals with multiple oxidation states
• -    some metal forms more than one cation
• -    use Roman numeral in name

                      • PbCl2

                  • Pb2+ is cation

           • PbCl2 = lead(II) chloride
          Naming Binary Compounds
•   -   Compounds between two nonmetals
•   -   First element in the formula is named first.
•   -   Second element is named as if it were an anion.
•   -   Use prefixes
•   -   Only use mono on second element -
           P2O5   = diphosphorus pentoxide
            CO2   = carbon dioxide
             CO   = carbon monoxide
           N2O    = dinitrogen monoxide
      Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.




24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
   Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
                 24.31 
           Mg           100  28.83%
                 84.32 
                 12.01 
            C           100  14.24%
                 84.32 
                 48.00 
            O           100  56.93%
                 84.32        100.00
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.

    molecular formula = (empirical
    formula)n [n = integer]
    molecular formula = C6H6 = (CH)6
    empirical formula = CH
Formulas     (continued)

Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).

Examples:
 NaCl       MgCl2    Al2(SO4)3   K2CO3
 Formulas    (continued)

 Formulas for molecular compounds MIGHT
 be empirical (lowest whole number ratio).


Molecular:   H2O     C6H12O6     C12H22O11


Empirical:   H2O       CH2O      C12H22O11
   Empirical Formula Determination
1. Base calculation on 100 grams of compound.
2. Determine moles of each element in 100
   grams of compound.
3. Divide each value of moles by the smallest of
   the values.
4. Multiply each number by an integer to obtain
   all whole numbers.
    Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?

           49.32 g C 1 mol C   4.107 mol C
                      12.01 g C 
            6.85g H 1 mol H   6.78 mol H
                       1.01 g H 
           43.84 g O 1 mol O   2.74 mol O
                      16.00 g O 
    Empirical Formula Determination
                (part 2)
Divide each value of moles by the smallest of the
values.
 Carbon:
          4.107 mol C
                       1.50
           2.74 mol O
          6.78 mol H
Hydrogen:              2.47
          2.74 mol O
           2.74 mol O
   Oxygen:             1.00
           2.74 mol O
     Empirical Formula Determination
                 (part 3)
 Multiply each number by an integer to obtain all
 whole numbers.

Carbon: 1.50     Hydrogen: 2.50        Oxygen: 1.00
        x 2                x 2                 x 2
           3                  5                   2
            Empirical formula: C3H5O2
   Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
  1. Find the formula mass of C3H5O2

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
   Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
  2. Divide the molecular mass by the
  mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  146
      2
   73
   Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
 3. Multiply the empirical formula by this
 number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  146
      2      (C3H5O2) x 2     C6H10O4
   73         =
Combination (Synthesis) Reactions

 Two or more substances combine to form a
 new compound.

            A + X  AX

    Reaction of elements with oxygen and sulfur
    Reactions of metals with Halogens
    Synthesis Reactions with Oxides
    There are others not covered here!
    Decomposition Reactions
 A single compound undergoes a reaction that
 produces two or more simpler substances
               AX  A + X
Decomposition of:
  Binary compounds   H2O(l )  2H2(g) + O2(g)
  Metal carbonates   CaCO3(s)  CaO(s) + CO2(g)
  Metal hydroxides   Ca(OH)2(s)  CaO(s) + H2O(g)
  Metal chlorates    2KClO3(s)  2KCl(s) + 3O2(g)
  Oxyacids           H2CO3(aq)  CO2(g) + H2O(l )
  Single Replacement Reactions
               A + BX  AX + B
               BX + Y  BY + X
Replacement of:
        Metals by another metal
        Hydrogen in water by a metal
        Hydrogen in an acid by a metal
        Halogens by more active halogens
The Activity Series of the Metals
 Lithium
             Metals can replace other metals
 Potassium
 Calcium     provided that they are above the
 Sodium      metal that they are trying to
 Magnesium
             replace.
 Aluminum
 Zinc
 Chromium    Metals above hydrogen can
 Iron        replace hydrogen in acids.
 Nickel
 Lead
 Hydrogen    Metals from sodium upward can
 Bismuth     replace hydrogen in water
 Copper
 Mercury
 Silver
 Platinum
 Gold
The Activity Series of the Halogens

    Fluorine      Halogens can replace other
    Chlorine      halogens in compounds, provided
    Bromine       that they are above the halogen
    Iodine        that they are trying to replace.

2NaCl(s) + F2(g)  ??? 2NaF(s) + Cl2(g)

MgCl2(s) + Br2(g)  ???   No Reaction
  Double Replacement Reactions

The ions of two compounds exchange places in an
aqueous solution to form two new compounds.
         AX + BY  AY + BX

One of the compounds formed is usually a
precipitate, an insoluble gas that bubbles out of
solution, or a molecular compound, usually water.
       Combustion Reactions
A substance combines with oxygen, releasing a large
amount of energy in the form of light and heat.

    Reactive elements combine with oxygen
               P4(s) + 5O2(g)  P4O10(s)
            (This is also a synthesis reaction)
    The burning of natural gas, wood, gasoline
          C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
           Stoichiometry
             “In solving a problem of this sort,
             the grand thing is to be able to
             reason backward. This is a very
             useful accomplishment, and a very
             easy one, but people do not practice
             it much.”


             Sherlock Holmes, in Sir Arthur Conan
             Doyle’s A Study in Scarlet

Stoichiometry - The study of quantities of
materials consumed and produced in chemical
reactions.
Review: Atomic Masses
      Elements occur in nature as mixtures of
    isotopes
     Carbon =      98.89% 12C
                    1.11% 13C
                   <0.01% 14C
     Carbon’s atomic mass = 12.01 amu
Review: The Mole

 The number equal to the number of
 carbon atoms in exactly 12 grams of pure
 12C.

 1 mole of anything = 6.022  1023 units
 of that thing
The Mole
Using Compound Masses
Review: Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole
of the compound.
       CO2 = 44.01 grams per mole
      H2O = 18.02 grams per mole
   Ca(OH)2 = 74.10 grams per mole
Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.

    C2H5OH + 3O2  2CO2 + 3H2O
           reactants              products
When the equation is balanced it has quantitative
significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles
of water
Mole Relations
Calculating Masses of Reactants and
              Products
1. Balance the equation.
2. Convert mass to moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of
   desired substituent.
5. Convert moles to grams, if necessary.
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed.
1. Identify reactants and products and write
the balanced equation.
      4 Al    + 3 O2            2 Al2O3
       a. Every reaction needs a yield sign!
       b. What are the reactants?
       c. What are the products?
       d. What are the balanced coefficients?
   Working a Stoichiometry Problem
   6.50 grams of aluminum reacts with an excess of
   oxygen. How many grams of aluminum oxide are
   formed?


              4 Al       +   3 O2  2Al2O3

6.50 g Al   1 mol Al     2 mol Al2O3 101.96 g Al2O3
                                                      = ? g Al2O3
            26.98 g Al   4 mol Al    1 mol Al2O3

 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3
                            Standard Molar
                                Volume
                           Equal volumes of all gases
                           at the same temperature
                           and pressure contain the
                           same number of
                           molecules.
                                 - Amedeo Avogadro


At STP (Standard Temperature and Pressure):

1 mole of a gas occupies 22.4 liters of volume
      Gas Stoichiometry #1
  If reactants and products are at the same
  conditions of temperature and pressure,
  then mole ratios of gases are also volume
  ratios.
    3 H2(g)   +   N2(g)            2NH3(g)

3 moles H2    + 1 mole N2        2 moles NH3
3 liters H2   + 1 liter N2       2 liters NH3
   Gas Stoichiometry #2
How many liters of ammonia can be
produced when 12 liters of hydrogen react
with an excess of nitrogen?
 3 H2(g)   +    N2(g)                2NH3(g)


      12 L H2      2 L NH3
                                 =   8.0   L NH3
                   3 L H2
       Gas Stoichiometry #3
   How many liters of oxygen gas, at STP, can
   be collected from the complete decomposition
   of 50.0 grams of potassium chlorate?
         2 KClO3(s)  2 KCl(s) + 3 O2(g)



50.0 g KClO3     1 mol KClO3    3 mol O2      22.4 L O2
               122.55 g KClO3   2 mol KClO3   1 mol O2



                                       = 13.7 L O2
       Gas Stoichiometry #4
   How many liters of oxygen gas, at 37.0C
   and 0.930 atmospheres, can be collected
   from the complete decomposition of 50.0
   grams of potassium chlorate?
         2 KClO3(s)  2 KCl(s) + 3 O2(g)

50.0 g KClO3     1 mol KClO3    3 mol O2
                                              = “n” mol O2
                                                  0.612
               122.55 g KClO3   2 mol KClO3       mol O2

                               L  atm
    nRT      (0.612mol)(0.0821         )(310 K)
 V                           mol  K          = 16.7 L
     P                  0.930 atm
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
Limiting Reagents - Combustion
     Solving a Stoichiometry Problem

1. Balance the equation.
2. Convert masses to moles.
3. Determine which reactant is limiting.
4. Use moles of limiting reactant and mole
   ratios to find moles of desired product.
5. Convert from moles to grams.

				
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