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

 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
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.

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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