Download Chapter 7 Chemical Formulas and Chemical Compounds

Chapter 7
Chemical Formulas and Chemical
Compounds
Chemists use chemical names and
formulas to describe the atomic
composition of compounds.
 A chemical formula indicates the relative number of
atoms of each element in a chemical compound.
 A molecular formula indicates the number of atoms of
each element contained in a SINGLE molecule of the
compound.
 A formula unit represents the simplest ratio of the cations
and anions in an ionic compound.
 C8H18
Octane- a straight chain saturated
hydrocarbon containing 8 (oct) carbon atoms.
 Try this- C5H12
 Al2(SO4)3 Aluminum sulfate. An IONIC compound
containing a cation and a polyatomic anion.
 The polyatomic anion is defined by the parentheses and the
number of these polyatomic ions are given outside of the
parentheses.
 If there is only 1 polyatomic anion- the parentheses are not
used; you have to look for the polyatomic ion.
Ex: LiBrO3 lithium bromate
 How many Al, S, and O atoms are found in aluminum sulfate?
MONATOMIC IONS
 Ions formed from a single atom.
 This DOES NOT mean there is only one ionization.
 Ex: N-3. A single nitrogen atom takes on 3 electrons.
 Group trends
 – main group elements: groups 1-3 form cations (+1, +2,
+3) while groups 15-17 form anions (-3, -2, -1)
 Group 14, as you move down the table, will form cations
when they lose the 2 e- from their p-orbital.
 D-block elements usually form +2, +3 , but can also form
+1, +4.
 The same elements can form more that one ion. Example:
Chromium can form a +2 (chromium II) or a +3 (chromium III)
ion.
See table 7-1 pg 205 for other examples.
Naming Monatomic Ions
 Cations are simple…just the name of the element.
 Calcium  calcium ion
 Potassium  ?
 Anions- the ending is dropped and -ide is added
to the root.
 chlorine  chloride ion
 Nitrogen  ?
The Stock System of naming [Iron (III)] will be discussed
later.
Binary Ionic Compounds
 Compounds composed of two different elements.
 The total number of positive and negative charges must
be equal.
 To find the subscripts in a formula the + and - charges are “crossed
over”
Al3+ O2 2. Use the absolute value of each atom’s charge as the subscript for
the other ion.
Al2O3
 3. Check the subscripts by multiplying (charge X subscript) to make
sure the charges are balanced.
Al23+ O32- = +6, -6
 4. If necessary, divide the subscripts by the largest common
denominator to get the smallest whole number ratio.
 1. Write the symbols side by side.
Naming Binary Ionic Compounds
 Nomenclature- a system of naming.
 The ratio of ions is not indicated in the name…it is
assumed you will know (or be able to figure it out).
 Remember… cations come first in ionic
compounds and naming is the same as for
monatomic ions.
 The name of the cation is the name of the element.
 The name of the anion is found by dropping the ending
and adding –ide.
Try this- CuBr2
The STOCK SYSTEM of Naming
 Some elements can form cations with two or more
different charges and these must be distinguished.
 There is no element that commonly forms more than
one monatomic anion.
 Name of the cation + Roman numeral indicating
charge
 Examples: Iron (II) , Iron (III).
 Then just put the name of the anion.
 FeCl2  Iron (II) chloride
 Fe2O3  Iron (III) oxide
Try this - FeS
Compounds containing Polyatomic Ions
 The only common polyatomic cation is ammonium - NH4+
 Most anions CONTAIN OXYGEN, making them OXYANIONS.
 In some cases more than one oxyanion can form from the same two
elements. Ex: ClO- , ClO2- , ClO3- , ClO4-
 The most common form of the anion is given the ending –ate.
 In this case ClO3chlorate
 The form of the anion with ONE LESS oxygen is given the ending –ite.
 ClO2- = chlorite
 The form of the anion with ONE LESS oxygen than the –ite is given the
PREFIX hypo–
 ClO- = hypochlorite
 The form of the anion with ONE MORE oxygen than the –ate is given
the PREFIX per–
 ClO4-
= perchlorate
Naming Compounds containing
Polyatomic ions
 The naming is the same as naming binary compounds.
 Cation followed by polyatomic anion.
 AgNO2
silver nitrite
 AgNO3
silver nitrate
See Table 7-2 on page 210
 Try this :
 K2SO4  ?
 Ca(ClO3)2  ?
NAMING BINARY MOLECULAR COMPOUNDS
 Two different nomenclature systems are used.
 1. Stock System - System that is used now.
 2. The prefix system uses prefixes.
 See table 7-3 pg 212 for a list of prefixes.
PREFIXES
 The LESS electronegative element is given first, and is given a
prefix only if there is more than one atom of that element.
 The second element is named by combining the prefix, the root
of the element and the ending –ide.
 Ex: CCl4 = carbon tetrachloride
 Try these: N2O3 = ?
 As2O5 = ?
ACIDS and SALTS
 In the classical naming system, acids are named according to their anions.
 BINARY ACIDS usually contain hydrogen and a halogen
 To name binary acids the anion ending is dropped and replaced with a new suffix
(and sometimes prefix).
 For example, HCl has chloride as its anion, so the -ide suffix is dropped and
is replaced with -ic … taking the form hydrochloric acid.
 OXYACIDS – acids that contain Hydrogen, Oxygen, and another (usually
nonmetal) element.
 To name oxyacids, you must first be able to recognize them by the general
formula HaXbOc, with X representing an element other than hydrogen or
oxygen.
 In some cases, adding H ions to polyatomic ions results in oxyacids.
 SO4 = sulfate the “ate” is dropped and “ic” is added to the root name* when H+ ions are .
 Ex: H2SO4 = sulfuric acid
 SO3 = sulfite the “ite” is dropped and “ous” is added
 H2SO3 = sulfurous acid
 More fun with acids later
 See table 7-4 on page 214 for some common acids
Home Fun
 Pg 215
q 2-4
7-2
OXIDATION NUMBERS
 Each atom in a compound has an apparent
charge. This apparent charge, called the oxidation
number, represents the charge that an atom would
have if electrons were transferred completely to the
atom with the greater attraction for them in a given
situation.
 These oxidation numbers can be used to predict the
ratio by which atoms will combine when they form
compounds.
 The oxidation number indicates the general
distribution of electrons among the bonded atoms in
molecular compounds or a polyatomic ion.
Rules for assigning Oxidation Numbers
 1. Pure elements have oxidation numbers of 0.
 2. The more electronegative element in a binary molecular compound is
assigned the number equal to the charge of the anion it would form.
 3. Fluorine always has an oxidation number of -1 when in a compound.
 4. Oxygen has an oxidation number of -2 in most compounds.
 Exceptions include peroxides (-1) and compounds in which oxygen forms compounds
with halogens (+2).
 5. Hydrogen has an oxidation number +1 in all compounds containing
elements that are more electronegative, and -1 when combined with metals.
 6. The algebraic sum of the oxidation numbers of all atoms in a neutral
compound is equal to zero.
 7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic
ion is equal to the charge on the ion.
 8. Oxidation numbers can be assigned to atoms in ionic compounds
 See page 216
Practice- Assign oxidation numbers to each
atom in the following compounds or ions:
 HF
 H = +1, F = -1
 H2SO4
 H = +1, O = -2,
 4O x -2 = -8
 2H x +1 = +2
 -8+2 = +6, so sulfur has
an oxidation number of
+6, since the sum of all the
oxidation numbers must
equal zero.
Using Oxidation Numbers for Formulas
and Names
 use oxidation numbers in the stock system
 When naming a binary compound, regardless of whether it is ionic or




molecular, follow the following steps:
1. Write the name of the element represented by the first symbol in the formula.
2. Write the name of the element represented by the second symbol in the
formula, but change the ending of the element's name to "ide".
3. Check a reference table to determine the number of positive oxidation
numbers that the first element forms. If it only forms one then you are done.
4. If the first element shows more than one oxidation number, then use the stock
system. Determine the oxidation number that the first element is showing and
write that roman numeral in-between the two elemental names.
Home fun
 – pg 219 q 1,2
 – pg 236 q 27-29
 Do them now.
Ch7-3
USING CHEMICAL
FORMULAS
Finding formula mass, molar mass
and percent composition.
FORMULA MASSES
 The sum of the average atomic masses of all the
atoms represented in the formula of any
molecule, formula unit or ion.
 Finding the formula mass for water, H2O:
 2H atoms x 1.01 = 2.02 amu
 1O atom x 16.00 = 16.00 amu

18.02 amu
 Try these: (NH4)2CO3
96 amu

C9H8O4
180.16 amu

Fe2(SO4)3
399.88 amu
Molar Masses
 The mass in grams of one
mole of a substance
 Same as a formula mass,
just change the units.
 Try these:
 PO43- =
94.97g /mol
 Ba(OH)2 =171.35g /mol
 PO43 Ba(OH)2
 MgCl2
 MgCl2
= 95.21g /mol
MOLAR MASS as a CONVERSION FACTOR
 Amount in moles X molar mass = mass in grams
 Mass in grams /molar mass = number of moles
 Number of moles X Avagadro’s = number of molecules
 Number of molecules/ Avagadro’s = number of moles
Try this:
How many moles are in 6.60g of ammonium sulfate?
How many molecules of ammonium sulfate are in 1.25 moles?
How many atoms are found in 1.25 moles of ammonium sulfate?
Percentage Composition- Finding the percentage,
by mass, of a particular element in a compound
 % composition = mass of the ELEMENT (or compound) in 1 mole of a substance
mass of 1 mole of the compound
 Try these:
 Pb in PbCl2
 Cl in PbCl2

 Water in ZnSO4 • 7H2O
 Ba in Ba(NO3)2
 N in Ba(NO3)2
 O in Ba(NO3)2


Pb = 74.51%
Cl = 25.49%
H2O = 43.86%
Ba = 52.55%
N = 10.72%
O =36.73%
X 100
Home fun
 Pg 228 q 1-6
 Pg 236 30-35
Chapter 7 Section 4
Determining Chemical Formulas
 Empirical formulas – the symbols for the elements




combined in a compound with subscripts showing the
smallest whole number mole ratio of the elements in the
compound.
In molecular compounds an empirical formula makes no
reference to isomerisms, structure, or absolute number of
atoms.
Example – Glucose molecular formula is C6H12O6
the empirical formula for glucose is CH2O
For ionic componds the formula units are usually the
empirical formulas.
Calculating Empirical Formulas
 Given % composition – convert to mass by assuming 100.0g
sample of the compound.
 Ex: If you are given the following percentages in a compound:
B = 78.1%
H =21.9%
*Assume the mass of the sample is 100.0g and convert to grams
B = 78.1g
H = 21.9g
*Determine the number of moles in each using conversion
factors: 78.1gB/(10.81g/mol) = 7.22 moles B
21.9gH / (1.01g/mol) = 21.7 moles H
*Divide each number of moles by the smallest number of
moles to find the mole ratio:
7.22 mol B/7.22 = 1 mol B
21.7 mol H/7.22 = 3.01
*Put it together in an empirical formula: BH3
Try this:
 32.38% Na, 22.65% S, 44.99% O
Step 1: convert to grams
Step 2: Find the # of moles of each
element.
Step 3: Find the mole ratio.
Step 4: Determine the empirical
formula.
 32.38 g Na
 22.65g S
 44.99g O
 1.408 mol Na
 0.7063 mol S
 2.812 mol O
 1.993 mol Na: 1mol S:
3.981 mol O
 Na2SO4
A 10.150g sample of a substance known to contain only
phosphorous and oxygen contains 4.433g of phosphorous.
What is the empirical formula of the substance?
 Since we already know the




masses of the elements in the
compound we can skip step 1.
Find the mass of the oxygen.
Find the number of moles of
each.
Find the smallest whole number
ratio.
Since this is not a whole number
ratio we multiply each side by 2.
If that doesn’t work try 3.
 5.717g O
 0.1431 mol P
 0.3573 mol O
 1 mol P : 2.5 mol O
 2 mol P : 5 mol O
 P2O5
Practice
 1. Calculate the empirical formula
of the compound that contains 1.0
g S for each 1.5 g O.
 2. Calculate the empirical formula
of the compound containing 75.0%
C and 25.0% H.
 3. Calculate the empirical formula
of the compound containing 81.8%
C and 18.2% H.
 4. Determine the empirical
formula of the compound
containing 37.5% C, 12.5% H, and
50.0% O by weight.
 1. SO3
 2. CH4
 3. C3H8
 4. CH4O
Home Fun
 pg. 233 q 1-3
 pg 237 q. 36-37
Calculation of Molecular Formulas
 Molecular formulas- the actual formula of a molecular compound. Ex:
C6H12O6 (hexose), C5H10O5 (pentose)are different molecular
compounds that have the same empirical formula, CH2O.
 x(empirical formula mass) = molecular formula mass
 x is a whole number multiple by which the subscripts in the empirical
formula must be multiplied.
 If we know the empirical formula of a compound, all we need to do is
divide the molecular mass of the compound by the mass of the empirical
formula. (x).
 Multiply the empirical formula by this number
 It is also possible to do this with one of the elements in the formula;
simply divide the mass of that element in one mole of compound by the
mass of that element in the empirical formula.
 Example: if we know that the empirical formula of a
compound is HCN and we are told that a 2.016 grams of
hydrogen are necessary to make the compound, what is the
molecular formula?
 In the empirical formula hydrogen weighs 1.008
grams. Dividing 2.016 by 1.008 we see that the amount of
hydrogen needed is twice as much. Therefore the empirical
formula needs to be increased by a factor of two (2). The
answer is:
H 2C 2N 2.
PRACTICE
 1. Empirical formula C3H7,
molecular weight 86 g/mole
 2. Empirical formula S,
molecular weight 256 g/mole.
 3. Empirical formula CH,
molecular weight 26 g/mole
 4. Empirical formula NO2,
molecular weight 46 g/mole
 1. C6H14
 2. S8
 3. C2H2
 4. NO2
HOME FUN
 Pg. 233 q 4, 5
 Pg. 237 q 38, 39