Download Chapter 7 – Chemical Formulas and Chemical

Chapter 7 – Chemical Formulas and Chemical Compounds
7.1 Chemical Names and Formulas
Before we start talking about chemical formulas we need to talk about
polyatomic ions. These are groups of atoms bonded together covalently.
What makes them different is that they carry a charge as a group.
Ex: OH-, SO4-2, NH4+
These polyatomic atoms like many other formulas have common names that
do not state the chemical composition. The common name for dihydrogen
monoxide is water and sodium chloride is table salt. The most important
thing about chemical formulas is to state the composition.
The chemical formula of an ionic compound can tell the number of ions in
the formula unit.
Ex: NaCl tells us that there is one cation of Na and one anion of Cl to make
up table salt, making sodium chloride.
NH4OH tells is that there is one ion of ammonium and one ion of
hydroxide, making ammonium hydroxide.
When a single atom forms an ion it is known as a monatomic ion. The
periodic table can be set as a guide for ion formation.
G1- +1, G2- +2, G13- +3, G15- -3, G16- -2, G17- -1
Table 1 from pg 221 shows the monatomic ions for the transition metals.
Naming Monatomic Ions
To name cation just use the name. K+ is potassium, Na+ is sodium
To name anions we will replace the -ine eding with the –ide ending.
Chlorine is chloride, Fluorine is fluoride.
Binary Ionic Compounds
A compound consisting of 2 elements is known as a binary ionic compound.
The total number of positive charges equal the number of negative
charges.
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To write a binary ionic formula:
1. Write the two ions with the charge side by side with the cation listed first.
2. Cross the charges so that the ion charges are neutral.
3. Name the compound.
Ex: Na+ + F-  NaF
Ba+2 + Cl-  BaCl2
sodium fluoride
barium chloride
For compounds containing transition metals the charge will be expressed in
its name. The charges will be represented by roman numerals in parentheses.
We will follow the same rules for naming these compounds.
Ex: Fe+3 + Cl-  FeCl3
Copper (II) oxide
Aluminum (III) sulfide
iron (III) chloride
Cu+2 + O-2  CuO
Al+3 + S-2  Al2S3
Compounds Containing Polyatomic Ions
Most polyatomics are oxyanions, polyatomic ions that contain oxygen.
For naming oxyanions the compound with the most oxygens ends with –ate,
next highest, -ite, and the fewest has the prefix hypo-. An ion that has one
more oxygen than the –ate uses a prefix of per-.
ClOClO2ClO3ClO4-
hypochlorite
chlorite
chlorate
perchlorate
To name these compounds name the cation first then name the anion. Table
2 on pg 226 gives common polyatomics and their names.
Ex: Mg+2 + ClO2-  Mg(ClO2)2 magnesium chlorite
Ag+2 + NO3-  Ag(NO3)2 silver (II) nitrate
Naming Binary Molecular Compounds
The old system of naming molecular compounds is based on using prefixes.
These prefixes are on pg 228. The prefix mono- means one, di- means two,
tri- means three, and etc.
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Rules for naming molecular binary compounds:
1. The element with the smallest group number is listed first. If both are
in the same group the element from the greatest period is listed first.
This element is given a prefix only if it is greater than one.
2. The second element is named with the appropriate prefix and with a
new ending of –ide.
3. The o or a at the end of the prefix is usually dropped with added to the
root word.
Acids and Salts
Acids are another type of binary compound. There are binary acids,
those that contain H and usually a halogen, and oxyacids, those that
contain H, O, and another element.
Table 5 on pg 230 states some common acids.
Salts are composed of cations and anions from an acid. NaCl forms from
the anion from hydrochloric acid. Salts are also any compound formed
from the halogens.
7.2 Oxidation Numbers
The electrons around an atom illustrate the electron distribution for that
element. The oxidation numbers give the general distribution of electrons
among bonded atoms in a compound.
Assigning Oxidation Numbers:
General Rule: shared electrons belong to the most electronegative element in
each bond.
Atoms of pure element have an oxidation number of zero. Ex: O2, P4,
H2
2. The most electronegative element in a binary molecular compound is
assigned the number equal to its negative charge of the anion. The
less electronegative element is given the number of the positive
charge equal to its cation.
3. F has an oxidation number of -1 in all compounds because it is the
most electronegative element.
1.
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4. O has an oxidation number of -2 in almost all compounds. Exceptions
are in H2O2 where is carries a -1 and compounds with F and it is +2.
5. H has an oxidation number of +1 with elements with higher
electronegativities than it and -1 with all 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 a polyatomic ion is
equal to the charge of that ion.
8. Rules 1-7 apply to covalent compounds, but oxidation numbers can be
assigned to ionic compounds.
Examples: for monatomic ions the oxidation number is equal to the
ionic charge. Ca+2 = +2 or I- = -1
Compound HF: F is the most electronegative element in the
polar covalent compound. It will take the number of -1. H is bonded to a
more electronegative than it, so it will have a number of +1. These
charges make the neutral atom equal to zero. Also the electrons should
go to the F because of its electronegativites, and it is a – so this is correct.
Compound H2O: The most electronegative element is O. It
has a number of -2, which means it takes charge of the electrons. The H
will have a number of +1 because of the difference in electronegativity.
Sample Problem E
Assign oxidation numbers to each atom:
UF6: F has a number of -1, they make a total charge of -6 so U must
equal a number to make the sum of the numbers zero. U will have a
number of +6.
H2SO4: O will have a number of -2, all together it has a charge of -8. H
has a number of +1 because of electronegativities. This makes a number
of +2. If the sum of the compound must be zero, S will have a charge of
-6.
ClO3-: O has a number of -2, so it will equal -6. The sum of the ion
must equal -1 because of the charge it carries. So Cl will equal +5.
Roman numeral can also be used instead of the prefix system to show the
charge.
PCl3 phosphorous trichloride
phosphorous(III) chloride
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7.3 Using Chemical Formulas
The chemical formula can be used to tell us the atoms present and the
number of each atom that is present in a compound.
Earlier we used the formula to calculate the mass of a particular compound.
For this, we use the mass of each atom present and then add each of them
together to get a total mass as a molecule or the molecular mass.
Ex: H2O H: 1.008 X 2 = 2.016
O: 15.99 X 1 = 15.99 total: 16.02 amu
NaCl Na: 23.0 amu
Cl: 35.45 amu
total: 58.45 amu
The formula mass of any molecule, formula unit, or ion is the sum of the
average atomic masses of all atoms represented in its formula.
Sample Problem F
Find the formula mass of potassium chlorate (KClO3).
K: 39.10 amu
Cl: 35.45 amu
O: 15.99 amu X 3= 48.00 amu
Formula mass: 122.55 amu
Molar Mass
In chaper 3 we learned that each of the elements on the periodic table could
be measured per mole. Such as 1 mole of H is 1.008 g/mol. Just as the
elements can be measured per mole molecules can as well. 1 mole of water
has 2 H and 1 O present. So H2O is 1 mole of water, weighing 18.02 g/mol.
The molar mass of a compound will be numerically equal to the formula
mass. From sample problem F the molar mass of KClO3 is 122.55 g/mol.
Sample Problem G
One mole of barium nitrate contains exactly one mole of Ba2+ and 2
moles of NO3- ions. The two moles of NO3- contains 2 moles of N atoms
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and 6 mole of O atoms. Therefore, the molar mass of Ba(NO3)2 is
calculated as follows.
Ba: 137.33 g/mol X 1 mole = 137.33 g
N: 14.01 g/mol X 2 moles = 28.02 g
0: 15.99 g/mol X 6 moles= 96.00 g
total molar mass: 261.35 g/mol
Molar mass as a Conversion Factor
This can be used to relate the mass of a compound to the grams of a given
substance.
Amt of moles X molar mass = mass in grams
Sample Problem H
What is the mass in grams of 2.50 mol of O2 gas?
2.0 moles X 16.00 g/mol= 32.00 g in the gas
2.50 moles of 32.00 g of O2 gas = 80.0 g O2 gas
Sample Problem I
Ibuprofen, C13H18O2, is the active ingredient in many nonprescription
pain relievers. It molar mass is 206.31 g/mol. A) If the tablets in a
bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen
are in the bottle? B) How many molecules of ibuprofen are in the bottle?
C) What is the total mass in grams of carbon in 33 g on ibuprofen?
A) grams to moles: 33 g C13H18O2 X
1 mol
= 0.16 mol
206.31g C13 H18O2
C13H18O2
 6.022 X 1023 molecules 
22
  9.6 X 10 molecules C13H18O2
1 mol


B)  0.16 mol C13 H18O2  
C) 13 moles of C present:

  12.01 g C 
13 mol

  25 g C
 mol C13 H18O2   mol C 
 0.16 mol C13 H18O2  
Percent Composition
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The percent composition of a compound could be helpful to know if there
was ever a case that we needed to find out how much of one element was
present in a certain compound.
To calculate this we must use:
mass of element in a sample compound
x100  % element in cmpd
mass of sample of compound
because the mass percent is the same no matter the sample size we can
also use:
mass of element in1mol of cmpd
X 100  % element in cmpd
molar mass of cmpd
The mass percent of each element in a compound is the percent
composition of the compound.
Sample Problem J
Find the percent composition of copper (I) sulfide, Cu2S.
There are 2 moles of Cu and 1 mole of S present.
2 mole Cu X 63.55 g Cu/mol Cu = 127.1 g Cu
1 mole C X 32.07 g S/mol S = 32.07 g S
molar mass= 159.2 g Cu2S
127.1g Cu
32.07 g S
x100  79.85% Cu
x100  20.15% S
159.2 g Cu2 S
159.2 g Cu2 S
Sample Problem K
As some salts crystallize from a water solution, they nond water
molecules in their crystal structure. Sodium carbonate forms such a
hydrate, in which 10 water molecules are present for every formula
unit of sodium carbonate. Find the mass percent of water in sodium
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carbonate decahydrate, Na2CO3•10 H2O, which has a molar mass of
286.19 g/mol.
There are 10 moles of water present:
 18.02 g H 2O 
10 mol H 2O 
  180.2 g H 2O
mol
H
O

2

180.2g H 2O
x 100  62.97% H 2O
286.19 g Na2CO3 10 H 2O
7.4Determining Chemical Formulas
An empirical formula consists of the symbols for the elements combined in
a compound, with subscripts showing the smallest whole-number mole ratio
of the different atoms in the compound.
To calculate we will need to convert the percent composition to the mass
composition. If we assume that the sample contains 100 g then the percent
can be used as a mass also.
Ex: 78.1% B can be 78.1 g B and 21.9% H can be 21.9 g H.
We can then go from the amount of grams to the amount of moles by using
the molar mass of an element. If we calculate this for B and H we get:
 1 mol B 
78.1 g B 
  7.22 mol B
 10.81 g B 
 1 mol H
21.9 g H 
 1.008 g H

  21.7 mol H

This would give us a mole ratio of 7.22:21.7. Is this the smallest possible
ratio? No. To find this we till divide the ratios by the smallest individual
ratio, in this case it is 7.22.
7.22 mol B 21.7 mol H
:
1 mol B :3.01 mol H
7.22
7.22
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We will use the whole number forms of the ratios, so the ratio tells us that in
the simplest form of the compound formed between B and H that there will
be 1 mol of B present and 3 moles of H present.
Formula: BH3
Sample Problem L
Quantitative analysis shows that a compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this
compound.
32.38% Na = 32.38 g Na
22.65% S = 22.65 g S
44.99 % O = 44.99 g O
 1 mol Na 
32.38 g Na 
  1.408 mol Na
 22.99 g Na 
 1 mol S 
22.65 g S 
  0.7063 mol S
 32.07 g S 
 1 mol O 
44.99 g O 
  2.812 mol O
15.99
g
O


The smallest ratio is 0.7063 mol S, so each ratio will be divided by this
amount.
1.408 mol Na 0.7063 mol S 2.812 mol O
:
:
0.7063
0.7063
0.7063 = 1.993 mol Na: 1 mol S: 3.981 mol O
if we take these to the whole number then we get, 2mol Na:1mol S:4 mol O
Empirical Formula: Na2SO4
If the mass is already given in the question then a step has been completed
for you. Go directly into solving for moles of each element.
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Calculating Molecular Formulas
The empirical formula tells us the simplest possible ratio of element in a
compound, but the molecular formula will state the actual formula of a
compound.
The empirical formula plays a part in the calculations for the molecular
formula. We must know the mass of the empirical and experimental masses
to solve.
The relationship between the two can be seen by:
x(empirical formula mass) = molecular formula mass
x is a whole number multiple that indicates the factor that must be used to
multiply to get the molecular mass. To obtain this whole number factor we
must use the experimental mass/ empirical mass. Then multiply the
empirical formula by this number.
Sample Problem N
In sample Problem M, the empirical formulas of a compound of
phosphorous and oxygen was fond to be P2O5. Experimentally the
molar mass of this compound is 283.39 g/mol. What is the compounds
molecular formula?
Experimental mass: 283.39 g/mol
Empirical mass: 141.94 g/mol
283.39 g / mol
2.0001 2(P2O5)= P4O10
141.94 g / mol
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