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Chemical Formulas
CHEMICAL FORMULAS:
CATIONS – POSITIVE IONS
ANIONS- NEGATIVE IONS
Cations are always written 1st
Anions are always written 2nd
Ions can be a single element of a
combination of several elements
Na+ is a cation as is NH4+
Cl- is anion as is SO4-2
SO4-2 is the sulfate ion. It is poly atomic.
You do not
have right to change the charge or the number of atoms making it
up. There is one sulfur atom and 4 oxygen atoms in a sulfate ion.
You do get to chose how many total ions can be in a chemical
formula. You may have more that one present and this is
represented by placing parenthesis around the polyatomic ion and
using a subscript to represent the number of Ions used.
An aluminum cation has a plus 3
Example:
Al2(SO4)3
charge and is represented by Al+3. The Sulfate anion is represented
by SO4-2.
A correct chemical formula must have an equal number of positive
and negative charges resulting in a neutral chemical structure.
In the formula Al2(SO4)3 there are two positive ions each with a plus
three charge resulting in positive charge with a value of 6. There is
also three negative ions each with a value of negative 2, this results
in a total negative charge of minus 6. When these ions are joined
together the net result is zero charge.
When any chemical formula is correctly written the net charge will
be zero.
Another way to think about this is to let the Cations be X and the Anions
be Y. The chemical formula would then be represented by XY. XY will
have zero charge, but will be composed of a positive charge and a
negative charge.
If
X+3 = Al+3
and Y-2 = SO4-2
then my formula is
X2Y3 or Al2(SO4)3
The correctly written formula will only have the subscripts, no charge
values will be shown. Parenthesis are only present when more than one
polyatomic ion needs to represented, and will always be followed by a
subscript. Again you cannot change the value of the subscript of a
polyatomic ion, you can change the number of ions.
Magnesium permanganate is composed of a Magnesium cation (Mg++)
The X value then is +2 while
and the permanganate anion (MnO4-).
the Y value is -1, therefore it will be necessary to have 2 units of Y for
each unit of X to obtain the required neutral value of the formula ---XY2 would be the form that the formula will take, therefore XY2 =
Mg(MnO4)2
The following would be incorrect:
MgMnO6
MgMn O
MgMn O
(Mg)1(MnO4)2 - Mg(MnO6) -
Mg+2(MnO )
-2
Naming Acids
When the anion does NOT contain Oxygen:
Use the prefix hydro + root of the anion’s name – ic
+ the word acid
Examples: HCl - hydrochloric acid; HBrhydrobromic acid
When the anion contains Oxygen:
The name will depend on the name of the
polyatomic anion. DO NOT use the prefix hydro.
Examples: H2SO4 the anion is sulfate, therefore the
acid name will end in ic – Sulfuric acid. H2SO3 the
anion is sulfite, therefore the name of the acid will
end in ous – sulfurous acid.
ATE → IC
ITE → OUS
LEARN THESE
PREFIXES WHICH
ARE USED TO NAME
BINARY MOLECULAR
COMPOUNDS
MATH OF CHEMICAL FORMULAS
A chemical formula that is correctly written yields a large amount of information
about the substance.
H2SO4 Hydrogen Sulfate
This is sulfuric acid when in an aqueous solution
one molecule of has: 2 atoms of hydrogen, 1 atom of sulfur, and 4 atoms of oxygen
The formula mass can be determined by adding up the masses of each atom - this information
Is obtained from the atomic chart.
Hydrogen = 1 amu (atomic mass unit)
Sulfur = 32 amu
Oxygen = 16 amu
Therefore
(2x1)+32+4(16) = 98 amu
The mole concept allows us to convert this information in to usable mass units
One mole of sulfuric acid has a mass in grams equal to the formula mass.
One mole of sulfuric acid has a mass of 98 grams
For the molar mass of any substance, find the formula mass in amu’s and
change the unit to grams.
H2SO4 has 2 atoms of Hydrogen
1 atom of Sulfur
4 atoms of Oxygen
One mole of H2SO4 has : 2 moles of hydrogen atoms
1 mole of sulfur atoms
4 moles of oxygen atoms
The percent composition of a substance is found by dividing the formula mass of the
substance by the atomic mass of the atoms that make it up.
For the % composition of Hydrogen
The formula mass of H2SO4 = 98 u
divide the mass of the hydrogen (2 u) by the formula mass (98u) times 100 %
2u/98u x 100% =2.04 %
For the % composition of Sulfur divide the mass of sulfur by the formula mass times 100%
32u/98u x 100% =32.6%
For the % composition of Oxygen divide the mass of oxygen by the formula mass times 100%
4(16u)/98u x 100% =65.3 %
When you know the percent composition it is then possible to determine the amount of
individual elements in a given mass of the substance.
For Example: if you had a 1000 grams of sulfuric acid what would be the mass of each of
the elements present.
Simply multiply 1000 grams times the % percent composition.
For Hydrogen
1000g x 0.0204=20.4 g of hydrogen per 1000 g H2SO4
For Sulfur
1000g x 0.326 = 326 g of sulfur per 1000g H2SO4
For Oxygen
1000g x 0.653 = 653 g of sulfur per 1000g H2SO4
From this mass of H2SO4 it is possible to determine the number of moles present
and the number of atoms. We determined that 1 mole of sulfuric acid was 98 grams
If we divide 1000g by 98 g/mole
1000g/98g/mole = 10.20 moles
When we know the number of moles we can find the number of molecules and number of atoms
10.2 moles of sulfuric acid will have (10.20 x 6.022 x 1023 =) 6.03 x 1024 molecules of acid
We can also determine the number of atoms present by multiplying the above value by the number
of atoms present in the formula. Hydrogen would be twice this value, sulfur would be equal to this
value, and oxygen would be four times it.
CALCULATION OF EMPIRICAL 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.
This is done by finding the number of moles for each element in the compound. When this has been
determined you can divide each value by the smallest value and then round to the nearest whole
number.
To make this determination you will first have to know either the percentage of each element present
or the mass of the sample compound and the mass of its elements. If you are given the percentages you
can assume a mass of 100 grams. This will allow you to compute the molar masses in this imaginary
sample. Each element would have a mass in grams equal to its percentage.
I
Sample one from your text:
Quantitative analysis shows that a compound contains 32.88% sodium, 22.65% sulfur, and
44.99% oxygen.
Your plan to solve this should be :
Percent composition  mass composition  composition in moles  smallest whole-number
ratio.
If we have a 100.0 g sample of this substance we will have:
32.38 g Na, 22.65 g S, 44.99 g O
Composition in moles :
32.38 22.65 44.99 . .
. =1.408 mol Na
= 0.7063 mol S
2.812
Composition in moles :
32.38 22.65 44.99 . .
. =1.408 mol Na
= 0.7063 mol S
2.812
Now divide each by the smallest value to determine the simplest ratio
.
.
:
.
.
:
.
.
= 1.993 mol Na: 1.0 mol S: 3.981 mol O
If you round these to nearest whole number it will give a 2:1:4 ratio
Therefore the empirical formula will be Na2SO4.
Sample problem M page 247 in your text
A sample of a compound known to contain only phosphorus and oxygen has a mass of
10.150 grams and the phosphorus content is 4.433 g Determine the empirical formula
Given: sample mass = 10.150 grams
mass of phosphorus = 4.433g
unknown: empirical formula
PLAN: mass composition  composition in moles  smallest whole-number ratio of atoms
Find mole composition:
10.150 4.433 5.717 mass composition 4.433g P, 5.717g O
.
.
.
.
.
/
/
= 0.3573 mol O
Next determine the simplest whole-number mole ratio
.
.
:
.
.
1 mol P:2.497 mol O
Since the molar value for Oxygen is not close to a whole number we will increase the ration
by factor of 2 and see if we have a workable set of values
2(1 mol P:2.497 mol O) = 2 mol P: 5 mol O
This gives an empirical formula of
P2O5
Checking the oxidation numbers for Phosphorus and Oxygen will show that this is a reasonable
Formula.
Calculation of Molecular Formulas
The relationship between a compound’s empirical formula and its molecular formula
can be written as follows.
x(empirical formula)=molecular formula
The number represented by x is a whole-number multiple indicating
the factor by which the subscripts in the empirical formula must
be multiplied to obtain the molecular formula.
The formula masses have a similar relationship.
x(empirical formula mass) = molecular formula mass
To determine the molecular formula of a compound, you must know the
compounds formula mass.
For diborane - empirical formula BH3 experimentation shows that the formula
mass of diborane is 27.67 amu. The formula mass for the empirical formula is
13.84 amu.
A rewrite of the formula mass-molecular formula mass relationship
x(empirical formula mass) = molecular formula mass
yields.
molecular formula mass
X=
empirical formula mass
27.67 amu
=2.000
13.87 amu
Therefore the molecular formula for diborane is B2H6
2(BH3)=B2H6
X=
Determine the molecular formula for a compound with the empirical formula of CH2.
The experimental molar mass is 28g/mol.
Empirical molar mass= 12+2(1)=14 amu
X=
molecular molar mass
=
empirical molar mass
2(CH2)= C2H4
/
/
=2.00
C2H4 is the molecular formula