Chemical formulas are not simply abbreviations for words. They represent
precise quantities.
•Elements
What 2 “bits” of information do formulas give?
•Number of atoms
of each element
Water for example is H2O.
It indicates that 1 molecule of water consists of 2 hydrogen atoms and 1
oxygen atom.
How could you represent 2 molecules of water?
Use a coefficient
2 H 2O
This formula shows 4 hydrogen atoms and 2 oxygen atoms.
In daily life, pieces of matter are often measured either by
counting them or by massing them; the choice is determined by
convenience. If you buy eggs, you buy them by the dozen—that
is, by number. Eggs are easy to count out. So are oranges and
lemons.
Other items, though countable, are more conveniently sold by
mass. A dozen peanuts is too small a number to buy, and several
hundred are too difficult to count. You buy peanuts by the pound
or kilogram—that is, by mass.
Chemists, too, are interested in quantity—the quantity of an
element or compound, which, like grocery items, can be measured
by number or by mass. Although you can easily mass a sample
of a substance, yet the number of atoms or molecules in it is
much too large to count. Nevertheless, chemists are interested
in knowing such numbers.
In the laboratory, when we “run” a reaction we want to know the number of
atoms, molecules, or formula units in a substance because these are the
entities that react with each other. However, these entities are much too
small to count individually, so chemists use a unit called the mole to count
by massing them.
The mole was derived from the Latin word moles, meaning “heap” or
“pile.” The mole, whose symbol, mol, is the SI base unit for measuring the
amount of substance.
A mole is the amount of substance that is equal to the number of carbon
atoms in exactly 12 g of the carbon-12 isotope.
One mole always contains the same number of particles, no matter
what the substance.
1 mole = 6.022 x 1023 particles
This value is referred to as Avogadro’s number (Amadeo Avogadro) in
honor of the man who conceived the basic idea.
One mole each of various substances - Clockwise from top
left: 1-octanol (C8H17OH); mercury(II) iodide (HgI2); methanol
(CH3OH); sulfur (S8).
The central relationship between the mass of one atom and
the mass of one mole of atoms is that the mass of an element
is expressed in amu.
The mass of one mole of atoms of an element is the same
numeric value but measured in grams.
1 atom of sulfur has a mass of 32.07 amu.
1 mole of sulfur atoms has a mass of 32.07 g.
So,
1 mole S
= 6.022 x 1023 atoms S
= 32.07 g S
1 mole of Fe2+
= 6.022 x 1023 ions
= 55.85 g Fe2+
1 mole H2O
= 6.022 x 1023 molecules H2
= 18.02 g H2O
1 mole NaCl
= 6.022 x 1023 formula units NaCl
= 58.44 g NaCl
How can I get the mass of one unit of a
compound?
•Add up the individual masses of the elements.
•If we are talking about the mass of an atom we call it the atomic mass.
•If a compound is a molecule, we call this the molecular mass.
•If a compound is a formula unit, we call it the formula mass.
g-formula mass = 1 MOLE = 6.022 x 1023
“particles”
1 mole
of a substance
elements
= Avogadro’s constant
(6.022 x 1023)
charged
particles
=
covalent
compounds
mass of substance
in grams
ionic
compounds
A 1-carat diamond has a mass of 0.200 g.
How many moles of carbon?
How many atoms of carbon?
HINT: If moles of your substance
does not appear in your given or in
the desired units, then you must do
at least 2 conversion factors.
USE FLM!!!!!!!
Start with your given, identify units of answer, and write
conversion factors.
0.200 gg C
0.200
C
1 mole C
12.01 g C
0.200
0.200 gg CC
==
11 mole
mole CC
6.022 x 1023 atoms C
12.01
12.01 gg CC
1 mole C
0.0167 moles C
atoms
= x 1022 atoms
==1.00
atoms C
CC
A ring is constructed out of 3.06 x 1022 atoms Au.
How many grams?
How many moles?
3.06 x 1022 atoms Au
1 mole Au
6.022 x
3.06 x 1022 atoms Au
1023 atoms
== 0.0508 moles
molesAu
Au
Au
1 mole Au
196.97 grams Au
23 atoms Au
6.022 x 1023
1 mole Au
=
grams
gramsAu
Au
Au
10.0 grams
A raindrop contains
about 0.050 g of water.
How many molecules of water?
How many moles of O?
How many atoms of H?
How many molecules of water?
0.050 grams H2O
1 mole H2O
6.022 x 1023 molecules H2O
18.02 grams H2O
1 mole H2O
=
= 1.7 x 1021 molecules H2O
How many moles of O?
0.050 grams H2O
1 mole H2O
1 mole O
18.02 grams H2O
1 mole H2O
=
= 0.0028 moles O
How many atoms of H?
0.050 grams H2O
0.050 grams H2O
1 mole H2O
2 mole H
6.022 x 1023 atoms H
18.02 grams H2O
1 mole H2O
1 mole H
1 mole H2O
6.022 x 1023
molecules H2O
2 atoms H
18.02 grams H2O
1 mole H2O
1 molecule H2O
= 3.3 x 1021 atoms H
=
=
One teaspoon of table sugar (sucrose) is
approximately 15 g of C12H22O11.
How many moles of sugar is this?
15 g C12H22O11
1 mole C12H22O11
342.34 g C12H22O11
=
How many atoms of carbon are in this 15 g sample?
15 g C12H22O11
1 mole C12H22O11
6.022 x 1023
molecules C12H22O11
12 atoms C
342.34 g C12H22O11
1 mole C12H22O11
1 molecule
C12H22O11
What mass of CaCl2 would be needed to furnish 1.00
mole of Cl1- ions?
1.00 mole Cl1- ions
1 mole CaCl2
2 moles
Cl1- ions
110.98 g CaCl2
1 mole CaCl2
=
Compare the mass of each element present in one mole of a
compound to the total mass of one mole of a compound.
EXAMPLE: Find the % composition of Al2(SO4)3
Al
S
O
= 2 x 26.98 = 53.96
= 3 x 32.07 = 96.21
= 12 x 16.00 = 192.00
342.17
EXAMPLE: Find the % composition of Al2(SO4)3
Al = 2 x 26.98 = 53.96
S
= 3 x 32.07 = 96.21
O
= 12 x 16.00 = 192.00
% Al =
%S=
%O=
342.17
53.96
x 100%
342.17
96.21
x 100%
342.17
192.00
342.17
x 100%
=
15.77 % Al
=
28.12 % S
=
56.11 % O
Check the sum of the %’s. They should sum to very near 100.00 %
EXAMPLE: Find the mass percent (% composition)
of Ferrous nitrate
Fe
N
O
% Fe =
%N=
%O=
= 1
= 2
= 6
55.85
179.87
28.02
179.87
96.00
179.87
x 55.85
x 14.01
x 16.00
=
=
=
55.85
28.02
96.00
179.87
x 100%
=
31.05 % Fe
x 100%
=
15.58 % N
x 100%
=
53.37% O
100.00 %
If you had a 100.0 g sample of ferrous nitrate,
how many grams of iron are there?
31.05 g Fe
15.58 g N
Nitrogen?
If you had a 540.0 g sample of ferrous nitrate,
how many grams of iron are there?
540.0 x (.3105 % Fe) =
540.0 g Fe(NO3)2
Nitrogen?
31.05 g Fe
100 g Fe(NO3)2
540.0 x (.1558 % N) =
540.0 g Fe(NO3)2
15.58 g N
100 g Fe(NO3)2
167.7 g Fe
=
84.13 g N
=
Calcium fluoride 48.67% F,
while sodium fluoride 45.25% F
We can also do the reverse and use percent composition
data to determine empirical and molecular formulae.
Empirical formulas simplest whole number ratio of
atoms in a compound.
Molecular formulas show the actual numbers of atoms
of each element in a compound.
There are two ways to determine empirical formulas.
Experimental Data
% composition data
Need to know mass of
each element in the
laboratory sample
Assume the amount
to be 100 g
EXAMPLE:
40.0% C
6.71% H
53.3% O
1. Assume 100 g sample
40.0 g C
6.71 g H
53.3 g O
2. Convert to moles
40.0 g C
1 mole6.71
C g H
=
12.01 g C
= 3.33 moles C
1 mole H53.3 g O
=
1.01 g H
= 6.64356 moles H
1 mole O
16.00 g O
= 3.33 moles O
=
= 3.33 moles C
= 6.64356 moles H
= 3.33 moles O
3. Find the mole ratio
(divide by the smallest number of moles.)
3.33 moles C
=
3.33 moles
=1C
6.64356 moles H
=
3.33 moles
3.33 moles O
=
3.33 moles
=2H
=1O
If the mole ratio ends in
.20 (x5) .25 (x4) .33 or .66 (x3)
.50 (2)
Otherwise round to one of the above decimals or to the nearest
4.
Empirical
CH2O or CHOH
whole
number. Formula
EXAMPLE: A binary compound consisting only of
nitrogen and oxygen with a mass
percent of 37.00% N
37.00 g N
1 mole N
14.01 g N
2.6409707 mole N
2.6409707 mole N
2.6409707 mole
=1N
=
=
=
63.00g O
1 mole O
=
16.00 g O
3.9375 mole O
3.9375 mole O
2.6409707 mole
= 1.490929 O
=
=
= 1.490929 O
=1N
multiply by 2 to eliminate and make whole number ratios.
N2O3
EXAMPLE:
37.70 g Na
37.70% Na,
1 mole22.95
Na g Si
=
22.99 g Na
1.63984 mol Na
0.8170167 mol
= 2.007 Na
=
=
22.95% Si,
1 mole Si
39.34g O
=
28.09 g Si
0.8170167 mol Si
0.8170167 mol
= 1 Si
Na2SiO3
=
=
39.34% O
1 mole O
=
16.00 g O
2.45875 mol O
0.8170167 mol
= 3.009 O
=
=
Molecular formulas show the actual number of atoms of
each element in a molecule, as well as, the ratio of
atoms.
In order to solve these types of problems, 3 pieces of
information are needed.
•molecular mass of the compound
•empirical mass of the compound
•number of empirical units present
in the molecular formula
EXAMPLE: What is the formula of an unknown
substance with a percent composition
of 5.90% H and 94.10% O. It also has a
molecular mass of 34.00 g/mol.
5.90% H and 94.10% O means 5.90 g H and 94.10 g O
5.90 g H
1 mole H
1.01 g H
5.841584 mole H
5.841584 mole H
5.841584 mole
=
94.10 g O
1 mole O
16.00 g O
=
5.88125 mole O
=
=
5.88125 mole O
5.841584 mole
=
= 1.00 H
= 1.006790 O
=
Empirical Formula is:
HO
The empirical mass is 17.01 g (1.01 + 16.00)
We need to determine how much bigger the molecular formula is
compared to the empirical.
34.00
17.01
=
34.00
17.01
=
2
Empirical Formula is: HO
Molecular Formula is: H2O2
Hydrates are crystals that contain water molecules in
their crystal structure.
The crystal has crystallized from a water solution with
molecules of water adhering to the particles of the
crystal.
EXAMPLE: NiSO3 . 6 H2O
1 crystal of NiSO3 with 6 molecules of H2O
To calculate the formula of a hydrate, we need:
•mass of the hydrated sample
•mass of dry sample (anhydrous, w/o water) after
the H2O has been driven off.
•Mole ratio of the anhydrous compound to the
H2O driven off.
Finding the formula of a hydrate is very similar to
finding empirical formulas.
EXAMPLE: We have a 10.407 g sample of
hydrated barium iodide. The sample is
heated to drive off the water. The dry
sample has a mass of 9.520 g.
What is the formula of the hydrate?
The difference between the initial mass and that of the
dry sample is the mass of water that was driven off.
Mass of hydrate 10.407 g
Mass of dry sample - 9.520 g
Mass of water
0.887 g
9.520 g BaI2
1 mole BaI2
= 0.887 g H2O 1 mole H2O =
391.13 g BaI2
18.02 g H2O
0.02434 moles BaI2 =
0.0492 moles H2O
=
0.02434 moles BaI2
=
0.02434 moles
0.0492 moles H2O
0.02434 moles
=
= 1 BaI2
= 2 H2O
Formula: BaI2 . 2 H2O