CHEMISTRY NOTES: 11/10/14

Consider the formula H2O
o The molecules of water will always have 2 hydrogens and 1 oxygen atom in them.
o
Since the water molecules always has the ratio of 2 H atoms :1 O atom, and you are allowed to multiply
the numbers in a ratio by the same factor, the following combinations could be written and still have the
2:1 ratio of H:O:
H20O10
H2000O1000
H2(6.022 x 1023 ) O6.022 x 1023
o
Since 6.022 x 1023 is the number of items in 1 mole, the last ratio written above could be written:
H2 molO1 mol
o
Therefore the subscripts in a formula tells us:
 the ratio of the atoms in the particle
 the mole ratio of the elements in the particle

There are two different types of formulas we can write for compounds:
o molecular formulas → shows all the atoms in a particle of a substance
o empirical formulas → shows the ratio of the different atoms in a substance

Examples of the different formula types for the same compounds:
molecular formulas

empirical formulas
C6H12O6
CH2O
C6H6
CH
H2O
H2O
P4O10
P2O5
Since the subscripts in a formula tells the mole ratio of the elements in the substance, given the right data we can
determine the empirical formula of a compound if we can find the mole ratio of the elements. To do this, you need
to follow the following steps:
STEPS TO DETERMINE EMPIRICAL FORMULAS
1) Determine the mass, in grams, for each element in the substance.
2) Determine the moles of each element.
3) Divide each mole number for the elements by the smallest mole number to find the mole ratio
between the elements.
4) Get the smallest whole number ratio for the elements.
5) Write the empirical formula
PROBLEM: A compound was found to contain 2.91 g N and 22.09 g Cl. What is its empirical formula?
Step 1
Step 2
You were given the masses of the elements in this
problem.
Divide the masses by the molar masses of the
elements.
2.91 g N
22.09 g Cl
1 mol N
2.91 g N (
) = 0.208 mol N
14.01 g N
22.09 g Cl (
Step 3
Step 4
Step 5
Divide each mole number by the smaller mole
number—0.208. Also, I like to show this as shown—
you don’t need to do this if you don’t want to, you can
just do it with the work shown in step 2.
1 mol Cl
) = 0.6231 mol Cl
35.45 g Cl
N0.208 Cl0.6231
0.208
N1Cl3
NCl3
0.208
⇨
N1.00 Cl3.00
PROBLEM: A compound consist of 44.87% K, 18.40% S and 36.72% O. What is the empirical formula?
Step 1
Since these are percents, you can assume whatever
total mass of the compound you want. The easiest
way is to assume 100 g of sample and, thus, the
percentages will be the grams for each element.
44.87 g K
18.40 g S
36.72 g O
1 mol K
44.87 g K (
) = 1.148 mol K
39.10 g K
Step 2
Divide the masses by the molar masses of the
elements.
18.40 g S (
1 mol S
) = 0.5737 mol S
32.07 g S
36.72 g O (
Step 3
Divide each mole number by the smaller mole
number—0.5737.
Step 4
Step 5
1 mol O
) = 2.295 mol O
16.00 g O
⇨
K 1.148 S0.5737 O 2.295
0.5737
0.5737
0.5737
K2.001 S1.000 O4.000
K2S1O4
K2SO4
PROBLEM: If 3.4972 g Fe is oxidized to form 5.0000 g of an iron oxide, what is the empirical formula?
Step 1
Step 2
You were given the mass of Fe. The Law of
Conservation of Matter states that whatever you start
with in a reaction you will still have at the end, only
rearranged. Therefore, the 3.4972 g of Fe is still
present in the end except it is part of the 5.0000 g of
the iron oxide. So, to find the mass of O, simply
subtract.
Divide the masses by the molar masses of the
elements.
3.4972 g Fe
5.0000 g – 3.4972 g = 1.5028 g O
1 mol Fe
3.4972 g Fe (
) = 0.06262 mol Fe
55.85 g Fe
1.5028 g O (
Step 3
Step 4
Step 5
Divide each mole number by the smaller mole
number—0.06262.
To get the whole number ratio, you will need to
multiply the ratio by some number to get it—it will be
whatever the denominator is for the fraction. In this
case, multiply the ratio by 2
1 mol O
) = 0.09392 mol O
16.00 g O
Fe0.06262 O0.09392
0.06262
0.06262
Fe2.000O3.000
Fe2O3
⇨
Fe1.000 O1.500