Empirical Formulas and
Molar Mass:
Part 1-Determination of an
Empirical Formula
1
Objectives
• -Explain what empirical formulas are
• -Be able to determine empirical
formulas using charges or using
experimental data
2
Empirical Formulas
• Empirical formula (“formula unit”) -the
lowest whole number ratio of atoms in an
ionic compound
• There are two ways to determine the
empirical formula for a compound
1. Using charges (use your ions and this is
easy) C
CRISS
O
S
S
2. Mathematically (yippee!!!)
3
Method #1 – Charges
EX: Write the empirical formula
for the compound formed by
Na & P
c. K & Ne
Na3P
Sr & Cl
SrCl2
No compound
d. Cu & Cl
CuCl or CuCl2
4
Method #2 –
Mathematically
• Step 1:Use the info given in the
problem and DA with the atomic mass
of the element (from periodic tableround to 3 SDs) to find moles
• Step 2:Take all the mole values and
divide them by the SMALLEST one to
figure out a ratio
• Step 3: Use the answers as subscripts
in the empirical formula
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Law of Conservation of Mass
Chemical Equation
Na + Cl 
NaCl
Reactants  Products
Mass of Reactants = Mass of Products
Example:
Na + Cl  NaCl
5.0g + 5.0g = ?
10.0g
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Atomic Mass of an Element =
Mole
• Each element has a unique atomic mass and
this is a standard for each element
• Atomic mass = mass of 1 mole of an element
• The atomic masses are from the periodic
table and we use grams
1 mole O = 6.02 x 1023 atoms O = 15.999 g O
1 mole Mg= 6.02 x 1023 atoms Mg= 24.305 g Mg
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Empirical Formulas
7.06 g of silver combine with an excess of
fluorine to produce 8.30 g of a compound
Silver
7.06g
+
Fluorine
1.24g

Ag?F?
8.30g
Found by subtracting!
7.06 g Ag X
𝟏 𝒎𝒐𝒍𝒆 𝑨𝒈
𝟏𝟎𝟖 𝒈 𝑨𝒈
.𝟎𝟔𝟓𝟒
= .0654 moles Ag
.𝟎𝟔𝟓𝟑
= 1.00 or 1
Ag1F1
1.24 g F X
𝟏 𝒎𝒐𝒍𝒆 𝑭
𝟏𝟗.𝟎 𝒈 𝑭
= .0653 moles F
.𝟎𝟔𝟓𝟑
.𝟎𝟔𝟓𝟑
= 1.00 or 1
ANSWER = AgF
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Empirical Formulas
A compound contains 24.58% K, 35.81% Mn,
and 40.50% O. Find the empirical formula
(assume working with 100 grams of the
compound and change percentages to grams)
24.58 g K X
35.81 g Mn X
40.50 g O
X
1 𝑚𝑜𝑙𝑒 𝐾
39.1 𝑔 𝐾
= .629 mole K
.𝟔𝟐𝟗
.𝟔𝟐𝟗
= 1.00 or 1
1 𝑚𝑜𝑙𝑒 𝑀𝑛
= .652 moles Mn
54.9 𝑔 𝑀𝑛
.𝟔𝟓𝟐
.𝟔𝟐𝟗
= 1.04 or 1
1 𝑚𝑜𝑙𝑒 𝑂
16.0 𝑔 𝑂
𝟐.𝟓𝟑
.𝟔𝟐𝟗
= 4.02 or 4
= 2.53 moles O
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ANSWER = KMnO4
Uneven Empirical Formulas
• When figuring empirical formulas
mathematically, sometimes the resulting
numbers don’t come out so clean
• You can’t just assume and round how you
choose
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.05 Rule
• This 0.05 rule allows for experimental error
that occurs causing varied number values:
– If a value is within .05 of a whole number (+0.05 or
- 0.05), then the value may be rounded
– Ex: 1.96 can be rounded to 2
• 1.07 cannot be rounded to 1
• 3.02 could be rounded to 3
• 1.93 cannot be rounded to 2
• If one value is not within .05 of a whole
number, all the values must be multiplied by
11
an integer so all values fall within .05 of whole
numbers
Uneven Empirical Formulas
4.35 g sample of zinc is combined with an excess
of the element phosphorus. 5.72 g of compound
are formed. Calculate the empirical formula.
Zinc + Phosphorous  Zn?P?
4.35g
1.37g Found by
5.72g
4.35 g Zn X
𝟏 𝒎𝒐𝒍𝒆 𝒁𝒏
𝟔𝟓.𝟒 𝒈 𝒁𝒏
= .0665 moles Zn
subtracting!
.𝟎𝟔𝟔𝟓
.𝟎𝟒𝟒𝟐
= 1.50 X 2 = 3.00 or 3
Not within .05 of a
whole number
𝟏 𝒎𝒐𝒍𝒆 𝑷
𝒈𝑷
1.37 g P X 𝟑𝟏.𝟎
= .0442 moles P .𝟎𝟒𝟒𝟐 = 1.00 X 2 = 2.00 or 2
.𝟎𝟒𝟒𝟐
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ANSWER = Zn3P2
Let’s Do It!!!
A compound is found to contain 72.3%
Fe and 27.7% O by weight. Calculate
the empirical formula.
Assume in 100 g of compound there
would be 72.3 g Fe and 27.7 g O
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72.3g Fe
1 mole Fe
X ——————
55.8 g Fe
1.30 mole
= 1.00
1.30 mole
= 1.30 mole Fe
X 3 = 3.00 = 3
1 mole O
27.7g O X —————— = 1.73 mole O
16.0 g O
1.73 mole =1.33 X 3 = 3.99 = 4
1.30 mole
Fe3O4
14
Objectives
• -Explain what empirical formulas are
• -Be able to determine empirical
formulas using charges or using
experimental data
15