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EMPIRICAL FORMULAS
From Percentages to Formulas
CONSTANT COMPOSITION
WHAT DOES THAT MEAN?
 Compounds have constant composition
 Definite proportions of elements chemically combined within them
 A chemist could measure this in two different ways:
 Count numbers of constituent atoms

Chemical formula with subscripts would result

Example

Formula AlCl3 means one atom of aluminum is bonded with three atoms of chlorine
 Percentages (by mass) of its elements
 Up until now, you’ve only worked with chemical formulas
 You believed a chemical formula was just the ratio of number of atoms in a
compound
 Chemical formulas are also the ratio of MOLES of atoms in a compound!

In 1 MOLE of CO2, there is 1 MOLE of carbon and 2 MOLES of oxygen
TYPES OF CHEMICAL FORMULAS
THE EMPIRICAL FORMULA
 Defined as the lowest whole number ratio of elements in a compound
 Up until now, you’ve been working with molecular formulas!
 The actual ratio of elements in a compound
 C2H4 and C3H6 are both molecular formulas
 Both have empirical formulas of CH2
 Empirical formulas and molecular formulas can be the same!
 H2 O
HOW TO CALCULATE EMPIRICAL FORMULAS
GIVEN MOLECULAR FORMULAS
 Look at molecular formula and find the lowest whole number ratio of
atoms
 Example 1
 Molecular: C6H12O6
 Empirical: CH2O
 Example 2
 Molecular: CH4N
 Empirical: CH4N
CONSTANT COMPOSITION AND MASS
PERCENT
 It is useful to know how much an element contributes
to the total molar mass of a compound
 So, how do you calculate mass percent?
 Obtained by comparing the MASS OF EACH ELEMENT present in 1
mole of the compound to the TOTAL MASS of 1 mole of the
compound
 A pure compound should show the same percent mass
of each element consistently
 So given a formula , you should be able to figure out
the percent mass of each element
BASIC STEPS OF DETERMINING MASS
PERCENT FROM FORMULA
Write correct formula of compound with subscripts
2. Calculate mass of each element in 1 mole of the
compound
3. Get molar mass of compound
4. Find the fraction of the total mass contributed by each
element and convert it to a percentage
1.
 Set up a ratio, or fraction, with molar mass of element on top (numerator) and molar
mass of compound on bottom (denominator)
5.
Sum the individual mass percent values to make sure they add up to
100%!
PRACTICE!

Carvone is a substance that occurs in two forms having different arrangements of the atoms but the same
molecular formula of C10H14O. One type of carvone gives caraway seeds their characteristic smell, and the
other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in
carvone.

Step 1: Write correct formula of compound with subscripts
C10H14O

Step 2: Find the masses of each element in 1mole of carvone:
𝑀𝑎𝑠𝑠 𝑜𝑓 𝐶 𝑖𝑛 1 𝑚𝑜𝑙𝑒 = 10 𝑚𝑜𝑙 ×
12.01 𝑔
= 120.1 𝑔
1 𝑚𝑜𝑙
𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻 𝑖𝑛 1 𝑚𝑜𝑙𝑒 = 14 𝑚𝑜𝑙 ×
1.008 𝑔
= 14.11 𝑔
1 𝑚𝑜𝑙
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑂 𝑖𝑛 1 𝑚𝑜𝑙𝑒 = 1 𝑚𝑜𝑙 ×

16.00 𝑔
= 16.00 𝑔
1 𝑚𝑜𝑙
Step 3: Get molar mass of compound
𝑀𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙 𝐶10 𝐻14 𝑂 = 120.1 + 14.11 + 16.00 = 150.2 𝑔
EXAMPLE - CONTINUED
 Step 4: Find the fraction of the total mass contributed by each element and convert
it to a percentage:
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝐶 =
120.1 𝑔 𝐶
× 100% = 79.96%
150.2 𝑔𝐶10 𝐻14 𝑂
14.11 𝑔 𝐻
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝐻 =
× 100% = 9.394%
150.2 𝑔𝐶10 𝐻14 𝑂
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝑂 =
16.00 𝑔 𝑂
× 100% = 10.65%
150.2 𝑔𝐶10 𝐻14 𝑂
 Step 5: Sum the individual mass percent values to make sure they add up to 100%!
STEPS TO USING MASS PERCENTS TO
DETERMINE EMPIRICAL FORMULAS
1.
Pretend that you have a 100 gram sample of the
compound

2.
3.
Convert the grams to moles for each element
Write the number of each element as a subscript in a
chemical formula

4.
In other words, change the % to grams
Keep each number as a decimal at this point!
Divide each subscript by the smallest number
 This will give the empirical formula
5.
Multiply the result by some integer to get rid of any
fractions

May not be necessary
PRACTICE!
 Calculate the empirical formula of a compound composed of 38.67 % C,
16.22 % H, and 45.11 % N
 Step 1: Pretend that you have a 100 gram sample of the compound
 Step 2: Convert the grams to moles for each element
EXAMPLE (CONTINUED)
 Step 3: Write the number of each element as a subscript in a chemical
formula
C3.22H16.09N3.22
 Step 4: If we divide all of these by the smallest subscript, it will give us
the empirical formula
MORE PRACTICE!
 Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its
empirical formula? What is the empirical mass?
HOW TO CONVERT BETWEEN EMPIRICAL
FORMULAS AND MOLECULAR FORMULAS
 Since the empirical formula is the lowest ratio, the actual molecule
would has a bigger mass
 Molecular formula can always be obtained by multiplying by some whole number
 To do so, follow the steps below:
1.
Divide the actual molecular molar mass (usually given in the problem) by empirical
molar mass

2.
Gives whole number
Multiply empirical formula by the whole number to get the molecular formula
 Look back at the previous problem. Caffeine has a molar mass of 194
g/mol. What is its molecular formula?
PRACTICE!
 A compound is known to be composed of 71.65 % Cl, 24.27% C, and
4.07% H. Its molecular molar mass is known to be 98.96 g/mol. What
is its molecular formula?
A REAL-WORLD DETERMINATION
OF EMPIRICAL FORMULA
REAL-WORLD DETERMINATION OF
EMPIRICAL FORMULA
 Combustion analysis is one of
the most common methods for
determining empirical formulas
 A weighed compound is burned
in oxygen and its products are
analyzed by a gas chromatogram
 It is particularly useful for
analysis of hydrocarbons!
 Products are CO2 and H2O
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MORE ON COMBUSTION ANALYSIS
 Combustion Analysis
 The technique of finding the mass composition of an unknown sample (X) by examining the
products of its combustion
X + O2 → CO2 + H2O
 Example Data
 0.250 g of compound X produces 0.686 g CO2 and 0.562 g H2O
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DETERMINING EMPIRICAL FORMULAS
FROM COMBUSTION ANALYSIS
X + O2 → CO2 + H2O
 Step 1: Find the mass of C & H that must have been present in X

Multiply masses of products by percent composition (decimal form) of the
products
C: 0.686 g x (12.01 g/44.01 g) = 0.187 g C
H: 0.562 g x [(2 x 1.008 g)/18.02 g]= 0.063 g H
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COMBUSTION ANALYSIS
X + O2 → CO2 + H2O
 Step 2: Add masses of products and compare to mass of original
compound
0.187 g C + 0.063 g H = 0.250 g total

Based on original compound mass of 2.90 g, compound X must contain only C and H!!
 Step 3: Find the number of moles of C and H
C: 0.187 g x mole/12.01 g = 0.0156 moles C
H: 0.063 g x mole/1.008 g = 0.063 moles H
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COMBUSTION ANALYSIS
X + O2 → CO2 + H2O
 Step 4: Write the formula using the mole numbers as subscripts
C.0156H.063
 Step 5: Divide by smallest number of moles
C: 0.0156/0.0156 = 1
H: 0.063/0.0156 = 4

If these numbers are fractions, multiply each by the same whole number
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COMBUSTION ANALYSIS
X + O2 → CO2 + H2O
 Step 6: Rewrite formula with new mole whole numbers

You now have the empirical formula!
CH4
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