Transcript Empirical Formulas and Molecular Formulas


shows the smallest whole-number ratio of
the atoms in the compound

CH only means it is a 1:1 ration between
carbon and hydrogen
 could be C2H2 (ecetylene), or C6H6
(benzene), or C8H8 (styrene)

What is the empirical formula of a compound
that’s mass is 25.9% N and 74.1% O?
 if there were 100g of this compound then
25.9g would be nitrogen and 74.1g oxygen
 change to moles in order to convert the % to a
number
25.9g N x 1 mol N
14.0 g N
 74.1g O x 1 mol O
16.0 g O

= 1.85 mol N
= 4.63 mol O
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a formula is not just the ratio of atoms, it is also
the ratio of moles
this leaves a ratio of N1.85O4.63 which is not a
whole number
 therefore, divide each by the lowest quantity
(in this case, 1.85)
 now is N1O2.5

again, still not a whole number, but if you
were to double it, it would be N2O5
 the empirical formula is N2O5
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Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and 45.11 %N.
Assume 100 g so…
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38.67 g C x
1mol C
= 3.220 mole C
12.01 g C
16.22 g H x
1mol H = 16.09 mole H
1.01 g H
45.11 g N x 1mol N
= 3.219 mole N
14.01 g N
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3.220 mole C
16.09 mole H
3.219 mole N
• C3.22H16.09N3.219
If we divide all of these by the smallest
number (3.22), it will give us whole numbers and
therefore the empirical formula

C1H5N1 is the empirical formula
Compound
Water
Hydrogen Peroxide
Glucose
Methane
Ethane
Octane
Molecular Formula
Empirical Formula
H 2O
H2O2
C6H12O6
CH4
C2H 6
H2O
HO
CH2O
CH4
CH3
C4 H 9
C8H18

a molecular formula is the same as, or a multiple
of, the empirical formula

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it is based on the actual number of atoms of each
type in the compound
the following have the same ratio of elements and
therefore the same empirical formula– CH2O,
however, many molecules can have that ratio, each
having a different molecular formula
C2H4O2 is the molecular formula of ethanoic acid
 CH2O is the molecular formula of methanal
 C6H12O6 is the molecular formula of glucose
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divide the experimental molar mass by the
mass of one mole of the empirical formula
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this results in the multiplier to convert to the
molecular formula
calculate the molecular formula of a
compound whose molar mass is found to be
60.0g and has an empirical formula of CH4N
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experimental molar mass --- 60.0 g
empirical molar mass --------- 30.0 g
2 (CH4N) = C2H8N2
=2
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A compound is known to be composed of 71.65 %
Cl, 24.27% C and 4.07% H. Its molar mass is
known is known to be 98.96 g. What is its
molecular formula?
need to calculate the empirical formula first
assume 100 g so…

71.65 g Cl x

24.27 g C x

4.07 g H x
1mol Cl = 2.02 mole Cl
35.5 g Cl
1mol C = 2.02 mole C
12.0 g C
1mol H
= 4.07 mole H
1.0 g H

Cl2.02C2.02H4.07
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divide by lowest (2.02 mol )
Cl1C1H2 is the empirical formula
this would give an empirical formula mass of
48.5 g
recall the problem asked for the molecular
formula of a compound with a molar mass of
98.96 g which is twice that of 48.5 g
therefore, Cl2C2H4 is the molecular formula