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3–2

• molecular formula = (empirical formula) n
[ n = integer]
Ex. (CH
5
N) n or (CH
2
O) n
To be able to specify the exact formula of the molecule involved, the molecular formula, we must know the molar mass.
• molecular formula = C
6
H
6
• empirical formula = CH
= (CH)
6
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3–3

Molecular Formula Determination Method 1
• Obtain the empirical formula
• Compute the mass corresponding to the empirical formula.
• Calculate the ratio
Molar Mass/Empirical formula mass
• The integer from the previous step represents the number of empirical formula in one molecule. When the empirical formula subscripts are multiplied by this integer, the molecular formula results.
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3–4

Molecular Formula Determination Method 2
• Using the mass percentages and the molar mass, determine the mass of each element present in one mole of compound.
• Determine the number of moles of each element present in one mole of compound.
• The integers from the previous step represent the subscripts in the molecular formula.
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3–5

Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition
Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C ,
9.63% H and 27.79% O . Its experimentally determined molecular mass (molar mass) is 230 amu . What are the empirical and molecular formulas of dibutyl succinate?
Step 1: Determine the mass of each element in a 100g sample.
C 62.58 g H 9.63 g O 27.79 g
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3–6

Step 2: Convert masses to amounts in moles.
n
C
=
62 .
58 g C
×
1 mol C
12 .
011 g C
=
5 .
210 mol C n
H
=
9 .
63 g H
×
1 mol H
1 .
008 g H
=
9 .
55 mol H n
O
=
27 .
79 g O
×
1 mol O
15 .
999 g O
=
1 .
737 mol O
Step 3: Write a tentative formula.
C
5.21
H
9.55
O
1.74
Step 4: Convert to small whole numbers.
C
2.99
H
5.49
O
C = 5.21/1.74 = 2.99; H = 9.55/1.74 = 5.49; O = 1.74/1.74 = 1
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Step 5: Convert to a small whole number ratio since H
5.49
.
Multiply
×
2 to get C
5.98
H
10.98
O
2
The empirical formula is C
6
H
11
O
2
Step 6: Determine the molecular formula.
(6 mol C x 12.01 amu C/mol C + 11 mol H x 1.008 amu H/mol + 2 mol O x 15.998 amu)
= 115.14 amu
Calc ratio = Molar Mass/ Empirical Formula Mass
Molecular formula mass exp determined is 230 amu.
Empirical formula mass is 115 amu.
= 2
(C
6
H
11
O
2
) n=2
The molecular formula is C
12
H
22
O
4
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3–8

• A white power is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compounds empirical and molecular formulas?
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The dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo?
1. C
6
H
4
NO
2. C
8
H
3
NO
3. C
8
H
5
NO
4. I know this should be whole numbers for each atom, but I do not know how to accomplish that.
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Choice 3 is the smallest whole number ratio of the atoms that make up a molecule of indigo. The percentage must be converted to a mass, then the mass is converted to moles of the atoms and finally, the smallest is divided into the others to obtain the proper ratio.
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3–11

Without further information, which of the following is more likely to be both a molecular formula and an empirical formula for a compound? Briefly explain why. H
2
O
2
; C
6
H
6
; C
2
H
6
O
1. H
2
O
2
. The whole numbers for each atom are reduced as far as possible and still retain the identity of the molecule.
2. C
6
H
6
. The whole number of atoms are equal for each element.
3. C
2
H
6
O . One element shows only one atom so the formula ratio could not be further reduced, but the molecular formula could just have one atom.
4. None of these, it is not possible for a formula to be both empirical and molecular.
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3–12

Choice 3 shows the correct choice and proper justification. The whole numbers are reduced to their lowest value so C
2
H
6
O could be the empirical formula. The other choices could still be reduced to smaller whole numbers and maintain the same ratio.
Section 3.6: Determining the Formula of a Compound
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3–13