3.6 Determining the Formula of a Compound

3.6 Determining the
Formula of a
Compound
Molecular compounds
Copyright © Houghton Mifflin Company. All rights reserved.
3–2
Formulas
• molecular formula = (empirical formula)n
[n = integer]
Ex. (CH5N)n or (CH2O)n
To be able to specify the exact formula of the molecule
involved, the molecular formula, we must know the
molar mass.
• molecular formula = C6H6 = (CH)6
• empirical formula = CH
Copyright © Houghton Mifflin Company. All rights reserved.
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.
Copyright © Houghton Mifflin Company. All rights reserved.
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.
Copyright © Houghton Mifflin Company. All rights reserved.
3–5
EXAMPLE using Method 1
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
Copyright © Houghton Mifflin Company. All rights reserved.
O 27.79 g
3–6
EXAMPLE continue
Step 2: Convert masses to amounts in moles.
1 mol C
= 5.210 mol C
12.011 g C
1 mol H
nH = 9.63 g H ×
= 9.55 mol H
1.008 g H
1 mol O
nO = 27.79 g O ×
= 1.737 mol O
15.999 g O
nC = 62.58 g C ×
Step 3: Write a tentative formula.
C5.21H9.55O1.74
Step 4: Convert to small whole numbers.
C2.99H5.49O
C = 5.21/1.74 = 2.99; H = 9.55/1.74 = 5.49; O = 1.74/1.74 = 1
Copyright © Houghton Mifflin Company. All rights reserved.
3–7
EXAMPLE continue
Step 5: Convert to a small whole number ratio since H5.49.
Multiply × 2 to get C5.98H10.98O2
The empirical formula is C6H11O2
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.
=2
Empirical formula mass is 115 amu.
(C6H11O2)n=2
The molecular formula is C12H22O4
Copyright © Houghton Mifflin Company. All rights reserved.
3–8
Learning Check
• 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?
Copyright © Houghton Mifflin Company. All rights reserved.
3–9
QUESTION
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.
2.
3.
4.
C6H4NO
C8H3NO
C8H5NO
I know this should be whole numbers for each atom, but I do not
know how to accomplish that.
Copyright © Houghton Mifflin Company. All rights reserved.
3–10
ANSWER
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.
Copyright © Houghton Mifflin Company. All rights reserved.
3–11
QUESTION
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. H2O2; C6H6; C2H6O
1. H2O2 . The whole numbers for each atom are reduced as far as
possible and still retain the identity of the molecule.
2. C6H6. The whole number of atoms are equal for each element.
3. C2H6O . 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.
Copyright © Houghton Mifflin Company. All rights reserved.
3–12
ANSWER
Choice 3 shows the correct choice and proper justification. The
whole numbers are reduced to their lowest value so C2H6O 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
Copyright © Houghton Mifflin Company. All rights reserved.
3–13