1. Chapter 7 Chemical Quantities
7.5
Molecular Formulas
Brightly colored dishes are made of melamine.
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2. Relating Molecular and Empirical Formulas
A molecular formula
is equal or a multiple of its empirical formula
has a molar mass that is the product of the empirical formula mass multiplied by a small integer
molar mass = a small integer
empirical mass
is obtained by multiplying the subscripts in the empirical formula by the same small integer
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3. Diagram of Molecular and Empirical Formulas
A small integer links
a molecular formula and its empirical formula
a molar mass and its empirical formula mass
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4. Some Compounds with Empirical Formula CH2O
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5. Calculating a Molecular Formula from an Empirical Formula
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6. Finding the Molecular Formula
Determine the molecular formula of a compound that
has a molar mass of g and an empirical
formula of CH.
STEP 1 Calculate the empirical formula mass.
Empirical formula mass of CH = g
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
78.11 g = ~ 6
13.02 g
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7. Finding the Molecular Formula (continued)
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula.
Multiply each subscript in C1H1 by 6.
Molecular formula = C1x6 H1x6 = C6H6
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8. Learning Check
A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is its molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
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9. Solution STEP 1 Calculate the empirical formula mass.
C3H4O3 = g/EF
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
176.1 g (molar mass) = 2
88.06 g (empirical formula mass)
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula.
molecular formula = 2 x empirical formula
C3x2H4x2O3x = C6H8O (2)
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10. Molecular Formula
A compound contains 24.27% C, 4.07% H, and 71.65% Cl. The molar mass is about 99 g. What are the empirical and molecular formulas?
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11. Solution STEP 1 Calculate the empirical formula mass.
24.27 g C x 1 mol C = mol of C
12.01 g C
4.07 g H x 1 mol H = mol of H
1.008 g H
71.65 g Cl x 1 mol Cl = mol of Cl
35.45 g Cl
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12. Solution (continued) 2.021 mol C = 1 mol of C 2.021
4.04 mol H = 2 mol of H
2.02 mol Cl = 1 mol of Cl
Empirical formula = C1H2Cl1 = CH2Cl
Empirical formula mass (EM) CH2Cl = g
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13. Solution (continued)
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
Molar mass = 99 g = 2
Empirical formula mass g
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. 2 x (CH2Cl)
C1x2H2x2Cl1x2 = C2H4Cl2
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14. Learning Check
A compound is 27.4% S, 12.0% N, and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?
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15. Solution STEP 1 Calculate the empirical formula mass.
In 100 g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.
27.4 g S x 1 mol S = mol of S
32.07 g S
12.0 g N x 1 mol N = mol of N
14.01 g N
60.6 g Cl x 1mol Cl = mol of Cl
35.45 g Cl
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16. Solution (continued)
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
0.854 mol S = mol of S
0.854
0.857 mol N = mol of N
1.71 mol Cl = mol of Cl
empirical formula = SNCl2
empirical formula mass = g
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17. Solution (continued)
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula.
Molar mass = 351 g = 3
Empirical formula mass g
Molecular formula = (SNCl2)3 = S3N3Cl6
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18. Concept Map
Basic Chemistry Copyright © 2011 Pearson Education, Inc.