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Chapter 7 Chemical Quantities

7.5 Molecular Formulas

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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 integerempirical mass

• is obtained by multiplying the subscripts in the empirical formula by the same small integer

Relating Molecular and Empirical Formulas

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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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Some Compounds with Empirical Formula CH2O

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Calculating a Molecular Formula from an Empirical Formula

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Determine the molecular formula of a compound thathas a molar mass of 78.11 g and an empiricalformula of CH.

STEP 1 Calculate the empirical formula mass.

Empirical formula mass of CH = 13.02 g

STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 78.11 g = 5.999 ~ 6

13.02 g

Finding the Molecular Formula

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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

Finding the Molecular Formula (continued)

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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

Learning Check

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STEP 1 Calculate the empirical formula mass.

C3H4O3 = 88.06 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 formulaC3x2H4x2O3x2 = C6H8O6 (2)

Solution

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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?

Molecular Formula

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STEP 1 Calculate the empirical formula mass.

24.27 g C x 1 mol C = 2.021 mol of C 12.01 g C

4.07 g H x 1 mol H = 4.04 mol of H 1.008 g H

71.65 g Cl x 1 mol Cl = 2.021 mol of Cl

35.45 g Cl

Solution

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Solution (continued)

2.021 mol C = 1 mol of C

2.021

4.04 mol H = 2 mol of H

2.021

2.02 mol Cl = 1 mol of Cl

2.021

Empirical formula = C1H2Cl1 = CH2Cl

Empirical formula mass (EM) CH2Cl = 49.48 g

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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 49.48 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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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?

Learning Check

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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 = 0.854 mol of S 32.07 g S

12.0 g N x 1 mol N = 0.857 mol of N 14.01 g N

60.6 g Cl x 1mol Cl = 1.71 mol of Cl 35.45 g Cl

Solution

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STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.

0.854 mol S = 1.00 mol of S0.854 0.857 mol N = 1.00 mol of N0.8541.71 mol Cl = 2.00 mol of Cl0.854

empirical formula = SNCl2

empirical formula mass = 116.98 g

Solution (continued)

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STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula.

Molar mass = 351 g = 3

Empirical formula mass 116.98 g

Molecular formula = (SNCl2)3 = S3N3Cl6

Solution (continued)

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Concept Map

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