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Derivation of molecular formulas

Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass    . As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n :

molecular or molar mass ( amu or g mol ) empirical formula mass ( amu or g mol ) = n formula units/molecule

The molecular formula is then obtained by multiplying each subscript in the empirical formula by n , as shown by the generic empirical formula A x B y :

( A x B y ) n = A nx B nx

For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

180 amu/molecule 30 amu formula unit = 6 formula units/molecule

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

(CH 2 O) 6 = C 6 H 12 O 6

Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.

Determination of the molecular formula for nicotine

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

Solution

Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

( 74.02 g C ) ( 1 mol C 12.01 g C ) = 6.163 mol C ( 8.710 g H ) ( 1 mol H 1.01 g H ) = 8.624 mol H ( 17.27 g N ) ( 1 mol N 14.01 g N ) = 1.233 mol N

Next, we calculate the molar ratios of these elements relative to the least abundant element, N.

6.163 mol C / 1.233 mol N = 5
8.264 mol H / 1.233 mol N = 7
1.233 mol N / 1.233 mol N = 1
1.233 1.233 = 1.000 mol N 6.163 1.233 = 4.998 mol C 8.624 1.233 = 6.994 mol H

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C 5 H 7 N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

40.57 g nicotine 0.2500 mol nicotine = 162.3 g mol

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

162.3 g/mol 81.13 g formula unit = 2 formula units/molecule

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

(C 5 H 7 N) 2 = C 10 H 14 N 2

Check your learning

What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

Answer:

C 8 H 10 N 4 O 2

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Key concepts and summary

The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.

What information do we need to determine the molecular formula of a compound from the empirical formula?

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Calculate the following to four significant figures:

(a) the percent composition of ammonia, NH 3

(b) the percent composition of photographic “hypo,” Na 2 S 2 O 3

(c) the percent of calcium ion in Ca 3 (PO 4 ) 2

(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca 2+ = 38.76%

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Determine the following to four significant figures:

(a) the percent composition of hydrazoic acid, HN 3

(b) the percent composition of TNT, C 6 H 2 (CH 3 )(NO 2 ) 3

(c) the percent of SO 4 2– in Al 2 (SO 4 ) 3

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Determine the percent ammonia, NH 3 , in Co(NH 3 ) 6 Cl 3 , to three significant figures.

% NH 3 = 38.2%

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Determine the percent water in CuSO 4 ∙5H 2 O to three significant figures.

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Determine the empirical formulas for compounds with the following percent compositions:

(a) 15.8% carbon and 84.2% sulfur

(b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen

(a) CS 2
(b) CH 2 O

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Determine the empirical formulas for compounds with the following percent compositions:

(a) 43.6% phosphorus and 56.4% oxygen

(b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O

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A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?

C 6 H 6

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Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?

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Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.

Mg 3 Si 2 H 3 O 8 (empirical formula), Mg 6 Si 4 H 6 O 16 (molecular formula)

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Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:

(a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O

(b) Saran; 24.8% C, 2.0% H, 73.1% Cl

(c) polyethylene; 86% C, 14% H

(d) polystyrene; 92.3% C, 7.7% H

(e) Orlon; 67.9% C, 5.70% H, 26.4% N

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A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.

C 15 H 15 N 3

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Key equations

  • % X = mass X mass compound × 100 %
  • molecular or molar mass ( amu or g mol ) empirical formula mass ( amu or g mol ) = n formula units/molecule
  • (A x B y ) n = A nx B ny
Practice Key Terms 2

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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