3.3: Percent Composition and Molecular Formula
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Deriving Empirical Formulas from Percent Composition
Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.
Example \(\PageIndex{4}\): Determining an Empirical Formula from Percent Composition
The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure \(\PageIndex{3}\)). What is the empirical formula for this gas?
Solution
Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:
\[\begin{align*}
27.29\,\%\,\ce C&=\mathrm{\dfrac{27.29\:g\: C}{100\:g\: compound}} \nonumber \\ \nonumber \\
72.71\,\%\,\ce O&=\mathrm{\dfrac{72.71\:g\: O}{100\:g\: compound}} \nonumber
\end{align*}\]
The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:
\[\begin{align*}
\mathrm{27.29\:g\: C\left(\dfrac{mol\: C}{12.01\:g}\right)}&=\mathrm{2.272\:mol\: C} \nonumber \\ \nonumber \\
\mathrm{72.71\:g\: O\left(\dfrac{mol\: O}{16.00\:g}\right)}&=\mathrm{4.544\:mol\: O} \nonumber
\end{align*}\]
Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:
\[\mathrm{\dfrac{2.272\:mol\: C}{2.272}=1} \nonumber\]
\[\mathrm{\dfrac{4.544\:mol\: O}{2.272}=2} \nonumber\]
Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO 2 .
Exercise \(\PageIndex{4}\)
What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?
- Answer
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\(CH_2O\)
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. Molar 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 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 :
\[\mathrm{\dfrac{molecular\: or\: molar\: mass\left(u\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(u\: or\:\dfrac{g}{mol}\right)}= \mathit n\: formula\: units/molecule}\]
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 :
\[\mathrm{(A_xB_y)_n=A_{nx}B_{nx}}\]
For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The averge empirical formula mass for this compound is approximately 30 u (the sum of 12 u for one C atom, 2 u for two H atoms, and 16 u for one O atom). If the compound’s molecular mass is determined to be 180 u, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:
\[\mathrm{\dfrac{180\:u/molecule}{30\:\dfrac{u}{formula\: unit}}=6\:formula\: units/molecule}\]
Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:
\[\ce{(CH2O)6}=\ce{C6H12O6}\]
Note that this same approach may be used when the molar mass (g/mol) instead of the average molecular mass (u) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.
Example \(\PageIndex{5}\): 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:
\[\begin{alignat}{2}
&\mathrm{(74.02\:g\: C)\left(\dfrac{1\:mol\: C}{12.01\:g\: C}\right)}&&= \:\mathrm{6.163\:mol\: C}\\
&\mathrm{(8.710\:g\: H)\left(\dfrac{1\:mol\: H}{1.01\:g\: H}\right)}&&= \:\mathrm{8.624\:mol\: H}\\
&\mathrm{(17.27\:g\: N)\left(\dfrac{1\:mol\: N}{14.01\:g\: N}\right)}&&= \:\mathrm{1.233\:mol\: N}
\end{alignat}\]
Next, we calculate the molar ratios of these elements relative to the least abundant element, \(\ce{N}\).
\(6.163 \: \text{mol} \: \ce{C}\) / \(1.233 \: \text{mol} \: \ce{N}\) = 5
\(8.264 \: \text{mol} \: \ce{H}\) / \(1.233 \: \text{mol} \: \ce{N}\) = 7
\(1.233 \: \text{mol} \: \ce{N}\) / \(1.233\: \text{mol} \: \ce{N}\) = 1
1.233/1.233 = \(1.000 \: \text{mol} \: \ce{N}\)
6.163/1.233 = \(4.998 \: \text{mol} \: \ce{C}\)
8.624/1.233 = \(6.994 \: \text{mol} \: \ce{H}\)
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 u/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:
\[\mathrm{\dfrac{40.57\:g\: nicotine}{0.2500\:mol\: nicotine}=\dfrac{162.3\:g}{mol}} \nonumber \]
Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:
\[\mathrm{\dfrac{162.3\:g/mol}{81.13\:\dfrac{g}{formula\: unit}}=2\:formula\: units/molecule} \nonumber\]
Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:
\[\ce{(C5H7N)2}=\ce{C10H14N2} \nonumber\]
Exercise \(\PageIndex{5}\)
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 an average molecular mass of 194.2 u?
- Answer
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C 8 H 10 N 4 O 2
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.
Key Equations
- \(\mathrm{\%X=\dfrac{mass\: X}{mass\: compound}\times100\%}\)
- \(\mathrm{\dfrac{molecular\: or\: molar\: mass\left(u\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(u\: or\:\dfrac{g}{mol}\right)}=\mathit n\: formula\: units/molecule}\)
- (A x B y ) n = A nx B ny
Glossary
- percent composition
- percentage by mass of the various elements in a compound
- empirical formula mass
- sum of average atomic masses for all atoms represented in an empirical formula
Contributors and Attributions
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Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110 ).