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3.2: Determining Empirical and Molecular Formulas

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    Learning Objectives
    • Compute the percent composition of a compound
    • Determine the empirical formula of a compound
    • Determine the molecular formula of a compound

    In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

    Percent Composition

    The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

    \[\mathrm{\%H=\dfrac{mass\: H}{mass\: compound}\times100\%}\]

    \[\mathrm{\%C=\dfrac{mass\: C}{mass\: compound}\times100\%}\]

    If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

    \[\mathrm{\%H=\dfrac{2.5\:g\: H}{10.0\:g\: compound}\times100\%=25\%}\]

    \[\mathrm{\%C=\dfrac{7.5\:g\: C}{10.0\:g\: compound}\times100\%=75\%}\]

    Example \(\PageIndex{1}\): Calculation of Percent Composition

    Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

    Solution

    To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

    \[\mathrm{\%C=\dfrac{7.34\:g\: C}{12.04\:g\: compound}\times100\%=61.0\%} \nonumber\]

    \[\mathrm{\%H=\dfrac{1.85\:g\: H}{12.04\:g\: compound}\times100\%=15.4\%} \nonumber\]

    \[\mathrm{\%N=\dfrac{2.85\:g\: N}{12.04\:g\: compound}\times100\%=23.7\%} \nonumber\]

    The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

    Exercise \(\PageIndex{1}\)

    A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

    Answer

    12.1% C, 16.1% O, 71.8% Cl

    Determining Percent Composition from Formula Mass

    Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its average formula mass and the average atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 u and three H atoms weighing a total of (3 × 1.008 u) = 3.024 u. The average formula mass of ammonia is therefore (14.01 u+ 3.024 u) = 17.03 u, and its percent composition is:

    \[\mathrm{\%N=\dfrac{14.01\:u\: N}{17.03\:u\:NH_3}\times100\%=82.27\%}\]

    \[\mathrm{\%H=\dfrac{3.024\:u\: H}{17.03\:u\:NH_3}\times100\%=17.76\%}\]

    This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example \(\PageIndex{2}\). As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.

    Example \(\PageIndex{2}\): Determining Percent Composition from a Molecular Formula

    Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?

    Solution

    To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

    \[\begin{align*}
    \%\ce C&=\mathrm{\dfrac{9\:mol\: C\times molar\: mass\: C}{molar\: mass\:\ce{C9H18O4}}\times100=\dfrac{9\times12.01\:g/mol}{180.159\:g/mol}\times100=\dfrac{108.09\:g/mol}{180.159\:g/mol}\times100} \nonumber\\
    \%\ce C&=\mathrm{60.00\,\%\,C} \nonumber
    \end{align*}\]

    \[\begin{align*}
    \%\ce H&=\mathrm{\dfrac{8\:mol\: H\times molar\: mass\: H}{molar\: mass\:\ce{C9H18O4}}\times 100=\dfrac{8\times 1.008\:g/mol}{180.159\:g/mol}\times 100=\dfrac{8.064\:g/mol}{180.159\:g/mol}\times 100} \nonumber\\
    \%\ce H&=4.476\,\%\,\ce H \nonumber
    \end{align*}\]

    \[\begin{align*}
    \%\ce O&=\mathrm{\dfrac{4\:mol\: O\times molar\: mass\: O}{molar\: mass\: \ce{C9H18O4}}\times 100=\dfrac{4\times 16.00\:g/mol}{180.159\:g/mol}\times 100=\dfrac{64.00\:g/mol}{180.159\:g/mol}\times 100} \nonumber \\
    \%\ce O&=35.52\% \nonumber
    \end{align*}\]

    Note that these percentages sum to equal 100.00% when appropriately rounded.

    Exercise \(\PageIndex{2}\)

    To three significant digits, what is the mass percentage of iron in the compound \(Fe_2O_3\)?

    

    Answer

    69.9% Fe

    Determination of Empirical Formulas

    As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

    \[\mathrm{1.71\:g\: C\times \dfrac{1\:mol\: C}{12.01\:g\: C}=0.142\:mol\: C}\]

    \[\mathrm{0.287\:g\: H\times \dfrac{1\:mol\: H}{1.008\:g\: H}=0.284\:mol\: H}\]

    Thus, we can accurately represent this compound with the formula C0.142H0.284. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

    \[\ce C_{\Large{\frac{0.142}{0.142}}}\:\ce H_{\Large{\frac{0.284}{0.142}}}\ce{\:or\:CH2}\]

    (Recall that subscripts of “1” are not written, but rather assumed if no other number is present.)

    The empirical formula for this compound is thus CH2. This may or not be the compound’s molecular formula as well; however, we would need additional information to make that determination (as discussed later in this section).

    Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

    \[\mathrm{Cl_{0.150}O_{0.525}=Cl_{\Large{\frac{0.150}{0.150}}}\: O_{\Large{\frac{0.525}{0.150}}}=ClO_{3.5}}\]

    In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl2O7 as the final empirical formula.

    Procedure

    In summary, empirical formulas are derived from experimentally measured element masses by:

    1. Deriving the number of moles of each element from its mass
    2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
    3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

    Figure \(\PageIndex{1}\) outlines this procedure in flow chart fashion for a substance containing elements A and X.

    A flow chart is shown that is composed of six boxes, two of which are connected together by a right facing arrow and located above two more that are also connected by a right-facing arrow. These two rows of boxes are connected vertically by a line that leads to a right-facing arrow and the last two boxes, connected by a final right facing arrow. The first two upper boxes have the phrases, “Mass of A atoms” and “Moles of A atoms” respectively, while the arrow that connects them has the phrase, “Divide by molar mass,” written below it. The second two bottom boxes have the phrases, “Mass of X atoms” and “Moles of X atoms” respectively, while the arrow that connects them has the phrase, “Divide by molar mass” written below it. The arrow that connects the upper and lower boxes to the last two boxes has the phrase “Divide by lowest number of moles” written below it. The last two boxes have the phrases, “A to X mole ratio” and “Empirical formula” respectively, while the arrow that connects them has the phrase, “Convert ratio to lowest whole numbers” written below it.
    Figure \(\PageIndex{1}\): The empirical formula of a compound can be derived from the masses of all elements in the sample.
    Example \(\PageIndex{3}\): Determining an Empirical Formula from Masses of Elements

    A sample of the black mineral hematite (Figure \(\PageIndex{2}\)), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

    3.2.1.png
    Figure \(\PageIndex{2}\): Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)

    Solution

    For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

    \[\begin{align*}
    \mathrm{34.97\:g\: Fe\left(\dfrac{mol\: Fe}{55.85\:g}\right)}&=\mathrm{0.6261\:mol\: Fe} \nonumber\\ \nonumber\\
    \mathrm{15.03\:g\: O\left(\dfrac{mol\: O}{16.00\:g}\right)}&=\mathrm{0.9394\:mol\: O} \nonumber
    \end{align*}\]

    Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

    \[\mathrm{\dfrac{0.6261}{0.6261}=1.000\:mol\: Fe} \nonumber\]

    \[\mathrm{\dfrac{0.9394}{0.6261}=1.500\:mol\: O} \nonumber\]

    The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

    \[\mathrm{2(Fe_1O_{1.5})=Fe_2O_3} \nonumber\]

    The empirical formula is \(Fe_2O_3\).

    Exercise \(\PageIndex{3}\)

    What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

    Answer

    \(N_2O_5\)

    Video \(\PageIndex{1}\): Additional worked examples illustrating the derivation of empirical formulas are presented in the brief video clip.

     

    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

    • \(\mathrm{\%X=\dfrac{mass\: X}{mass\: compound}\times100\%}\)
    • \(\mathrm{\dfrac{molecular\: or\: molar\: mass\left(u\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(u\: or\:\dfrac{g}{mol}\right)}=\mathit n\: formula\: units/molecule}\)
    • (AxBy)n = AnxBny

    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

    This page titled 3.2: Determining Empirical and Molecular Formulas is shared under a CC BY license and was authored, remixed, and/or curated by Scott Van Bramer.