When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure \(\PageIndex{2}\)). The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate. Show Figure \(\PageIndex{2}\): Combustion analysis apparatus One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure \(\PageIndex{3}\) and a typical combustion analysis is illustrated in Examples \(\PageIndex{3}\) and \(\PageIndex{4}\).
Example \(\PageIndex{3}\): Combustion of Isopropyl Alcohol What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O? Solution From this information quantitate the amount of C and H in the sample. \[ (0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2 \] Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this? \[ (0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C \] How about the hydrogen? \[ (0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O \] Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample. When we add our carbon and hydrogen together we get: 0.154 grams (C) + 0.034 grams (H) = 0.188 grams But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol: 0.255 grams - 0.188 grams = 0.067 grams oxygen This much oxygen is how many moles? \[ (0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O \] Overall therefore, we have:
Divide by the smallest molar amount to normalize:
Within experimental error, the most likely empirical formula for propanol would be \(C_3H_8O\)
Example \(\PageIndex{4}\): Combustion of Naphalene Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene. Given: mass of sample and mass of combustion products Asked for: empirical formula Strategy:
Solution: A Upon combustion, 1 mol of \(\ce{CO2}\) is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams: \[ mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C} \] \[ = 1.883 \times 10^{-2} \, g \, C \] \[ mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H} \] \[ = 1.264 \times 10^{-3} \, g \, H \] B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount: \[ moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C \] \[ moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H \] Dividing each number by the number of moles of the element present in the smaller amount gives \[H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250\] Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.
Learning Outcomes
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 CompositionThe 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: [latex]\%\text{H}=\dfrac{\text{mass H}}{\text{mass compound}}\times 100\%[/latex] [latex]\%\text{C}=\dfrac{\text{mass C}}{\text{mass compound}}\times 100\%[/latex] 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: [latex]\%\text{H}=\dfrac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%[/latex] [latex]\%\text{C}=\dfrac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%[/latex]
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? Check Your LearningA 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? Determining Percent Composition from Molecular or Empirical FormulasPercent 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 formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is: [latex]\begin{array}{c}\\ \%\text{N}=\dfrac{14.01\text{ amu N}}{17.03\text{ amu }{\text{ NH}}_{3}}\times 100\%=82.27\%\\ \%\text{H}=\dfrac{3.024\text{ amu N}}{17.03\text{ amu }{\text{ NH}}_{3}}\times 100\%=17.76\%\end{array}[/latex] 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 in the example problem below. 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.
Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition? Check Your LearningTo three significant digits, what is the mass percentage of iron in the compound Fe2O3?
Determination of Empirical FormulasAs 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: [latex]\begin{array}{l}1.17\text{g C}\times \dfrac{1\text{mol C}}{12.01\text{g C}}=0.142\text{mol C}\\ 0.287\text{g H}\times \dfrac{1\text{mol H}}{1.008\text{g H}}=0.284\text{mol H}\end{array}[/latex] Thus, we can accurately represent this compound with the formula C0.142H0.248. Per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript: [latex]\displaystyle{\text{C}}_{\frac{0.142}{0.142}}{\text{H}}_{\frac{0.248}{0.142}}\text{ or }{\text{CH}}_{2}[/latex] (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: [latex]{\text{C1}}_{0.150}{\text{O}}_{0.525}={\text{Cl}}_{\frac{0.150}{0.150}}{\text{O}}_{\frac{0.525}{0.150}}={\text{ClO}}_{3.5}[/latex] 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. In summary, empirical formulas are derived from experimentally measured element masses by:
Figure 1 outlines this procedure in flow chart fashion for a substance containing elements A and X.  Figure 1. The empirical formula of a compound can be derived from the masses of all elements in the sample.
A sample of the black mineral hematite (Figure 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? Figure 2. Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb) Check Your LearningWhat is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?
For additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip below. You can view the transcript for “Calculating Percent Composition and Empirical Formulas” here (opens in new window). Deriving Empirical Formulas from Percent CompositionFinally, 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.
The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure 3). What is the empirical formula for this gas? Figure 3. An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: “Dual Freq”/Wikipedia) Check Your LearningWhat is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O? Derivation of Molecular FormulasRecall 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 module on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced later. 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, ([latex]n[/latex]): [latex]\displaystyle\dfrac{\text{molecular or molar mass}\left(\text{amu or}\dfrac{\text{g}}{\text{mol}}\right)}{\text{empirical formula mass}\left(\text{amu or}\dfrac{\text{g}}{\text{mol}}\right)}=n\text{ formula units/molecule}[/latex] The molecular formula is then obtained by multiplying each subscript in the empirical formula by [latex]n[/latex], as shown below for the generic empirical formula [latex]A_{x}B_{y}[/latex]: [latex]{{\text{(A}}_{\text{x}}{\text{B}}_{\text{y}}\text{)}}_{\text{n}}={\text{A}}_{\text{nx}}{\text{B}}_{\text{nx}}[/latex] For example, consider a covalent compound whose empirical formula is determined to be CH2O. 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: [latex]\dfrac{180\dfrac{\text{amu}}{\text{molecule}}}{30\dfrac{\text{amu}}{\text{formula unit}}}=6\dfrac{\text{formula units}}{\text{molecule}}[/latex] Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula: [latex]{\text{(CH}}_{\text{2}}{\text{O)}}_{\text{6}}={\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}[/latex] 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.
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? Check Your LearningWhat 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?
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
Glossaryempirical formula mass: sum of average atomic masses for all atoms represented in an empirical formula percent composition: percentage by mass of the various elements in a compound |
What is the empirical formula of a compound that contains carbon and hydrogen only where carbon is 85.71 by mass?
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