M4Q4: Empirical Formula

Chem 103 Textbook Team; Chem 104 Textbook Team

M4Q4: Empirical Formula

Introduction

The section explores empirical formulas and the use of combustion to analyze a compound. This section includes worked examples, sample problems, and a glossary.

Learning Objectives for Empirical Formulas

Determine empirical and chemical formulas using composition by mass, mass spectrometry, combustion analysis data, and molar mass information.
| Determination of Empirical Formulas | Derivation of Molecular Formulas | Mass Spectrometry of Molecules | Combustion Analysis |

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 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:

1.71 g C × $\dfrac{1 \;\text{mol C}}{12.01 \;\text{g C}}$ = 0.142 mol C

0.287 g H × $\dfrac{1 \;\text{mol H}}{1.008 \;\text{g H}}$ = 0.285 mol H

Thus, we can accurately represent this compound with the formula C_0.142H_0.285. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

$\text{C}_{\frac{0.142}{0.142}} \; \text{H}_{\frac{0.285}{0.142}}$ 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:

Cl_0.150O_0.525 = $\text{Cl}_{\frac{0.150}{0.150}} \; \text{O}_{\frac{0.525}{0.150}}$ = 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.

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

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

Figure 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 1.** The empirical formula of a compound can be derived from the masses of all elements in the sample.

Example 1

Determining a Compound’s Empirical Formula from the Masses of Its Elements

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?

Two rounded, smooth black stones are shown. — **Figure 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:

34.97 g Fe × $\dfrac{1\;\text{mol Fe}}{55.85\;\text{g Fe}}$ = 0.6261 mol Fe

15.03 g O × $\dfrac{1\;\text{mol O}}{16.00\;\text{g O}}$ = 0.9394 mol O

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

$\dfrac{0.6261}{0.6261}$ = 1.000 mol Fe

$\dfrac{0.9394}{0.6261}$ = 1.500 mol O

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe₁O_1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

2 × Fe₁O_1.5 = Fe₂O₃

The empirical formula is Fe₂O₃.

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

Answer:

N₂O₅

For additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip.

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. Since the scale for percentages is 100, it is most convenient to assume a 100 g sample to calculate the mass of each element.

Example 2

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 3). What is the empirical formula for this gas?

A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one. — **Figure 3.** An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: “Dual Freq”/Wikimedia Commons)

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:

27.29% C = $\dfrac{27.29\;\text{g C}}{100\;\text{g compound}}$

72.71% O = $\dfrac{72.71\;\text{g O}}{100\;\text{g compound}}$

The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

27.29% C × $\dfrac{\text{mol C}}{12.01\;\text{g}}$ = 2.272 mol C

72.71% O × $\dfrac{\text{mol O}}{16.00\;\text{g}}$ = 4.544 mol O

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

$\dfrac{2.272\;\text{mol C}}{2.272\;\text{mol}}$ = 1 C

$\dfrac{4.544\;\text{mol O}}{2.272\;\text{mol}}$ = 2 O

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO₂.

Check Your Learning
What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?

Answer:

CH₂O

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 section on mass spectrometry of molecules from earlier in the semester). 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:

$\dfrac{\text{molecular or molar mass (amu or} \;\frac{\text{g}}{\text{mol}})}{\text{empirical formula mass (amu or} \;\frac{\text{g}}{\text{mol}})}$ = 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_xB_y:

(A_xB_y)_n = A_nxB_ny

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:

$\dfrac{180 \;\text{amu/molecule}}{30\;\frac{\text{amu}}{\text{formula unit}}}$ = 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:

(CH₂O)₆ = C₆H₁₂O₆

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.

Example 3

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) $\left(\dfrac{1\;\text{mol C}}{12.01\;\text{g C}}\right)$ = 6.163 mol C

(8.710 g H) $\left(\dfrac{1\;\text{mol H}}{1.01\;\text{g H}}\right)$ = 8.624 mol H

(17.27 g N) $\left(\dfrac{1\;\text{mol N}}{14.01\;\text{g N}}\right)$ = 1.233 mol N

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

$\dfrac{6.163\ \text{mol C}}{1.233\ \text{mol N}}$ ≅ 5

$\dfrac{8.264\ \text{mol H}}{1.233\ \text{mol N}}$ ≅ 7

$\dfrac{1.233\ \text{mol N}}{1.233\ \text{mol N}}$ ≅ 1

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C₅H₇N. 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:

$\dfrac{40.57 \;\text{g nicotine}}{0.2500 \;\text{mol nicotine}} = \dfrac{162.3 \;\text{g}}{1\;\text{mol}}$

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

$\dfrac{162.3 \;\text{g/mol}}{81.13 \;\frac{\text{g}}{\text{formula unit}}}$ = 2 formula units/molecule

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

(C₅H₇N)₂ = C₁₀H₁₄N₂

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:

C₈H₁₀N₄O₂

Mass Spectrometry

Previously, we have learned about how mass spectrometry can provide important information about the natural abundance of isotopes. Mass spectrometry can also provide experimental data on the empirical and molecular formula of a molecular substance.

Figure 4 and Figure 5 show the hard ionization and soft ionization mass spectrum of carbon dioxide, CO₂. When all atom-to-atom bonds are broken in CO₂during hard ionization, the result is one C⁺ ion and two O⁺ ions.

Hard ionization mass spectrum of CO2. The x-axis is labeled with "m/z" and there are three vertical lines at x=12, x=13, and x=16. The vertical line at 12 has a height of 50%, the line at 16 has a height of 100%, and the line at 13 is hardly registered. This indicates that there are twice as many O atoms (at x=16) as C atoms (at x=12). The small spike at x=13 is for the other common isotope of carbon, C-13. — **Figure 4.** Hard MS of carbon dioxide, CO₂.

Note that there are three discrete peaks visible in the mass spectra of CO₂: one major peak at 12 amu, one very small peak at 13 amu, and a major peak at 16 amu. (Remember that we are assuming that the charge, Z, is equal to +1. This means that if the peak is at m/Z = 12, then m must be 12 amu.) The peaks at 12 and 13 amu correspond to the two stable isotopes of the element carbon, ¹²C (98.9%) and ¹³C (1.1% ). Separate analysis shows that the element oxygen is composed of mainly the isotope ¹⁶O (99.76%), with just trace amounts of other isotopes. Thus, we can use the abundance of the two major peaks at 12 and 16 amu to obtain the ratio of carbon atoms to oxygen atoms in the molecule.

When subjected to soft ionization, the CO₂ molecule remains intact, and ionization results in one CO₂⁺ ion. Figure 5 shows that there are two discrete peaks visible in the mass spectra of CO₂⁺ ion: one major peak at 44 amu and one very small peak at 45 amu.

The soft mass spectrum of CO2 has an x-axis labeled as "m/z" and has two spikes: one at x=44, and another at x=45. The one at x=44 has a height of 100% and the one at x=45 is barely registered. — **Figure 5.** Soft MS of carbon dioxide, CO₂.

Processing the data from the hard MS and soft MS provides different information. From the hard MS, we can calculate the molar mass of the lowest ratio of atoms, i.e., the empirical formula. If there is one copy of carbon and two copies of oxygen, then the molecule has a molar mass of 44 amu (or 45 amu if the isotope of carbon in the molecule is the minor ¹³C). Since the molecule is not broken into its constituent parts in soft MS, the soft MS spectrum tells us the molecular weight of the molecule and therefore its molecular formula. The empirical formula mass derived from hard MS is the same value as the major peak in the soft MS, which tells us that the molecular formula is the same as the empirical formula, which is CO₂.

The CO₂⁺ ion at 44 amu contains one carbon-12 atom plus two oxygen-16 atoms, while the CO₂⁺ ion at 45 amu contains one carbon-13 atom plus two oxygen-16 atoms. As the abundance of carbon-12 is significantly greater than that of carbon-13, we expect the abundance of the ion at 44 amu to be much greater than that of the ion at 45 amu. Separate analysis shows the abundance to be 989 : 11 like that of the isotopes of the element carbon.

Example 4

More Complicated Mass Spectra

Use these mass spectra data to determine the chemical formula of an unknown compound. Assume your sample contains only the predominant isotopes of the atoms.

Two mass spectrum are shown, a hard and soft. The hard mass spectrum shows two signals at x=1 and x=12 that are both at a height of 3. The soft mass spectrum shows a single signal at x=78 at a height of 7.

Solution
The hard MS has peaks at 1 amu and 12 amu, and they are approximately the same height. Hydrogen has mass of 1 amu and carbon has mass of 12 amu, and they are in a 1:1 ratio.

The soft MS has a single peak at 78 amu. If there was one hydrogen atom and one carbon atom in the molecule, then the mass of the compound would be 13 amu. However, there is no peak at 13 amu. There must be multiple copies of the CH-unit in the molecule. Dividing the molecular weight (78 amu) by the empirical formula’s molecular weight (13 amu) gives the number of copies. There are 78/13 or six copies of the CH unit. Thus, the unknown compound has a formula of C₆H₆.

Check Your Learning
Draw both the hard MS and soft MS of water.

Answer:

Two mass spectrum are shown, a hard and soft. The hard mass spectrum shows two signals at x=1 and x=16. The signal at x=1 has a height of 8, and the signal at x=16 has a height of 4. The soft mass spectrum shows a single signal at x=18 and a heights of 8.

The formula for water is H₂O.

In the hard MS, the signal at 1 amu represents H⁺ ion, and the signal at 16 amu represents O⁺ ion. The ratio of the abundance of H⁺ ion to O⁺ ion is 2:1. Note that as long as the two peaks in the hard MS are represented in a 2:1 ratio it would be correct, e.g., drawing a 6:3 H:O ratio would be equally acceptable.

In the soft MS, there is a single peak at 18 amu for the H₂O⁺ ion (1 + 1 + 16 amu). The single peak can be drawn at any height.

Example 5

Mass Spectra of a Molecule Containing Bromine

Use these mass spectra data to determine the chemical formula of an unknown compound that contains bromine. The element bromine consists of two isotopes, ⁷⁹Br (50.7%) and ⁸¹Br (49.3%).

This hard mass spectrum of an unknown compound has signals at x=1, x=12, x=79, and x=81. The spike at x=1 has a height of 60%, the spike at x=12 has a height of 20%, and the spikes at x=79 and x=81 each have a height of 10%.

This soft mass spectrum of an unknown compound has signals at x=94 and x=96. These spikes are the same height.

Solution
The hard MS has peaks at 1 amu, 12 amu, 79 amu and 81 amu. Hydrogen has mass of 1 amu and carbon has mass of 12 amu, and they are in a 3:1 ratio.

The two peaks at 79 and 81 amu correspond to the single element, Br, that exists as two isotopes in very nearly equal abundances. As a single element, the sum of the two percentages (10%+10%) needs to be calculated for comparison with the C and H. This indicates a C:H:Br ratio of 20:60:20 or 1:3:1. and an empirical formula of CH₃Br.

The soft MS has two peaks at 94 and 96 corresponding to the molecular weights 94 amu and 96 amu of the two molecules CH₃⁷⁹Br and CH₃⁸¹Br respectively. The empirical formula and actual formula are therefore the same for this molecule, i.e., CH₃Br.

Combustion Analysis

The elemental composition of hydrocarbons and related compounds may also be determined via a method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure 4). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.

This diagram shows an arrow pointing from O subscript 2 into a tube that leads into a vessel containing a red material, labeled “Sample.” This vessel is inside a blue container with a red inner lining which is labeled “Furnace.” An arrow points from the tube to the right into the vessel above the red sample material. An arrow leads out of this vessel through a tube into a second vessel outside the furnace. An line points from this tube to a label above the diagram that reads “C O subscript 2, H subscript 2 O, O subscript 2, and other gases.” Many small green spheres are visible in the second vessel which is labeled below, “H subscript 2 O absorber such as M g ( C l O subscript 4 ) subscript 2.” An arrow points to the right through the vessel, and another arrow points right heading out of the vessel through a tube into a third vessel. The third vessel contains many small blue spheres. It is labeled “C O subscript 2 absorber such as N a O H.” An arrow points right through this vessel, and a final arrow points out of a tube at the right end of the vessel. Outside the end of this tube at the end of the arrow is the label, “O subscript 2 and other gases.” — **Figure 4.** This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.

Example 6

Combustion Analysis

Polyethylene is a hydrocarbon (with formula C_xH_y) polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO₂ and 0.00161 g of H₂O. What is the empirical formula of polyethylene?

Solution
The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

C_xH_y(s) + excess O₂(g) → x CO₂(g) + $\frac{y}{2}$ H₂O(g)

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described earlier. An outline of this approach is given in the following flow chart:

mol C = 0.00394 g CO₂ × $\dfrac{1 \;\text{mol CO}_2}{44.01 \;\text{g/mol}}$ × $\dfrac{1 \;\text{mol C}}{1 \;\text{mol CO}_2}$ = 8.95 × 10^-5 mol C

mol H = 0.00161 g H₂O × $\dfrac{1 \;\text{mol H}_2 \text{O}}{18.02 \;\text{g/mol}}$ × $\dfrac{2 \;\text{mol H}}{1 \;\text{mol H}_2 \text{O}}$ = 1.79 × 10^-4 mol H

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is

$\dfrac{\text{mol H}}{\text{mol C}}$ = $\dfrac{1.79 \times 10^{-4}\;\text{mol H}}{8.95 \times 10^{-5}\;\text{mol C}}$ = $\dfrac{2 \;\text{mol H}}{1 \;\text{mol C}}$

and the empirical formula for polyethylene is CH₂.

Check Your Learning
A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO₂ and 0.00148 g of H₂O in a combustion analysis. What is the empirical formula for polystyrene?

Answer:

CH

Example 7

Combustion Analysis of an Organic Compound Containing Oxygen

Combustion analysis of 23.1 mg of a white powder that contains only the elements C, H, and O yields 33.9 mg of CO₂ and 13.9 mg of H₂O. Determine the empirical formula of the white compound.

Solution

The mass of C and H in the white powder that produced the quantities of CO₂ and H₂O in the analysis must first be determined. This enables the amount of O in the compound to be calculated by subtraction, as this oxygen does not react in the combustion process.:

mass of C = 33.9 mg CO₂ × $\dfrac{1 \text{g}}{1000 \text{mg}}$ × $\dfrac{1 \text{mol CO}_2}{44.01 \text{g}}$ × $\dfrac{1 \text{mol C}}{1 \text{mol CO}_2}$ × $\dfrac{12.01 \text{g}}{1 \text{mol C}}$ × $\dfrac{1000 \text{mg}}{1 \text{g}}$ = 9.25 mg C

mass of H = 13.9 mg H₂O × $\dfrac{1 \text{g}}{1000 \text{mg}}$ × $\dfrac{1 \text{mol H}_2 \text{O}}{18.02 \text{g}}$ × $\dfrac{2 \text{mol H}}{1 \text{mol H}_2 \text{O}}$ × $\dfrac{1.008 \text{g}}{1 \text{mol H}}$ × $\dfrac{1000 \text{mg}}{1 \text{g}}$ = 1.54 mg H

Because the white powder consists of just three elements and we now know the mass of two of them, the mass of O is the remainder:

mass O = 23.1 mg – (9.25 mg + 1.54 mg) = 12.3 mg O

Now the moles of each element can be determined, and the empirical formula determined:

moles C: 9.25 mg C × $\dfrac{1\;\text{g}}{1000\;\text{mg}}$ × $\dfrac{1\;\text{mol C}}{12.01\;\text{g}}$ = 7.70 × 10^-4 mol C

moles H: 1.54 mg H × $\dfrac{1\;\text{g}}{1000\;\text{mg}}$ × $\dfrac{1\;\text{mol H}}{1.008\;\text{g}}$ = 1.53 × 10^-3 mol H

moles O: 12.3 mg O × $\dfrac{1\;\text{g}}{1000\;\text{mg}}$ × $\dfrac{1\;\text{mol O}}{16.00\;\text{g}}$ = 7.69 × 10^-4 mol O

Dividing each of these by the smallest number of moles (7.69 x 10^-4 mol) yields the empirical formula: CH₂O

This approach is outlined in the following flow chart:

Check Your Learning

Combustion analysis of 0.112 g of a compound that contains C, H, and O yields 0.0561 g of H₂O and 0.205 g of CO₂. The hard mass spectrum shows three peaks at 1, 12, and 16 amu, and the soft mass spectrum a single peak at 144 amu. Determine the empirical and actual formula of the compound.

Answer:

The hard mass spectrum indicates the compound consists of the elements C, H, and O.

Empirical Formula C₃H₄O₂, actual Formula C₆H₈O₄

Key Concepts and Summary

The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities. To determine an empirical formula, you must find the smallest whole-number ratio of atoms in a molecule. If you begin with the mass of each element in the compound, you must first convert these masses to moles before determining the ratio. 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. Combustion analysis can be used to identify an unknown hydrocarbon because the identity of the products are known to be CO₂ and H₂O.

Key Equations

%X = $\dfrac{\text{mass X}}{\text{mass compound}}$ × 100%
$\dfrac{\text{molecular or molar mass (amu or} \;\frac{\text{g}}{\text{mol}})}{\text{empirical formula mass (amu or} \;\frac{\text{g}}{\text{mol}})}$ = n formula units/molecule
(A_xB_y)_n = A_nxB_ny

Glossary

combustion analysis: analytical technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products

empirical formula: formula showing the composition of a compound given as the simplest whole-number ratio of atoms

empirical formula mass: sum of average atomic masses for all atoms represented in an empirical formula

molecular formula: formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in a molecule of the compound

percent composition: percentage by mass of the various elements in a compound

Chemistry End of Section Exercises

What information do we need to determine the molecular formula of a compound from the empirical formula?
Calculate the following to four significant figures:
1. the percent composition of ammonia, NH₃
2. the percent composition of photographic “hypo,” Na₂S₂O₃
3. the percent of calcium ion in Ca₃(PO₄)₂
Determine the percent ammonia, NH₃, in Co(NH₃)₆Cl₃, to three significant figures.
Determine the empirical formulas for compounds with the following percent compositions:
1. 15.8% carbon and 84.2% sulfur
2. 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen
A compound 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?
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.
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:
1. Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O
2. Saran; 24.8% C, 2.0% H, 73.1% Cl
3. polyethylene; 86% C, 14% H
4. polystyrene; 92.3% C, 7.7% H
5. Orlon; 67.9% C, 5.70% H, 26.4% N
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.
Deduce the empirical formula (element ratio) from the following hard ionization mass spectrum:

Given that the empirical formula for the compound is the same as the actual molecular formula, draw a representation of the soft ionization mass spectrum for this compound.
The hard and soft MS of a particular compound are given below. In a hard MS, all bonds in a molecule are broken and individual atoms are observed. In a soft MS, no bonds are broken, and the molecule is observed.
1. Based on the hard MS, what elements are present in this compound?
2. Based on the hard MS, what is the empirical formula of this compound?
3. Based on both the hard and soft MS, what is the molecular formula of this compound?
Draw both the hard and soft mass spectrum for phosphorus trifluoride. Both P and F are monotopic, i.e., exist as single isotopes.
A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B₂O₃. What are the empirical and molecular formulas of the compound? The general reaction is
B_xH_y + O₂ → B₂O₃ + H₂O (not balanced)
A compound has a molecular mass of about 130 amu, and contains only carbon and hydrogen. A 3.000-mg sample of this compound burns to give 10.3 mg of CO₂. Determine its empirical and molecular formulas.
Combustion analysis of 0.112 g of a compound that contains C, H, and O yields 0.0561 g of H₂O and 0.205 g of CO₂. The hard mass spectrum shows three peaks at 1, 12, and 16 amu, and the soft mass spectrum shows a single peak at 144 amu. Determine the empirical and actual formula of the compound.

Answers to Chemistry End of Section Exercises

You need a given mass and the molar amount of the compound to get the molar mass so it can be compared to the empirical mass
(a) % N = 82.24%, % H = 17.76%
(b) % Na = 29.08%, % S = 40.56%, % O = 30.36%
(c) % Ca²⁺= 38.76%
% NH₃ = 38.2%
(a) CS₂; (b) CH₂O
C₂H₄Cl₂

Left-click here to watch Exercise 5 problem solving video.
Mg₃Si₂H₃O₈ (empirical formula), Mg₆Si₄H₆O₁₆ (molecular formula)
(a) C₅H₈O₂; (b) CHCl; (c) CH₂; (d) CH; (e) C₃H₃N
C₁₅H₁₅N₃
CH₂Br₂. With two isotopes of Br and two Br atoms in each molecule there are four possible molecules: CH₂⁷⁹Br⁷⁹Br, CH₂⁷⁹Br⁸¹Br, CH₂⁸¹Br⁷⁹Br, and CH₂⁸¹Br⁸¹Br with masses of 172, 174, 174, and 176 amu respectively. The soft mass spectrum will therefore exhibit three peaks in a 1:2:1 ratio as two of the molecules have the same mass, which is twice as likely to occur as the other two possibilities:
(a) H, C, N, O
(b) H₁₀C₅N₂O₃(c) H₁₀C₅N₂O₃
For hard MS, 1:3 relative abundance of P:F with peaks at m/Z = 19 (F) and m/Z = 31 (P). For soft MS, peak at m/Z = 88 (PF₃) with 100% abundance.
The empirical formula is BH₃. The molecular formula is B₂H₆.
Empirical formula: C₅H₄, molecular formula: C₁₀H₈
The hard mass spectrum indicates the compound consists of the elements C, H, and O. Empirical formula: C₃H₄O₂, molecular formula: C₆H₈O₄

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