Empirical and Molecular Formulas

9.3 Determining Empirical and Molecular Formulas

Learning Objectives

To determine the empirical formula of a compound from its composition by mass.
To derive the molecular formula of a compound from its empirical formula.

Figure 9.3(a) Sea Snails

The analgesic drug ω-conotoxin (ziconotide) is a natural product derived from the sea snail Conus magus. Image Credit: By Richard Parker (Conus magus 56mm Uploaded by JoJan) [CC BY 2.0 (http://creativecommons.org/licenses/by/2.0)], via Wikimedia Commons

When a new chemical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, among many other curiousities, chemists want to know:

What elements are present in the compound?
How many atoms of each element are present?

Empirical versus Molecular Formula

In chapter 4, we started learning how to write names and formulas of compounds. At that level the names and formulas of compounds are simply a matter of applying conventional rules of nomenclature-a textbook activity. But if you are faced with a more real world problem, such as figuring out a formula given a sample, it's unlikely you'll be able to go straight to the type of formula we learned to write in chapter 4. You would likely start with an experiment that instead of giving you the molecular formula, gives the atoms in their lowest whole number ratio called the empirical formula. Consider acetic acid. In chapter 5, you learned the formula of acetic acid as HC₂H₃O₂. This is the molecular formula. Given a sample of acetic acid, and not knowing what the substance is, you'd likely first do something called combustion analysis. Combustion analysis of acetic acid gives the empirical formula of acetic acid, CH₂O. Notice if you divide the molecular formula of acetic acid by 2, you get CH₂O, the empirical formula. The molecular formula is the actual number of atoms of each element in the compound. The molecular formula is always some whole-number ratio of the empirical formula.

In addition to acetic acid, many known compounds have the empirical formula CH₂O, including formaldehyde and glucose. Formaldehyde is used to preserve biological specimens and has properties that are very different from the glucose circulating in our blood to provide energy to our cells. If all we knew about glucose was its empirical formula, we cannot know whether glucose is CH₂O, C₂H₄O₂, or any other (CH₂O)_n. We can, however, use the experimentally determined molar mass of glucose (180 g/mol) to resolve this dilemma.

First, we calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose,

Equation 9.3(eq1)

{F.W. CH}_{2} O = [1 mol C (\frac{12 .01 g}{1 mol C})] + [2 mol H (\frac{1 .008 g}{1 mol H})] + [1 mol O (\frac{16 .00 g}{1 mol O})] =

= 30.03 g/mol.

This is much smaller than the observed molar mass of 180 g/mol.

Second, we determine the number of formula units per mole. For glucose, we can calculate the number of (CH₂O) units—that is, the n in (CH₂O)_n—by dividing the molar mass of glucose by the formula mass of CH₂O:

Equation 9.3(eq2)

n = \frac{180 g}{30 {.03 g/CH}_{2} O} = 5.99 \approx {6 CH}_{2} O formula units

Each glucose contains six CH₂O formula units, which gives a molecular formula for glucose of (CH₂O)₆, which is more commonly written as C₆H₁₂O₆. The molecular structures of formaldehyde and glucose, both of which have the empirical formula CH₂O, are shown in Figure 9.3(b).

Figure 9.3(b) Structural Formulas and Ball-and-Stick Models of (a) Formaldehyde and (b) Glucose

Calculating Mass Percentages

In our example for acetic acid that we considered in a previous paragraph, we said combustion analysis gives the empirical formula. Well, that's not quite correct. Instead, combustion analysis gives the mass percentage. If you burn 100.0 grams of acetic acid, you'd get 40.00 grams of carbon, 6.714 grams of hydrogen and the remainder of the 100.0 g sample was oxygen. These are the mass percentages. By mass, acetic acid is 40.00% carbon, 6.714% hydrogen and 53.29% oxygen. The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass; that is, the percent compositionThe percentage of each element present in a pure substance. The percent composition of a chemical compound is constant by the law of definite proportions.—the percentage of each element present in a pure substance—is constant.

Rather than starting with combustion analysis data, most of the questions you'll run into for this section of material involve calculating the mass percentage given a molecular or empirical formula. In terms of the real world, having the formula before the percentage is analagous to putting the cart before the horse; however, working the problems in this seemingly backwards direction makes it easier to learn. Consider an example, sucrose (cane sugar).

Figure 9.3(c) Sugar

Sugar cane and a bowl of sugar. Image Credit: CSIRO [CC BY 3.0 (http://creativecommons.org/licenses/by/3.0)], via Wikimedia Commons

The molecular formula of sucrose is C₁₂H₂₂O₁₁. According to this molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. We can use this information to calculate the mass of each element in 1 mol of sucrose, which will give us the molar mass of sucrose. We can then use these masses to calculate the percent composition of sucrose. To three decimal places, the calculations are the following:

Equation 9.3(eq3)

\begin{array}{l} mass of C/mol of sucrose = 12 mol C \times \frac{12.011 g C}{1 mol C} = 144.132 g C \\ mass of H/mol of sucrose = 22 mol H \times \frac{1.008 g H}{1 mol H} = 22.176 g H \\ mass of O/mol of sucrose = 11 mol O \times \frac{15.999 g O}{1 mol O} = 175.989 g O \end{array}

Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon.

The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places:

\begin{array}{l} mass % C in sucrose = \frac{mass of C/mol sucrose}{molar mass of sucrose} \times 100 = \frac{144.132 g C}{342 .297 g/mol} \times 100 = 42.12 % \\ mass % H in sucrose = \frac{mass of H/mol sucrose}{molar mass of sucrose} \times 100 = \frac{22.176 g H}{342 .297 g/mol} \times 100 = 6.48 % \\ mass % O in sucrose = \frac{mass of O/mol sucrose}{molar mass of sucrose} \times 100 = \frac{175.989 g O}{342 .297 g/mol} \times 100 = 51.41 % \end{array}

You can check your work by verifying that the sum of the percentages of all the elements in the compound is 100%:

42.12% + 6.48% + 51.41% = 100.01%

If the sum is not 100%, you have made an error in your calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.

Example 9.3-1

Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C₁₄H₁₈N₂O₅.

Calculate the mass percentage of each element in aspartame.
Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.

Given: molecular formula and mass of sample

Asked for: mass percentage of all elements and mass of one element in sample

Strategy:

A Use atomic masses from the periodic table to calculate the molar mass of aspartame.

B Divide the mass of each element by the molar mass of aspartame; then multiply by 100 to obtain percentages.

C To find the mass of an element contained in a given mass of aspartame, multiply the mass of aspartame by the mass percentage of that element, expressed as a decimal.

Solution:

A We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:
$\begin{array}{r} 14C & (14 mol C)(12 .011 g/mol C) & = & 168 .154 g \\ 18H & (18 mol H)(1 .008 g/mol H) & = & 18 .114 g \\ 2N & (2 mol N)(14 .007 g/mol N) & = & 28 .014 g \\ + 5O & (5 mol O)(15 .999 g/mol O) & = & 79 .995 g \\ C_{14} H_{18} N_{2} O_{5} & molar mass of aspartame & = & 294 .277 g/mol \end{array}$
Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).

B To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:
$\begin{array}{l} mass % C = \frac{168 .154 g C}{294 .277 g aspartame} \times 100 = 57 .14% C \\ mass % H = \frac{18 .114 g H}{294 .277 g aspartame} \times 100 = 6 .16% H \\ mass % N = \frac{28 .014 g N}{294 .277 g aspartame} \times 100 = 9 .52% N \\ mass % O = \frac{79 .995 g O}{294 .277 g aspartame} \times 100 = 27 .18% O \end{array}$
As a check, we can add the percentages together:
57.14% + 6.16% + 9.52% + 27.18% = 100.00%
If you obtain a total that differs from 100% by more than about �1%, there must be an error somewhere in the calculation.
C The mass of carbon in 1.00 g of aspartame is calculated as follows:
$mass of C = 1 .00 g aspartame \times \frac{57.14 g C}{100 g aspartame} = 0.571 g C$

Exercise

Calculate the mass percentage of each element in aluminum oxide (Al₂O₃). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide.

Answer: 52.93% aluminum; 47.08% oxygen; 1.92 g Al

Example 9.3-2, below, takes us through a more realistic journey. We go from the percent composition, that we would get from combustion analysis, to first an empirical formula and then a molecular formula.

Example 9.3-2

Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol.

Given: percent composition and molar mass

Asked for: molecular formula

Strategy:

A Assume 100 g of caffeine. From the percentages given, use the procedure given in Example 6 to calculate the empirical formula of caffeine.

B Calculate the formula mass and then divide the experimentally determined molar mass by the formula mass. This gives the number of formula units present.

C Multiply each subscript in the empirical formula by the number of formula units to give the molecular formula.

Solution:

A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine.

\begin{matrix} moles C = 49 .18 g C \times \frac{1 mol C}{12 .011 g C} = 4 .095 mol C \\ moles H = 5 .39 g H \times \frac{1 mol H}{1 .0079 g H} = 5 .35 mol H \\ moles N = 28 .65 g N \times \frac{1 mol N}{14 .0067 g N} = 2 .045 mol N \\ moles O = 16 .68 g O \times \frac{1 mol O}{15 .9994 g O} = 1 .043 mol O \end{matrix}

To obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount:

\begin{array}{l} O: \frac{1 .043}{1 .043} = 1 .000 & C: \frac{4 .095}{1 .043} = 3 .926 & H: \frac{5 .35}{1 .043} = 5 .13 & N: \frac{2 .045}{1 .043} = 1 .960 \end{array}

These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C₄H₅N₂O.

B The molecular formula of caffeine could be C₄H₅N₂O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows:

\begin{array}{r} 4C & (4 atoms C)(12 .011 g/atom C) & = & 48 .044 g \\ 5H & (5 atoms H)(1 .0079 g/atom H) & = & 5 .0395 g \\ 2N & (2 atoms N)(14 .0067 g/atom N) & = & 28 .0134 g \\ + 1O & (1 atom O)(15 .9994 g/atom O) & = & 15 .9994 g \\ C_{4} H_{5} N_{2} O & formula mass of caffeine & = & 97.096 g \end{array}

Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives

\frac{196 g/mol}{97 {.096 g/C}_{4} H_{5} N_{2} O} = 2.02 \approx 2 C_{4} H_{5} N_{2} O empirical formula units

C There are two C₄H₅N₂O formula units in caffeine, so the molecular formula must be (C₄H₅N₂O)₂ = C₈H₁₀N₄O₂. The structure of caffeine is as follows:

Exercise

Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer.

Answer: C₂Cl₂F₄

Summary

The empirical formula of a substance can be calculated from the experimentally determined percent composition, the percentage of each element present in a pure substance by mass. In many cases, these percentages can be determined by combustion analysis. If the molar mass of the compound is known, the molecular formula can be determined from the empirical formula.

Key Takeaway

The empirical formula of a substance can be calculated from its percent composition, and the molecular formula can be determined from the empirical formula and the compound’s molar mass.

Conceptual Problems

What is the relationship between an empirical formula and a molecular formula?
Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.

Numerical Problems

What is the mass percentage of water in each hydrate?
1. H₃AsO₄·0.5H₂O
2. NH₄NiCl₃·6H₂O
3. Al(NO₃)₃·9H₂O
What is the mass percentage of water in each hydrate?
1. CaSO₄·2H₂O
2. Fe(NO₃)₃·9H₂O
3. (NH₄)₃ZrOH(CO₃)₃·2H₂O
Which of the following has the greatest mass percentage of oxygen—KMnO₄, K₂Cr₂O₇, or Fe₂O₃?
Which of the following has the greatest mass percentage of oxygen—ThOCl₂, MgCO₃, or NO₂Cl?
Calculate the percent composition of the element shown in bold in each compound.
1. SbBr₃
2. As₂I₄
3. AlPO₄
4. C₆H₁₀O
Calculate the percent composition of the element shown in bold in each compound.
1. HBrO₃
2. CsReO₄
3. C₃H₈O
4. FeSO₄
A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound?

The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds.

Compound	% Iron	% Oxygen
1	69.9	30.1
2	77.7	22.3
3	72.4	27.6

What is the mass percentage of water in each hydrate?
1. LiCl·H₂O
2. MgSO₄·7H₂O
3. Sr(NO₃)₂·4H₂O
What is the mass percentage of water in each hydrate?
1. CaHPO₄·2H₂O
2. FeCl₂·4H₂O
3. Mg(NO₃)₂·4H₂O

Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates?

Compound	Initial Mass (g)	Mass after Cooling (g)
NiSO₄·H₂O	2.08	1.22
CoCl₂·H₂O	1.62	0.88

Which contains the greatest mass percentage of sulfur—FeS₂, Na₂S₂O₄, or Na₂S?
Given equal masses of each, which contains the greatest mass percentage of sulfur—NaHSO₄ or K₂SO₄?
Calculate the mass percentage of oxygen in each polyatomic ion.
1. bicarbonate
2. chromate
3. acetate
4. sulfite
Calculate the mass percentage of oxygen in each polyatomic ion.
1. oxalate
2. nitrite
3. dihydrogen phosphate
4. thiocyanate
The empirical formula of garnet, a gemstone, is Fe₃Al₂Si₃O₁₂. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula?
A compound has the empirical formula C₂H₄O, and its formula mass is 88 g. What is its molecular formula?
Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula?

Answers

To two decimal places, the percentages are:
1. 5.97%
2. 37.12%
3. 43.22%
% oxygen: KMnO₄, 40.50%; K₂Cr₂O₇, 38.07%; Fe₂O₃, 30.06%
To two decimal places, the percentages are:
1. 66.32% Br
2. 22.79% As
3. 25.40% P
4. 73.43% C
Cr₂O₃.
To two decimal places, the percentages are:
1. 29.82%
2. 51.16%
3. 25.40%
NiSO₄ ·6H₂O and CoCl₂ · 6H₂O
NaHSO₄
1. 72.71%
2. 69.55%
3. 65.99%
4. 0%
C₄H₈O₂