3.2 The Concept of Empirical Formula and Molecular Formula
Empirical of “simplest” formulas
Empirical formulas give the relative numbers of the different elements in a sample of a compound, expressed in the smallest possible integers. The term empirical refers to the fact that formulas of this kind are determined experimentally; such formulas are also commonly referred to as simplest formulas.
Problem Example 2: Simplest formula from molecular formula
Glucose (the “fuel” your body runs on) is composed of molecular units having the formula C6H12O6. What is the empirical formula of glucose?
Solution: The glucose molecule contains twice as many atoms of hydrogen as carbons or oxygens, so we divide through by 6 to get CH2O.
Note: this simplest formula, which applies to all 6-carbon sugars, indicates that these compounds are “composed” of carbon and water, which explains why sugars are known as carbohydrates.
Figure 3.4 Carbohydrate Structure
Some solid compounds do not exist as discrete molecular units, but are built up as extended two- or three-dimensional lattices of atoms or ions. The compositions of such compounds are commonly described by their simplest formulas. In the very common case of ionic solids, such a formula also expresses the minimum numbers of positive and negative ions required to produce an electrically neutral unit, as in NaCl or CuCl2.
Problem Example 3: Molecular formula from ionic charges
a) Write the formula of ferric bromide, given that the ferric (iron-III) ion is Fe3+ and the bromide ion carries a single negative charge.
b) Write the formula of bismuth sulfide, formed when the ions Bi3+ and S2– combine.
Solution:
a) Three Br– ions are required to balance the three positive charges of Fe3+, hence the formula FeBr3.
b) The only way to get equal numbers of opposite charges is to have six of each, so the formula will be Bi2S3.
What formulas don’t tell us
The formulas we ordinarily write convey no information about the compound’s structure— that is, the order in which the atoms are connected by chemical bonds or are arranged in three-dimensional space. This limitation is especially significant in organic compounds, in which hundreds if not thousands of different molecules may share the same empirical formula.
The compounds ethanol and dimethyl ether both have the simplest formula C2H6O. The structural formulas reveal the very different nature of these two molecules:
Formulas can be made to convey structural information
It is often useful to write formulas in such as way as to convey at least some information about the structure of a compound. For example, the formula of the solid (NH4)2CO3 is immediately identifiable as ammonium carbonate, and essentially a compound of ammonium and carbonate ions in a 2:1 ratio, whereas the simplest or empirical formula N2H8CO3 obscures this information.
Similarly, the distinction between ethanol and dimethyl ether can be made by writing the formulas as C2H5OH and CH3–O–CH3, respectively. Although neither of these formulas specifies the structures precisely, anyone who has studied organic chemistry can work them out, and will immediately recognize the –OH (hydroxyl) group which is the defining characteristic of the large class of organic compounds known as alcohols. The –O– atom linking two carbons is similarly the defining feature of ethers.
Formulas imply molar masses
Several related terms are used to express the mass of one mole of a substance.
- Molecular weight This is analogous to atomic weight: it is the relative weight of one formula unit of the compound, based on the carbon-12 scale. The molecular weight is found by adding atomic weights of all the atoms present in the formula unit. Molecular weights, like atomic weights, are dimensionless; i.e., they have no units.
- Formula weight The same thing as molecular weight. This term is sometimes used in connection with ionic solids and other substances in which discrete molecules do not exist.
- Molar mass The mass (in grams, kilograms, or anyother mass unit) of one mole of particles or formula units. When expressed in grams, the molar mass is numerically the same as the molecular weight, but it must be accompanied by the mass unit.
Problem Example 4: Formula weight and molar mass
a) Calculate the formula weight of copper(II) chloride, CuCl2.
b) How would you express this same quantity as a molar mass?
Solution:
a) The atomic weights of Cu and Cl are, respectively 63.55 and 35.45; the sum of each atomic weight, multiplied by the numbers of each kind of atom in the formula unit, yields 63.55 + 2(25.35) = 134.45.
b) The masses of one mole of Cu and Cl atoms are, respectively, 63.55 g and 35.45 g; the mass of one mole of CuCl2 units is (63.55 g) + 2(25.35 g) = 134.45 g.
Mole ratios and mole fractions from formulas
The information contained in formulas can be used to compare the compositions of related compounds as in the following example:
Problem Example 5: mole ratio calculation
The ratio of hydrogen to carbon is often of interest in comparing different fuels. Calculate these ratios for methanol (CH3OH) and ethanol (C2H5OH).
Solution: the H:C ratios for the two alcohols are 4:1 = 4.0 for methanol and 6:2 (3.0) for ethanol.
Alternatively, one sometimes uses mole fractions to express the same thing. The mole fraction of an element M in a compound is just the number of atoms of M divided by the total number of atoms in the formula unit.
Problem Example 6: mole fraction and mole percent
Calculate the mole fraction and mole-percent of carbon in ethanol (C2H5OH).
Solution: The formula unit contains nine atoms, two of which are carbon. The mole fraction of carbon in the compound is 2/9 = .22. Thus 22 percent of the atoms in ethanol are carbon.
Percent composition and elemental masses from formulas
Since the formula of a compound expresses the ratio of the numbers of its constituent atoms, a formula also conveys information about the relative masses of the elements it contains. But in order to make this connection, we need to know the relative masses of the different elements.
Problem Example 7: mass of each element in a given mass of compound
Find the masses of carbon, hydrogen and oxygen in one mole of ethanol (C2H5OH).
Solution: Using the atomic weights (molar masses) of these three elements, we have
carbon: (2 mol)(12.0 g mol–1) = 24 g of C
hydrogen: (6 mol)(1.01 g mol–1) = 6 g of H
oxygen: (1 mol)(16.0 g mol–1) = 16 g of O
The mass fraction of an element in a compound is just the ratio of the mass of that element to the mass of the entire formula unit. Mass fractions are always between 0 and 1, but are frequently expressed as percent.
Problem Example 8: mass fraction and mass percent of an element in a compound
Find the mass fraction and mass percentage of oxygen in ethanol (C2H5OH)
Solution: Using the information developed in the preceding example, the molar mass of ethanol is (24 + 6 + 16)g mol–1 = 46 g mol–1. Of this, 16 g is due to oxygen, so its mass fraction in the compound is (16 g)/(46 g) = 0.35 which corresponds to 35%.
Finding the percentage composition of a compound from its formula is a fundamental calculation that you must master; the technique is exactly as shown above. Finding a mass fraction is often the first step in solving related kinds of problems:
Problem Example 9: mass of an element in a given mass of compound
How many tons of potassium are contained in 10 tons of KCl?
Solution: The mass fraction of K in KCl is 39.1/74.6=.524; 10 tons of KCl contains(39.1/74.6) × 10 tons of K, or 5.24 tons of K. (Atomic weights: K = 39.1, Cl = 35.5. )
Note that there is no need to deal explicitly with moles, which would require converting tons to kg.
Problem Example 10: mass of compound containing given mass of an element
How many grams of KCl will contain 10 g of potassium?
Solution: The mass ratio of KCl/K is 74.6 ÷ 39.1; 10 g of potassium will be present in (74.6/39.1) × 10 grams of KCl, or 19 grams.
Mass ratios of two elements in a compound can be found directly from the mole ratios that are expressed in formulas.
Problem Example 11: Mass ratio of elements from formula
Molten magnesium chloride (MgCl2) can be decomposed into its elements by passing an electric current through it. How many kg of chlorine will be released when 2.5 kg of magnesium is formed? (Mg = 24.3, Cl = 35.5)
Solution: Solution: The mass ratio of Cl/Mg is (35.5 ×2)/24.3, or 2.9; thus 2.9 kg of chlorine will be produced for every kg of Mg, or (2.9 × 2.5) = 7.2 kg of chlorine for 2.5 kg of Mg
(Note that is is not necessary to know the formula of elemental chlorine (Cl2) in order to solve this problem.)
Simplest formulas from experimental data
As was explained above, the simplest formula (empirical formula) is one in which the relative numbers of the various elements are expressed in the smallest possible whole numbers. Aluminum chloride, for example, exists in the form of structural units having the composition Al2Cl6; the simplest formula of this substance is AlCl3.
Simplest formulas from atom ratios
Some methods of analysis provide information about the relative numbers of the different kinds of atoms in a compound.
The process of finding the formula of a compound from an analysis of its composition depends on your ability to recognize the decimal equivalents of common integer ratios such as 2:3, 3:2, 4:5, etc.
Problem Example 12: Simplest formula from mole ratio
Analysis of an aluminum compound showed that 1.7 mol of Al is combined with 5.1 mol of chlorine. Write the simplest formula of this compound.
Solution: The formula Al1.7Cl5.1 expresses the relative numbers of moles of the two elements in the compound. It can be converted into the simplest formula by dividing both subscripts by the smaller one, yielding AlCl3 .
Simplest formulas from mass composition
More commonly, an arbitrary mass of a compound is found to contain certain masses of its elements. These must be converted to moles in order to find the formula.
Problem Example 13: Simplest formula from combustion masses
When 10.0 g of a certain organic compound containing only C, H, and O undergoes combustion in the presence of excess O2, 9.56 g of CO2 and 3.92 g of H2O are formed. Find the simplest formula of this substance.
Solution: Begin by calculating the moles of the two combustion products:
CO 2: (9.56 g) / (44 g mol–1) = .217 mol
H 2O: (3.92 g) / (18 g mol–1) = .218 mol (containing 2 × .218 mol = .436 mol of hydrogen.)
We can now write a preliminary formula of the compound as C.217H.436Ox , leaving the value of x to be determined. The easiest way to do this is by calculating the difference between the 10.0-g mass of the unknown compound and the total mass of carbon plus hydrogen in the combustion products. The latter quantities work out as follows: C: (.217 mol × 12 g mol–1) = 2.60 g;
H: (.436 mol × 1.01 g mol–1) = .440 g. The mass of oxygen in the compound is
(10.0 g) – (.440 + 2.60) g = 6.96 g, corresponding to (6.96 g / 16 g mol–1) =
.435 mol. Inserting this quantity into the preliminary formula gives C.217H.436O.435 . Allowing for experimental- and roundoff error, this reduces to CH2O2 .
Comment: This problem may at first seem very complicated, but it’s really just a combination of a number of almost trivially-simple calculations, carried out in a logical sequence. Your ability to construct this sequence is an essential part of solving practical problems of these kinds. it’s worth taking the time to work through this exerciset on your own, perhaps starting with 5.0 g of the compound, which will produce proportionally smaller quantites of products.
Problem Example 14: Simplest formula from element masses
A 4.67-g sample of an aluminum compound was found to contain 0.945 g of Al and 3.72 g of Cl. Find the simplest formula of this compound. Atomic weights: Al = 27.0, Cl=35.45.
Solution: The sample contains (.945 g)/(27.0 g mol–1) = .035 mol of aluminum and (3.72 g)(35.45) = 0.105 mol of chlorine. The formula Al.035Cl.105 expresses the relative numbers of moles of the two elements in the compound. It can be converted into the simplest formula by dividing both subscripts by the smaller one, yielding AlCl3.
Simplest formulas from mass ratios
The composition of a binary (two-element) compound is sometimes expressed as a mass ratio. The easiest approach here is to treat the numbers that express the ratio as masses, thus turning the problem into the kind described immediately above.
Problem Example 15: Simplest formula from element mass ratio
A compound composed of only carbon and oxygen contains these two elements in a mass ratio C:H of 0.375. Find the simplest formula.
Solution: Express this ratio as 0.375 g of C to 1.00 g of O.
moles of carbon: (.375 g)/(12 g/mol) = .03125 mol C;
moles of oxygen: (1.00 g)/(16 g/mol) = .0625 mol O
mole ratio of C/O = .03125/.0625 = 0.5;
this corresponds to the formula C0.5O, which we express in integers as CO2.
Simplest formulas from percent composition
The composition-by-mass of a compound is most commonly expressed as weight percent (grams per 100 grams of compound). The first step is again to convert these to relative numbers of moles of each element in a fixed mass of the compound. Although this fixed mass is completely arbitrary (there is nothing special about 100 grams!), the ratios of the mole amounts of the various elements are not arbitrary: these ratios must be expressible as integers, since they represent ratios of integral numbers of atoms.
Problem Example 16: Simplest formula from mass-percent composition
Find the simplest formula of a compound having the following mass-percent composition. Atomic weights are given in parentheses.
36.4 % Mn (54.9), 21.2 % S (32.06), 42.4 % O (16.0)
Solution: 100 g of this compound contains:
Mn: (36.4 g) / (54.9 g mol–1) = 0.663 mol
S: (21.2 g) / (32.06 g mol–1) = 0.660 mol
O: (42.4 g) / (16.0 g mol–1) = 2.65 mol
The formula Mn .663S.660 O 2.65 expresses the relative numbers of moles of the three elements in the compound. It can be converted into the simplest formula by dividing all subscripts by the smallest one, yielding Mn 1.00S1.00 O 4.01 which we write as MnSO4.
Note: because experimentally-determined masses are subject to small errors, it is usually necessary to neglect small deviations from integer values.
Problem Example 17: Simplest formula from mass-percent composition
Find the simplest formula of a compound having the following mass-percent composition. Atomic weights are given in parentheses.
27.6 % Mn (54.9), 24.2 % S (32.06), 48.2 % O (16.0)
Solution: A preliminary formula based on 100 g of this compound can be written as
Mn (27.6 / 54.9) S(24.2 / 32.06) O(42.4 / 16.0) or Mn.503S.754 O3.01
Dividing through by the smallest subscript yields Mn 1S1.5 O 6 . Inspection of this formula suggests that multiplying each subscript by 2 yields the all-integer formula Mn2S3O12.
More on elemental analysis
Elemental analysis in the laboratory
One of the most fundamental operations in chemistry consists of breaking down a compound into its elements (a process known as analysis) and then determining the simplest formula from the relative amounts of each kind of atom present in the compound. In only a very few cases is it practical to carry out such a process directly: thus heating mercury(II) sulfide results in its direct decomposition: 2 HgS → 2Hg + O2. Similarly, electrolysis of water produces the gases H2 and O2 in a 2:1 volume ratio.
Most elemental analyses must be carried out indirectly, however. The most widely used of these methods has traditionally been the combustion analysis of organic compounds. An unknown hydrocarbon CaHbOc can be characterized by heating it in an oxygen stream so that it is completely decomposed into gaseous CO2 and H2O. These gases are passed through tubes containing substances which absorb each gas selectively. By careful weighing of each tube before and after the combustion process, the values of a and b for carbon and hydrogen, respectively, can be calculated. The subscript c for oxygen is found by subtracting the calculated masses of carbon and hydrogen from that of the original sample.
Right: Illustration from Mark Bishop’s An Introduction to Chemistry: an online textbook.
This page also contains detailed examples of empirical formula calculations.
Figure 3.5 Combustion Analysis Experiment
Figure 3.6 Combustion Analysis Instrument
← Since the 1970s, it has been possible to carry out combustion analyses with automated equipment. This one can also determine nitrogen and sulfur.
For analyses of compounds containing elements other than C, H, and O, spectroscopic methods based on atomic absorption and inductively-coupled plasma atomic absorption are now widely used.
The analytical balance
Measurements of mass or weight have long been the principal tool for understanding chemical change in a quantitative way. Balances and weighing scales have been in use for commercial and pharmaceutical purposes since the beginning of recorded history, but these devices lacked the 0.001-g precision required for quantitative chemistry and elemental analysis carried out on the laboratory scale.
The classic equal-arm analytical balance and set of calibrated weights
Figure 3.7 Equal arm Balance
Figure 3.8 Weights for Balance
It was not until the mid-18th century that the Scottish chemist Joseph Black invented the equal arm analytical balance. The key feature of this invention was a lightweight, rigid beam supported on a knife-edged fulcrum; additional knife-edges supported the weighing pans. The knife-edges greatly reduced the friction that limited the sensitivity of previous designs; it is no coincidence that accurate measurements of combining weights and atomic weights began at about this time.
Analytical balances are enclosed in a glass case to avoid interference from air currents, and the calibrated weights are handled with forceps to prevent adsorption of moisture or oils from bare fingers.
Anyone who was enrolled in college-level general chemistry up through the 1960’s will recall the training (and tedium) associated with these devices. These could read directly to 1 milligram and allow estimates to ±0.1 mg. Later technical refinements added magnetic damping of beam swinging, pan brakes, and built-in weight sets operated by knobs. The very best research-grade balances achieved precisions of 0.001 mg.
Beginning in the 1970’s, electronic balances have come into wide use, with single-pan types being especially popular. A single-pan balance eliminates the need for comparing the weight of the sample with that of calibrated weights. Addition of a sample to the pan causes a displacement of a load cell which generates a compensating electromagnetic field of sufficient magnitude to raise the pan to its original position. The current required to accomplish this is sensed and converted into a weight measurement. The best research-grade electronic balances can read to 1 microgram, but 0.1-mg sensitivities are more common for student laboratory use.
Figure 3.9 Digital Analytical balance
An empirical formula expresses the relative ratios of the atoms in a compound or a molecule. The ratio of the atoms within the compound and molecule is held for the atomiclevel as well the molar level.
MgCl2 is made of 1 atom of Mg (magnesium atom) and 2 atoms of Cl (chlorine atom) à (atomic level)
MgCl2 is made of 1.0 mole of Mg and 2.0 moles of Cl à (molar level)
Knowing the molar ratio of each element within a molecule or a compound, one can determine the empirical formula of such molecule or compound.
Steps how to calculate the empirical formula from known mass percent (%):can be seen below:
Figure 3.10 Concept Map for Empirical Formula Determination
The videos below illustrates with examples the concept of the empirical formula:
Calculation of the percent composition from the molecular formula:
Examples:
Calculate the % composition in 3 significant figures:
- the percent composition of ammonia, NH3
- the percent composition of Na2S2O3
- Determine the percent water in CuSO4∙5H2O to three significant figures.
1.The percent composition of ammonia, NH3
First one has to determine the molar mass of the NH3 which is:
1N + 3H = [1x 14.0] g/mole + [3 x 1.00] g/mole = 17.0 g/mole
N% = { [1 x 14.0 g/mole] / [17.0 g/mole] } x 100% = 82.4 %
H% = { [3 x 1.00 g/mole] / [17.0 g/mole] } x 100% = 17.6 %
2.The percent composition of Na2S2O3
First one has to determine the molar mass of the Na2S2O3 which is:
[2 Na + 2 S + 3 O] = [2 x 23.0] g/mole + [2 x 32.0] g/mole + [3 x 16.0] g/mole = 158.0 g/mol
Na% = { [2 x 23.0 g/mole] / [158.0 g/mole] } x 100% = 29.1%
S% = { [2 x 32.0 g/mole] / [158.0 g/mole] } x 100% = 40.5%
O% = { [3 x 16.0 g/mole] / [158.0 g/mole] } x 100% = 30.4 %
3.Determine the percent water in CuSO4∙5H2O to three significant figures.
First one has to determine the molar mass of the CuSO4∙5H2Owhich is:
CuSO4 molar mass:
[1 Cu + 1 S + 4 O] = [1 x 63.5 g/mole] + [1 x 32.0 g/mole] + [4 x 16.0 g/mole] = 159.6 g/mole
5 H2O molar mass:
5 x [2 H + 1 O] = 5{ [2 x 1.00 g/mole] + [1 x 16.0 g/mole] } = 5 x 18.0 g/mole = 90.0 g/mole
Total molar mass of CuSO4∙5H2O = [159.6 g/mole + 90.0 g/mole] = 249.6 g/mole
H2O% = { [90.0 g/mole] / [249.6 g/mole] } x 100% = 36.1 %
Calculation of the percent composition from the masses of the elements:
Example:
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?
C% = { [3.01 g C] / [24.81 g compound] } x 100% = 12.1 %
O% = { [4.00 g O] / [24.81 g compound] } x 100% = 16.1%
Cl% = { [17.81 g Cl] / [24.81 g compound] } x 100% = 71.8%
How to Calculate the Empirical Formula from the mass percent of each element or actual mass of each element:
Example:
Chloroform has percent composition of chloroform is 10.06% carbon, 0.85% hydrogen, and 89.09% chlorine. Determine its empirical formula
- Assume that the sample 100 g (corresponds to 100%):
Then:
- C = 10.06 g
- H = 0.85 g
- Cl = 89.09 g
- Convert numbers of grams of each into moles by dividing the grams by the atomic mass of each element in the molecule or the compound:[use 3 significant figures]
C = 10.06 g / [1 mole / 12.0 g] = 0.838 moles
H = 0.85 g / [1 mole / 1.00 g] = 0.850 moles
Cl = 89.09 g / [1 mole / 35.5 g] = 2.51 moles
- Divide number of moles of each element by the smallest number of moles [in this problem 0.838 moles is the smallest number of moles]:
C = 0.838 moles / 0.838 moles = 1.00
H = 0.850 moles / 0.838 moles = 1.01 rounded as 1.00
Cl = 2.51 moles / 0.838 moles = 2.995 = rounded as 3.00
- The empirical formula is C1H1Cl3 or CHCl3 [Chloroform]
Example:
3.23 g of a compound has 0.728 grams of phosphorous and 2.50 grams chlorine. What is the empirical formula of this compound?
Convert the grams of each element into moles by dividing each element by its atomic mass:
P = 0.728 g / [mole / 31.0 g] = 0.0235 moles [3 sig. figs]
Cl = 2.50 g / [mole / 35.5 g] = 0.0704 moles [3 sig. figs]
The ratio of the elements within this compound is:
P0.0235Cl0.0704
Divide by smallest number of moles:
P = 0.0235 moles / 0.0235 moles = 1.00
Cl = 0.0704 moles / 0.0235 moles = 2.995 = 3.00
The empirical formula of this compound is:
P1Cl3 or PCl3
Calculating Molecular Formulas for a Compound by Using its Empirical Formula
What is the difference between Empirical Formula and Molecular Formula?
Empirical Formula: The smallest whole number ratio of elements within a compound or a molecule or a formula unit (ion).
Molecular Formula: The actual number of the atoms of each element within a compound or a molecule or a formula unit (ion).
The relationship between Molecular Formula (molar mass) and Empirical Formula (empirical formula mass) is given below:
Molecular or molar mass (amu or g/mole) =[n formula units/molecule] x [empirical formula mass (amu or g/mole]
The video below illustrates this relationship:
Example:
The empirical formula of hexane C3H7. The molecular weight is 86.2 amu.
What is the molecular formula of hexane?
Empirical formula weight of hexane C3H7 = 3 C + 7 H = [3 x 12.0] + [7 x 1.00] = 43.0 amu.
Following the formula given above:
86.2 amu / 43.0 amu = n = 2.00
Molecular formula = 2 x [C3H7] = C6H14
Example:
Reference: http://www.chem.uiuc.edu/rogers/Text6/Tx65/tx65fr.html
The compound dioxane contains only carbon, hydrogen, and oxygen. When 0.956 g dioxane is burned, 1.91 g carbon dioxide and 0.782 g water are formed. In another experiment, it was determined that 6.04×10-3 mol dioxane weighs 0.532 g. What is the molecular formula of dioxane?
1. Calculate the mass of carbon, hydrogen, and oxygen in 0.956 g dioxane.
2. Using the data from step 1 above, calculate the empirical formula of dioxane.
The empirical formula is:
C0.043H0.087O0.022
3. Calculate the molecular weight of dioxane.
4. Calculate the molecular formula of dioxane.
The empirical formula weight is 2(12.0) + 4(1.01) + 16.0 = 44.0. The molecular weight is 88.08. The ratio of molecular weight to empirical weight is:
Thus the molecular of dioxane formula is: C4H8O2.
Example:
The compound ethylene glycol is often used as an antifreeze. It contains 38.7% carbon, 9.75% hydrogen, and the rest oxygen. The molecular weight of ethylene glycol is 62.07 g. What is the molecular formula of ethylene glycol?
1. Calculate the empirical formula. Assume 100 g of the compound, which will contain 38.70 g carbon, 9.75 g hydrogen and the rest oxygen
? g O = 100 g – 38.70 g C – 9.75 g H = 51.55 g O.
2. Calculate the moles of each element present:
3. Next calculate the ratio of molecular weight to empirical formula weight. The molecular weight is given. The empirical formula is CH3O, so the empirical formula weight is 12.01 + 3(1.008) + 16.00 = 31.03.
Therefore the molecular formula is twice the empirical formula: C2H6O2