At the heart of chemistry are substances— elements or compounds— all of which have a definite composition which is expressed by a chemical formula. In this unit you will learn how to write and interpret chemical formulas both in terms of moles and masses, and to go in the reverse direction, in which we use experimental information about the composition of a compound to work out a formula.

 

How to read and write formulas

The formula of a compound specifies the number of each kind of atom present in one molecular unit of a compound. Since every unique chemical substance has a definite composition, every such substance must be describable by a chemical formula.

The well-known alcohol ethanol is composed of molecules containing two atoms of carbon, five atoms of hydrogen, and one atom of oxygen. What is its molecular formula?

Answer: C2H5O

Note that:

  • The number of atoms of each element is written as a subscript;
  • When only a single atom of an element is present, the subscript is omitted.
  • In the case of organic (carbon-containing) compounds, it is customary to place the symbols of the elements C, H, and O in this order at the left end of the formula.

Formulas of elements

The symbol of an element is the one- or two-letter combination that represents the atom of a particular element, such as Au (gold) or O (oxygen). The symbol can be used as an abbreviation for an element name (it is easier to write "Mb" instead of "molybdenum"!) In more formal chemical use, an element symbol can also stand for one atom, or, depending on the context, for one mole (Avogadro's number) of atoms of the element.

Different molecular forms of the same element (such as O2 and O3) are called allotropes.

Some of the non-metallic elements exist in the form of molecules containing two or more atoms of the element. These molecules are described by formulas such as N2, S6, and P4. Some of these elements can form more than one kind of molecule; the best-known example of this is oxygen, which can exist as O2 (the common form that makes up 21% of the molecules in air), and also as O3, an unstable and highly reactive molecule known as ozone. The soccer-ball-shaped carbon molecules sometimes called buckyballs have the formula C60.

Formulas of ions

Ions are atoms or molecules that carry an electrical charge. These charges are represented as superscripts in the ionic formulas. Thus:

Cl– the chloride ion, with one negative charge per atom
S2– the sulfide ion carries two negative charges
CO22– the carbonate ion— a molecular ion
NH4+ the ammonium ion

Note that the number of charges (in units of the electron charge) should always precede the positive or negative sign, but this number is omitted when the charge is ±1.

 

Empirical 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.

Sucrose (ordinary table sugar) is composed of molecular units having the formula C12H22O11. What is the empirical formula of sucrose?

Answer: CH2O

(Note: this simplest formula, which applies to all sugars, shows that this can be considered a compound of carbon and water, which explains why sugars are known as carbohydrates.)

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.

What formulas don’t tell us

The formulas we ordinarily write convey no information about the compound's structure— that is, order in which the atoms are connected by chemical bonds or are arranged in three-dmensional 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 natue of these two molecules:

More complex formulas

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 diethyl ether which both have the empirical formula C2H6O 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 and 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 any other 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.

What is the formula weight of copper(II) chloride, CuCl2?

Answer: the atomic weights of Cu and Cl are, respectively 63.55 and 35.45;
63.55 + 2(25.35) = 134.45.

What is the molar mass of copper(II) chloride, CuCl2?

Answer: the masses of Cu and Cl are, respectively, 63.55 g and 35.45 g;
(63.55 g) + 2(25.35 g) = 134.45 g.

Interpreting formulas in terms of mole ratios and mole fractions

The information contained in formulas can be used to compare the compositions of related compounds as in the following example:

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.

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.

Interpreting formulas in terms of masses of the elements

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.

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.

Find the mass fraction and mass percent 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:

How many tons of potassium are contained in 10 tons of KCl?

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.

How many grams of KCl will contain 10 g of potassium?

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.

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: 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

The simplest 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.

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.

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.

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 simplist 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.

In a student lab experiment, it was found that 0.5684 g of magnesium burns in air to form 0.9426 g of magnesium oxide. Find the simplest formula of this compound. Atomic weights: Mg = 24.305, O=16.00.

Solution: The mass of oxygen in the compound is found by difference: (.9426 – .5684)g = .3724 g.

The sample contains (.5684 g)/(24.305 g mol–1) = .0234 mol of aluminum and (.3724 g) / (16.00 g mol–1) = 0.0233 mol of chlorine. The formula Mg .0234O.0233 expresses the relative numbers of moles of the two elements in the compound. Neglecting the small difference due to experimental error, it is apparent that the mole ratio of the two elements is 1:1 and the simplest formula is MgO.

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.

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.

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.

It frequently happens that dividing the number of moles of each element by the smallest value does not yield an integer formula. In such cases the proper formula can almost always be found my multiplying through by a common factor as in the example below.

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 Mn 2S3 O12 .

 

What you need to know

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.

  • Explain why the symbol of an element often differs from the formula of the element..
  • Define an ion, and explain the meaning of its formula.
  • Find the simplest ("empirical") formula of a substance from a more complex molecular formula. Explain the meaning of the formula of an ionic solid such as NaCl.
  • Define molecular weight, formula weight, and molar mass. Calculate any of these from any chemical formula.
  • Given a chemical formula, express the mole ratios of any two elements, or the mole fraction of one of its elements.
  • Find the percentage composition of a compound from its formula.
  • Calculate the mass ratio of any two elements present in a compound from its formula.
  • Find the simplest formula of a binary comound from the mole ratio of its two elements, expressed as a decimal number.
  • Find the simplest formula of a binary comound from the mass ratio of its two elements.
  • Find the simplest formula of a compound from its mass- or percentage composition.

 

Concept Map

 

Creative Commons License
This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License.