## Chemical Formulas and Their Arithmetic

Read this text, which reviews how to write chemical formulas, how to determine information about the compound such as molar mass, mole ratios, and percent composition.

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.

#### Problem Example 1: Writing a Molecular 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?

Solution: Just write the symbol of each element, following by a subscript indicating the number of atoms if more than one is present. Thus: 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 if present,) O, N in this order in 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 HCO32– the hydrogen carbonate (bicarbonate) 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.

#### Formulas of Extended Solids and Minerals

In solid CdCl2, the Cl and Cd atoms are organized into sheets that extend indefinitely. Each atom is surrounded by six atoms of the opposite kind, so one can arbitrarily select any CdCl2 unit as the molecular unit. One such unit is indicated by the two red-colored bonds in the diagram, but it does not constitute a discrete molecule of CdCl2.

Many apparently simple solids exist only as ionic solids (such as NaCl) or as extended solids (such as CuCl2) in which no discrete molecules can be identified. The formulas we write for these compounds simply express relative numbers of the different kinds of atoms in the compound in the smallest possible integer numbers. These are identical with the empirical or simplest formulas that we discuss further on.

Many minerals and most rocks contain varying ratios of certain elements and can only be precisely characterized at the structural level. Because these are usually not pure substances, the formulas conventionally used to describe them have limited meanings.

For example the common rock olivine, which can be considered a solid solution of Mg2SiO4 and Fe2SiO4, can be represented by (Mg,Fe)2SiO4. This implies that the ratio of the metals to SiO 4 is constant, and that magnesium is usually present in greater amount than iron.

#### 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.

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 Do Not 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 any other 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: 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.)