Moles and Molarity

AIMS AND OBJECTIVES

This introductory chapter describes the simple ideas of atoms and molecules, types of chemical formula and their molecular weight for students who have not studied chemistry before. Chemical equations and balanced chemical equations are introduced through the reactions used in an introductory practical laboratory course. The concepts of molarity and molar solutions are introduced through solving volumetric problems, to enable the student to start a laboratory course in practical Inorganic Chemistry.

STATES OF MATTER

Chemistry is the science and study of the material world. It is generally accepted that there are three states of matter, solid, liquid and gaseous, and the chemicals that make up the materials of the world involve the chemical elements or molecules.

ELEMENTS, ATOMS AND MOLECULES

The physical state of an element relates to the three states of matter, and the precise state for an element is largely determined by the temperature. Thus at room temperature the element iron is a solid, bromine is a liquid and fluorine is a gas.

In the gaseous state at room temperature helium (He) is a mono-atomic gas, and the formula of the element helium is written as He. However, the gaseous form of hydrogen and oxygen at room temperature involves diatomic molecules, namely, H2 and O2. This difference is largely determined by the individual electron configuration of the elements, and their ability to form bonds to each other, rather than remain (in the gaseous state) as atomic species of the elements.

The way in which the elements of the Periodic Table react together is largely determined by the electron configuration of the individual elements as this determines the ratio in which two elements combine to form a molecule:

Atom 1 + Atom 2 [right arrow] Molecule H + Cl [right arrow] HCl 2 Atom H + 1 Atom O [right arrow] 1 Molecule H2O

The number of atoms of each element in a molecule determines the ratio of the elements in the molecule and is referred to as the stoichiometry of the molecule. In the molecule of HCl the ratio of H:Cl is 1:1, and the molecule has a stoichiometry of 1:1. In H2O the ratio of H:O is 2:1, and its stoichiometry is 2:1.

ELEMENTS, MIXTURES AND COMPOUNDS (MOLECULES)

An element consists of only one type of atom, i.e. helium, hydrogen or iron. A mixture may contain more than one type of substance that can be physically separated into its components, whereas a compound contains more than one type of element, usually with a definite stoichiometry, and cannot be separated into its elements by any simple physical method. Thus the element iron may be obtained as a magnetic black powder that can be mixed with yellow sulfur to give a blackish yellow mixture, from which the iron metal can be separated by means of a magnet. However, if the mixture is heated, a reaction occurs to give a black solid of FeS, iron(II) sulfide, on cooling, from which the iron present cannot be separated by the use of a magnet. The black solid FeS is referred to as a compound of Fe and S which has lost the properties of the elemental Fe and S and has unique properties of its own. Similarly, molecules of H2 and O2 react to give molecules of water, H2O:

2H2(g) + O2(g) [right arrow] 2H2O(1)

but while H2 and O2 are gases at room temperature, H2O is a liquid. In these new compounds the compound elements are said to have reacted chemically together to give a new compound, FeS and H2O, respectively, with definite stoichiometries between the atoms, namely, 1:1 in FeS and 2:1 in H2O.

SIMPLE CHEMICAL NAMES

The most simple compounds are those which contain only two elements, one metallic and one non-metallic (explained later). The metal is given the full element name, and the non-metal has the ending -ide.

Thus: NaCl sodium chloride

MgO magnesium oxide

CaS calcium sulfide

BN boron nitride

If the stoichiometry of the two elements is not 1:1, prefixes are used thus:

1:1 mono – carbon monoxide CO

1:2 di – carbon dioxide CO2

1:3 tri – sulfur trioxide SO3

1:4 tetra – carbon tetrachloride CCl4

1:5 penta – phosphorus pentachloride PCl5

1:6 hexa – sulfur hexafluoride SF6

Note: where more than one atom is present the number is written as a post-subscript.

Compounds with more than two elements cannot end in -ide and for those where the third element is oxygen, the endings -ite or -ate are used:

magnesium sulfide MgS

magnesium sulfite MgSO3

magnesium sulfate MgSO4

CATIONS AND ANIONS

In compounds such as NaCl, the lattice is made up of cations (positively charged species) of Na+ and anions (negatively charged species) of Cl-, Na+Cl-, such that the formula, NaCl, has an overall neutral charge. In Na2SO4 the overall neutral charge is maintained, but the compound contains two Na+ cations to one SO42- anion, with the latter referred to as an oxyanion, in this case a sulfate oxyanion. In aqueous solution the oxyanions occur as discrete species, in the case of the sulfate anion with a 2 – negative overall charge.

TYPES OF CHEMICAL FORMULA

In chemistry, different types of chemical formula are used to give different types of information.

(a) Empirical Formula: this is the simplest whole number ratio of the atoms in a molecule; thus in ethanoic acid the empirical formula is CH2O.

(b) Molecular Formula: this is the actual number of atoms making up the molecule; thus in ethanoic acid the molecular formula is C2H4O2i.e. twice the empirical formula.

(c) Structural Formula: this shows the various ways of representing the actual arrangement of atoms in the molecule, i.e.

[FORMULA NOT REPRODUCIBLE IN ASCII]

ATOMIC WEIGHT

The atomic weight or relative atomic mass of an element is the mass of one atom of that element relative to that of the most abundant form of carbon taken as 12 units. On this scale the atomic weight of hydrogen is 1, oxygen is 16, and copper is 63.54a.m.u. Table 1.1 lists the atomic weights of the first 18 elements of the Periodic Table.

On this scale the molecular formula of ethanoic acid, C2H4O2 has a molecular weight of:

2C(12) + 4H(1) + 20(16), i.e. (24 + 4 + 32 = 60)

namely, 60 atomic mass units (a.m.u.), i.e. the gram mole or molecular weight of ethanoic acid is 60. The gram mole of ethanoic acid is then 60 g and contains:

24 gram atoms of carbon

4 gram atoms of hydrogen

32 gram atoms of oxygen

Total: 60 grams.

AVOGADRO'S NUMBER

As the gram mole of a molecule (60 for ethanoic acid) is defined relative to the gram atom of carbon = 12 g, the actual number of atoms in 12 g carbon has been experimentally determined as 6.022 x 1023 atoms. This is referred to as Avogadro's Number, and is the number of atoms in the gram atomic weight of any element, i.e. 19 g fluorine, 32 g sulfur or 63.5 g copper. It then follows that the number of molecules in the gram molecular weight of a molecule (1 gram mole) is also 6.022 x 1023, Avogadro's Number. Thus one mole of ethanoic acid, 60 g, contains 6.022 x 1023 molecules of ethanoic acid. Equally, one mole of dihydrogen, H2, 2g, one mole of water, H2O, 18 g, and one mole of sulfuric acid H2SO4, 98 g, each contains 6.022 x 1023 molecules.

It also follows that 1 g of a molecule will contain Avogadro's Number divided by the gram molecule weight (1 mole) of the molecule:

[??] 1 g ethanoic acid contains 6.022 x 1023/60 molecules = 1.0037 x 1022 molecules

Likewise:

1 g hydrogen (0.5 1 gram mole) contains 3.011 x 1023 molecules 1 g sulfuric acid (1/98 1 gram mole) contains 6.145 x 1021 molecules.

EMPIRICAL FORMULA

This only expresses the relative number of atoms of each element in a compound. Nevertheless, it is the first step in the experimental determination of the molecular formula of a compound from its percentage composition.

1. Thus: A contains 42.9% C and 57.1% O; calculate its empirical/molecular formula

Atomic Wt. % %/At. Wt.Ratio

Carbon 12 42.9 42.9/12 = 3.58 1

Oxygen 16 57.1 57.1/16 = 3.58 1

[??] Empirical formula is C1 O1, or CO (carbon monoxide).

2. A contains 79.9% C and 20.1YO H:

Atomic Wt. % %/At. Wt.Ratio

Carbon 12 79.9 79.9/12 = 6.67 1

Hydrogen 1 20.1 20.1/1 = 20.1 3 CH3

3. A contains 37.5% C; 12.5% H; 50.0% O:

Atomic Wt. % %/At. Wt.Ratio

Carbon 12 37.5 37.5/12 = 3.12 1

Hydrogen 1 12.5 12.5/1 = 12.5 4

Oxygen 16 50.0 50.0/16 = 3.12 1 CH4O

4. A contains 43.7% P; 56.3% O:

Atomic Wt. % %/At. Wt.Ratio

Phosphorus 31 43.7 43.7/31 = 1.4 2

Oxygen 16 56.3 56.3/16 = 3.5 5 P2O5

5. Given the molecular formula of ethanoic acid, CH3CO2H what percentages of C, H and N are present?

CH3CO2H [equivalent to] C2H4O2 [equivalent to] 2 x CH2O

Empirical weight = 12 + 2 + 16 = 30 and the molecular weight = 24 + 4 + 32 = 60.

% C = 24/60 x 100 = 40.0% C
% H = 4/60 x 100 = 6.67% H
% O = 32/60 x 100 = 53.3% O

CHEMICAL EQUATIONS

Chemistry involves the study of the ways in which the elements and compounds react with each other. We have already seen:

[FORMULA NOT REPRODUCIBLE IN ASCII]

in which two pairs of elements react to form a compound. Some more complicated balanced equations are:

[FORMULA NOT REPRODUCIBLE IN ASCII]

Notice because of the balancing of charges, 1 mole of each of the reactants produces 2 moles of NaCl. Equally:

[FORMULA NOT REPRODUCIBLE IN ASCII]

Such reactions contain a great deal of information; thus in the reaction:

[FORMULA NOT REPRODUCIBLE IN ASCII]

could be represented alternatively:

[FORMULA NOT REPRODUCIBLE IN ASCII]

in a structural notation. It also contains more quantitative information:

1. 1 mole N2 reacts with 3 moles H2 to give 2 moles NH3;

2. 28g (1 mole) N2 reacts with 6g (3 moles) H2 to give 34g (2 moles) NH3;

3. 1 g N2 requires 6/28 g H2 for complete reaction to give 34/28 g NH3;

4. 1 g N2 in excess H2 will only yield 34/28 g NH3.

BALANCING CHEMICAL EQUATIONS

Such chemical equations must obey certain rules:

1. The reactants are written to the left-hand side, LHS, the products to the right-hand side, RHS, of the reaction arrow [right arrow].

2. Each side of the equation must have the same number of each kind of atoms, i.e. the equation must balance.

3. The common gaseous elements are shown as diatomic – H2, O2, N2, C12 – and solid elements as atoms – C, P, S, Cu or alternatively as C∞, P4, S8, Cu∞.

4. The overall ionic charges must be the same on each side of the equation.

For example, to balance the equation:

Al + HCl [right arrow] AlCl3 + H2.

steps 1–4 must be followed:

1. The products involve 3Cl, while the reactants involve only 1Cl [??] Al + 3HCl [right arrow] AlCl3 + H2

2. The reactants involve 3H, the products 2H [??] Al + 2 x 3HCl [right arrow] AlCl3 + 3H2

3. The reactants involve 6Cl, the products 3Cl [??] 2Al + 6HCl [right arrow] 2AlCl3 + 3H2

and the equation is now balanced.