3.5 Chemical Formulae

Chemical Formulae
■ Chemical Formulae

	►	To represent a chemical compound.
	►	The elements present in the compound which are denoted by their symbols.
	►	The relative numbers of each element present in the compound which are indicated by subscripts written after the symbols.

■ Example

	►	Several covalent compounds and their chemical formulae.
	►	Number of atoms of each element making up the compound.
		Example of electron configuration and valence electrons. Compound Chemical formula Number of each elements Oxygen ${\text{O}}_2$ 2 oxygen atom Chlorine ${\text{Cl}}_2$ 2 chlorine atom Carbon dioxide ${\text{CO}}_2$ 1 carbon atom and 2 oxygen atom Nitrogen dioxide ${\text{NO}}_2$ 1 nitrogen atom and 2 oxygen atom Sulphuric Acid ${\text{H}}_2{\text{SO}}_4$ 2 hydrogen atom, 1 sulphur atom and 4 oxygen atom Nitric Acid ${\text{HNO}}_3$ 1 hydrogen atom, 1 nitrogen atom and 3 oxygen atom

■ There are two types of chemical formula:

	(a)	Empirical formula.
	(b)	Molecular formula.

Empirical formula and Molecular formula
■ Empirical formula:

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Definition: Empirical formula: Shows the simplest ratio of the atoms to the elements that combine to form the compound.

■ Molecular formula:

	►	Definition: Molecular formula: Shows the actual number of atoms of the elements that combine to form the compound.
	►	The molecular formula is the multiple of the empirical formula. ${\text{Molecular formula = (Empirical formula)}}_{\mathrm{n}}$ , where n is an integer.
		Example : ○ Empirical formula of glucose, ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ = ${\text{CH}}_2\text{O}$ [Divide the number of atom with 6 to get simplest ratio of atoms] ○ Molecular formula of glucose = ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$

■ The following steps should be followed to obtain the empirical formula of a compound:

	►	Step 1: Find the mass or percentage of elements in compound.
	►	Step 2: Divide mass or percentage by relative atomic mass(RAM) to find number of moles.
	►	Step 3: Find smallest ratio by dividing the number obtained in Step 2 by the smallest number of all.
	►	Step 4: Write empirical formula

	Laboratory Activity 3.5.1: Determining the empirical formula of copper oxide
	Laboratory Activity 3.5.2: Determining the empirical formula of magnesium oxide

✍ Worked-example 3.5(a)

In an experiment, a metal X combines with 4.32g oxygen to form 13.66g of a metallic oxide of X. Find the empirical formula of the metallic oxide.[RAM: X = 52, O = 16] Solution Element X O Mass 13.66 – 4.32 = 9.34g 4.32g Number of moles $\begin{array}{c}\frac{9.34}{52}\text{=}0.18\end{array}$ $\begin{array}{c}\frac{4.32}{16}\text{=}0.27\end{array}$ Ratio $\begin{array}{c}\frac{0.18}{0.18}\text{=}1.00\end{array}$ $\begin{array}{c}\frac{0.27}{0.18}\text{=}1.50\end{array}$ Simplest ratio 2 3 Empirical formula of compound is ${\text{X}}_2{\text{O}}_3$

✍ Worked-example 3.5(b)

		A compound ${\text{C}}_{\mathrm{x}}{\text{H}}_{\mathrm{y}}{\text{O}}_{\mathrm{z}}$ contains 40% carbon and 53.3% oxygen. If the relative molecular mass of the compound is 180, find its (a) empirical formula (b) molecular formula [RAM: H, 1; C, 12; O = 16] Solution (a) Empirical formula Element C H O Mass 40% 100 - 53.5 - 40 = 6.7% 53.5% Number of moles $\begin{array}{c}\frac{40}{12}\text{=}3.33\end{array}$ $\begin{array}{c}\frac{6.7}{1.0}\text{=}6.7\end{array}$ $\begin{array}{c}\frac{53.3}{16.0}\text{=}3.33\end{array}$ Ratio $\begin{array}{c}\frac{3.33}{3.33}\text{=}1.0\end{array}$ $\begin{array}{c}\frac{0.27}{0.18}\text{=}1.5\end{array}$ $\begin{array}{c}\frac{3.33}{3.33}\text{=}1.0\end{array}$ Simplest ratio 1 2 1 Empirical formula of compound is ${\text{C H}}_2\text{O}$
		(b) Molecular formula Assume that the molecular formula of compound ${\text{C}}_{\mathrm{x}}{\text{H}}_{\mathrm{y}}{\text{O}}_{\mathrm{z}}$ is ${(\text{C}{\text{H}}_2\text{O})}_{\text{n}}$ The relative molecular mass of X is 180. Thus, $\mathrm{(12\; +\; (2}\times \mathrm{1)\; +\; 16)n\; =\; 180}$ $\mathrm{30n\; =\; 180}$ $\mathrm{n\; =\; 6}$ Thus the molecular formula of compound is ${(\text{C}{\text{H}}_2\text{O})}_{\text{6}}$ or ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$

Ionic Formulae
■ Method of writing ionic formulae

	►	Can be written if the charge of the cation(positively charged ion) and the anion(negatively charged ion) forming the ionic compound are known.
	►	Cation Anion Charge + Charge 2+ Charge 3+ Charge 4+ Charge 1- Charge 2- Charge 3- ${\text{H}}^{\mathrm{+}}$ ${\text{Na}}^{\mathrm{+}}$ ${\text{K}}^{\mathrm{+}}$ ${\text{Cu}}^{\mathrm{+}}$ ${\text{Ag}}^{\mathrm{+}}$ ${{\text{NH}}_4}^{\mathrm{+}}$ ${\text{Ca}}^{\mathrm{2+}}$ ${\text{Mg}}^{\mathrm{2+}}$ ${\text{Zn}}^{\mathrm{2+}}$ ${\text{Cu}}^{\mathrm{2+}}$ ${\text{Fe}}^{\mathrm{2+}}$ ${\text{Pb}}^{\mathrm{2+}}$ ${\text{Mn}}^{\mathrm{2+}}$ ${\text{Hg}}^{\mathrm{2+}}$ ${\text{Ni}}^{\mathrm{2+}}$ ${\text{Ba}}^{\mathrm{2+}}$ ${\text{Al}}^{\mathrm{3+}}$ ${\text{Fe}}^{\mathrm{3+}}$ ${\text{Cr}}^{\mathrm{3+}}$ ${\text{Pb}}^{\mathrm{4+}}$ ${\text{Sn}}^{\mathrm{4+}}$ ${\text{Mn}}^{\mathrm{4+}}$ ${\text{Cl}}^{\mathrm{-}}$ ${\text{Br}}^{\mathrm{-}}$ ${\text{I}}^{\mathrm{-}}$ ${\text{OH}}^{\mathrm{-}}$ ${\text{HCO}}^{\mathrm{-}}$ ${{\text{NO}}_3}^{\mathrm{-}}$ ${{\text{CO}}_3}^{\mathrm{2-}}$ ${{\text{SO}}_4}^{\mathrm{2-}}$ ${{\text{O}}_2}^{\mathrm{2-}}$ ${{\text{PO}}_4}^{\mathrm{3-}} \;$ ${{\text{PO}}_3}^{\mathrm{3-}}$ IMPORTANT NOTE: MEMORIZE ALL the charges of cation and anion.

■ To write the chemical formula of an ionic compound, the following steps can be used.

	(a)	Write the formulae of the ions involved in forming the compound and their charges.
	(b)	Then balances the positive and negative charges.
	(c)	Finally write the chemical formula of the ionic compound without the charges.

■ The animation below shows the method of writing chemical formulae.

Naming of Chemical compounds
■ Naming of chemical compound

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Based on the recommendation of the International Union of Pure and Applied Chemistry(IUPAC) .

■ Naming of ionic compound

	►	The name of the cation comes first, followed by the name of the anion. Cation Anion Name of ionic compound Sodium ion Chloride ion Sodium chloride Magnesium ion Carbonate ion Magnesium carbonate
	►	Roman numerals are used in naming of the metals than form more than one type of ions. Cation Anion Name of ionic compound iron(II) ions, ${\text{Fe}}^{\mathrm{2+}}$ Oxide ion Iron(II) oxide iron(III) ions, ${\text{Fe}}^{\mathrm{3+}}$ Oxide ion Iron(III) oxide

■ Naming of molecular compound

	►	In naming of the simple molecular compound, the name of the first element comes first, followed by the second element with an 'ide'.: ○ HCl – Hydrogen chloride ; HI – Hydrogen iodide ○ CO – Carbon monoxide ; ${\text{CO}}_2$ – Carbon dioxide.
	►	Greek prefix are used to show the number of atoms of each element in a compound.

← 3.4 Number of Moles and Its Molar Volume

3.6 Chemical Equations →

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