#### Empirical Formula of an Oxide, Chemistry tutorial

Introduction:

Chemical formulas point out the composition of compounds. A formula which provides only the simplest ratio of the relative number of atoms in a compound is the empirical formula or the simplest formula. The ratio generally comprises of small whole numbers. A formula which provides the actual numbers of each kind of atom in a compound is termed as the molecular formula. The numbers in a molecular formula will be whole number multiples of the numbers in an empirical formula. To find out the molecular formula of a compound, we require knowing both the empirical formula and the molar mass of the compound.

Theory:

A chemical formula is the simplest manner to express information regarding the atoms which comprise any given chemical compound. A chemical compound is comprised of atoms of two or more elements chemically joined in definite proportions. For illustration, H2O and CO are compounds while H2 is a diatomic element.

The atoms in a compound are held altogether via chemical bonds. The chemical formula helps recognize each and every constituent element via its chemical symbol and points out the number of atoms of each and every element in the compound. When a molecule contains more than one atom of a specific element, this number is pointed out by using a subscript after the chemical symbol. For illustration:

Copper (I) sulphide consists of a chemical formula of Cu2S. In this compound there are two atoms of copper and one atom of sulphur. Copper and sulphur are combined in a ratio of 2:1. This formula as well signifies that there are 2 moles of copper for each and every mole of sulphur.

The net mass of each and every element in a compound based on the number of its atoms. For illustration, in compounds, CO and CO2, the number of oxygen atoms are correspondingly, one and two. In another words, the mass of oxygen in CO2 is two times as much as in CO. Yet the other way to illustrate this is that one mole of carbon joins by one mole of oxygen in CO (that is, mole ratio: 1:1) and one mole of carbon joins by two moles of oxygen in CO2 (mole ratio: 1:2). This recommends that elements for all time combine in a definite proportions by mass or moles. Accordingly, the ratio of moles of the constituent elements in a compound is almost always a ratio of small, whole numbers.

The formula having the lowest possible whole number ratio is termed as the empirical formula. The empirical formula doesn't essentially point out the exact number of atoms in a single molecule. This information is represented by the molecular formula, which is always a simple multiple (n) of the empirical formula, that is, Molecular formula (MF) = Empirical formula (EF) x n. The value of 'n' can be derived by dividing the actual molar mass of the compound and empirical formula mass. The molar masses of compounds are obtained from different experimental methods comprising mass spectrometry method.

To comprehend the behavior of a chemical compound one should first know its chemical formula. The method of the discovery of chemical formula of a compound starts by the determination of the ratio of the individual elements in an acknowledged mass of the compound (that is, determination of empirical formula).

For most of the ionic compounds, empirical formula is similar as the compound's molecular formula. Though, for covalent compounds the empirical formula is not essentially similar as the molecular formula of the compound. For illustration, the empirical formula of water, H2O, is similar as its molecular formula. The empirical formula of hydrogen peroxide is HO; however its molecular formula is H2O2. To find out whether or not a covalent compound's empirical and molecular formulas are similar, we require knowing the molar mass of the compound.

To determine the empirical formula we should join the elements to form the compound under conditions which allow us to find out the mass of each element. From such data, the moles of each element might be found out. By dividing the moles obtained for each and every element by the smallest number of moles, we get quotients that are in a simple ratio of integers or they are simply changed to a simple ratio through multiplication method.

Sample computations: Determination of the empirical formula of a compound based on experimental data is illustrated below.

Illustration: A strip of aluminum weighing 0.690 g is ignited yielding an oxide which weighs 1.300 g. Compute the empirical formula of the compound formed. We will follow the given strategy:

a) Find out the mass of each and every element.

b) Convert masses of elements to their corresponding moles.

c) Divide each of the moles via the smallest mole number.

d) Multiply the mole ratios by whole number integers till a whole number is obtained for each and every element.

a) We should first find out the mass of aluminum and oxygen in the oxide formed. Given we began with 0.690 g of aluminum; it yielded 1.300 g of an oxide.

Mass of oxygen = Mass of product (aluminum oxide) - mass of aluminium

That is, Mass of oxygen = 1300 g - 0.690 g

= 0.610 g of oxygen

b) Now that we are familiar with the mass of aluminum and oxygen, we can compute the moles of each element

Moles of Al = Mass of Al x (1 mol Al/Gram atomic mass of Al)

= 0.690 g Al (1 mol Al/26.982g Al)

= 0.0256 mol Al

Moles of O = Mass of O x (1 mol O/Gram atomic mass of O)

= 0.610 g O (1 mol O/15.999 g O)

= 0.0381 mol O

c) As we now are familiar with the moles of each element, ratio of atoms might be found out by dividing the moles of each element by the smallest number of moles.

Aluminium = 0.0256 mol/0.0256 mol = 1.00

Oxygen = 0.0381 mol/0.0256 mol = 1.49

The ratio is 1.00 atom Al: 1.49 atoms O

This ratio can be modified to a whole number ratio by multiplying by a factor of 2.

That is, (1.00 atom: 1.49 atoms O) x 2

2 atom: 2.98 atoms O;

Simple rounding provides 2 atom Al: 3 atoms O

Thus the empirical formula of Aluminum oxide made Al2O3

Concept of the experiment:

In this experiment, after a known mass of magnesium is burned, the product will consist of magnesium oxide together with a small amount of magnesium nitride (Mg3N2). Next, we add water to convert this small amount of nitride to magnesium hydroxide, Mg(OH)2 with the liberation of ammonia, NH3 (the water will not react with the magnesium oxide). Moreover, heating will cause conversion of the hydroxide to the oxide by the loss of gaseous water. In summary, the given reactions occur:

Reaction of magnesium by molecular oxygen:

Mg (s) + O2 (g) → 2MgO (s)

Reaction of magnesium by molecular nitrogen:

3Mg (s) + N2 (g) → Mg3N2 (s)

Reaction of magnesium nitride by water:

Mg3N2 (s) + H2O (l) → Mg(OH)2 (s) + NH3 (g)

Reaction on heating magnesium hydroxide:

Mg(OH)2 (s) + Δ → MgO (s) + H2O (l)

In the end, all magnesium will have been transformed to magnesium oxide. We can then compute the mass of oxygen which is present in the oxide from its mass and the original mass of the magnesium. The laws of conservation of mass, and also the concept of a mole, will lead you to the technique by which you can find out the empirical formula of this oxide.

Materials and Equipment:

Balance, wire gauze, furnace, crucible, crucible tongs, sandpaper, magnesium ribbon, medicine dropper and tweezers.

Procedure:

1) Get a crucible, wash, rinse and dry it.

2) Heat the crucible for around 5 min in a furnace at around 200°C. Take away the crucible from the furnace and let the crucible to cool (5 to10 min).

CAUTION: Avoid burning your fingers. Don't touch the crucible at any time throughout this experiment.

3) While waiting, weigh around 0.2 g of magnesium ribbon (if it is not bright, clean the surface by sandpaper). Record the mass.

4) Whenever the crucible is cool, transfer it to the pan of a balance by employing crucible tongs as holding wire gauze under the crucible, but don't place the wire gauze on the pan. If you should wait to use the balance, don't put the crucible directly on the bench.

5) Evaluate and record the mass of the crucible.

6) Repeat the steps 2, 4 and 5 till two consecutive masses differ by no more than +0.01 g or any other precision which is stipulated via your laboratory instructor. Record the mean or average value of these two masses. You will make use of this outcome in the subsequent computations.

7) Fold the magnesium ribbon to a loose ball which will fit fully within the crucible.

8) Weigh and record the mass of the crucible by the magnesium.

9) Return the crucible to the furnace, by employing the crucible tong.

10) Heat the crucible by its contents in the furnace at around 800 °C for 45 min.

11) Take out crucible and place it on wire gauze and let it cool to room temperature. The contents must be white or slightly gray.

12) Add some drops of distilled water from a medicine dropper directly on the contents. The smell of ammonia might be obvious at this point.

13) Heat the crucible till the contents are dry. After that heat the crucible strongly for 8 to10 min in a furnace to transform the hydroxide to the oxide.

14) Permit the covered crucible and its contents to cool.

15) Weigh and record the new mass of the crucible by MgO.

16) Compute the empirical formula from your recorded masses.

17) Clean the crucible and all other items employed.

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