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### 2. Formulae.

posted Mar 14, 2016, 2:33 PM by Upali Salpadoru   [ updated Aug 29, 2019, 12:38 AM ]

What is a Chemical Formula?

A molecule, which is a cluster of atoms, can be described by the use of                                   atomic symbols. Chemical formulas name the atoms that have joined and their ratios.

A formula gives the elements          present and the combining ratios of      the Atoms.

Chlorine gas is di-atomic. So we write the formula of chlorine molecule as Cl2.

1.   The molecular formula in the figure   shows 2 molecules of water.

2.  The structural formula of CCl4  shows that the carbon atom is combined to four atoms of chlorine by co-valent bonds.Each short line depicts a pair of electrons.

Fig.1. Three different ways of showing Formulas.

2 H2O is the most common way to show the molecule of water.          A number  in front gives the number of molecules.  The subscript no .2 shows that there are 2 H atoms in each molecule.

How to calculate the percentage of the elements in a compound?

( according to mass)

Refer to the Periodic Table for the Relative Atomic masses.

Use the Relative atomic masses.  H = 1.08, O = 16.

Formula mass of H2O = 2x1.08 + 16,                   = 2. 16 + 16                                                                                                                                   = 18.16,

The percentage of Hydrogen =  (2.16 / 18.16) x 100.                                                           =                                                      =   11.89 %

The percentage of Oxygen   = 16/ 18.16 x 100                                                                                                                            =88.1 %

The formula of a compound is found by working this the other way.

How to determine the formula of a compound?

Let us take and examples:

Example.1

Fig.2 Magnesium burning inside a crucible. When magnesium burns, a white coloured solid, magnesium oxide is formed. The formula is shown here.

The equation for the reaction:

Magnesium + Oxygen = Magnesium oxide
Here are the readings of an experiment performed.
Mass of crucible                             =  10.5 g
Mass of crucible + Magnesium        =   13.5 g
Mass after burning                        =    15.5 g

 Mass ratio Mass of Magnesium          =   3.0 gMass of Oxygen combined =   2.0 g    Mg: O    =  3 : 2 Particle RatioAtomic mass of Mg= 24.Atomic mass of O  = 16Mg: O  = 3/24  :   2/16.           =  1/8  : 1/8           =  1 : 1Formula = Mg1 O1  =  MgO

Explanation.

Let us take an example. mass 10 g. mass 2 g

John has bought 500 g. of strawberries and 500 g. of grapes to be packeted and distributed among some children..
 A strawberry has a Mass of  10 g .      A grape has a Mass of  is 2 g.Can you find the number of fruits in each category? Calculating the number of fruits.Mass of straw berries  ........= 500gMass of one berry……………   =  10gNumber of berries=   500g ÷ 10 g                            =  50 fruits.Mass of one grape …………    = 2g  Number of grapes ……………= 500g ÷   2                                                 =  250 fruits

The number ratio of fruits strawberries : grapes =  50 : 250    =   1: 5
If you are to packet  according to availability  it should be in this ratio.
Each packet will have   =  strawberry 1 grapes  5

If we use S for strawberry and G for grapes,

The formula of a packet will be S1G5     This can be given as  S G5

This is similar to a chemical formula.

If you can get the mass ratio of the ingredients of a pure compound  and their molecular masses, it is a simple process to get the formula.

Let us take and example:

 Number of Particles =  Total Mass ÷  Mass of a particle.

Example 2.

Finding the percentage composition of Ammonium sulphate.

Molecular mass of (NH4)2SO4…………………………………=  132

 Element Mass proportion Percentage by mass N-14 14x2=28 28x100/132 = 21.21% H-  1 1x8=8 8 ÷ 132 x 100 =  6.06 % S-32 S=32 32 ÷  132 x 100=  24,24% o=16 4x16=  64 64÷ 132 x 100= 48.48 %

Use this Periodic Table to get the Relative atomic mass.

Some in alphabetical order for your convenience. (Without decimals)

 Ag Al Ar Au B Ba Be Br C Ca Cd Cl Cu Cr Fe H He Hg 108 27.0 40.0 197 10.8 137 9.0 80 12.0 40. 112 35.5 64 52.0 56 1.01 4.0 201

 I K Li Mg Mn N Na O P Pb S Sn Si Zn 127 39. 6.9 24.3 54.0 14.0 23.0 16.0 31.0 207 32. 119 26. 119

Example 3.

A chemical consisting of three elements  had the  following percentages by mass.

Na = 43.40 %     C = 11.30 %   O  =  45.3 %

Find the formula of the compound.

 Element Percentage.by Mass Mass÷  Mass of an atom simple ratio Na-23 43.40 % 43.4/ 23 = 1.89 2 C- 12 11.30 % 11.3/ 12 = 0.942 1 O- 16 45.3 % 45.3/ 16=2.83 3

Formula of the chemical is       =  Na2 C1 O3    (We don’t have to write the 1 for C.  )

Na2CO3         This is the formula of Sodium carbonate

Verification

The total of percentages must add up to 100.

Nitrogen… =  21.21 Hydrogen…=    6.06Sulphur ..=  24.24 Oxygen……=  48.48

Total =         99.99    gets rounded up to 100.

1.0  Determine the percentage composition of these compounds.

 Compound Molecule Molecular mass Percentages ExampleLead dioxide PbO2 207+ 16.x 2           =239 Pb. =   207 ÷ 239 x 100   =  86.6 % 2xO   =    32  ÷ 239 x 100 = 13.4 % 1.Calcium carbonate CaCO3 a. Ca =        b.  C  =       c.3xO =      d. 2. Potassium              permanganate. KMnO4 e. K =          f. Mn =        g.4x O  =     h. 3. Potassium ferri    cyanide K3Fe(CH)6 i. 3xK=       j. Fe  =       k. 6xC =      l.6xH  =      m. 4.Octane. C8H18 96+18=114 8xC  =    n.18xH  =    O.

2x15=30 marks

2.0   Find the formulae of these;  Mass ratio or the percentages by mass are given.

 Elements Ratio Number  Ratio Simple ratio Formula ExampleHg - 201O -16 Hg= 50 g  O = 4 g. Hg  50 ÷ 201= 0.25 O    4 ÷  16 =  0.25 0.25:0.25 = 1:1 HgO H -1 O- 16 H = 20 gO=320g. H = 20 ÷ 1 =20320 ÷ 16 =20 a. b. Cu-O- Cu = 1.6 O=  0.2 c. d. Ca-40. O - 16 H- 1 Ca=54.05% O= 43.24% H = 2.71% e. f. Pb-207 O- 16 Pb= 2.07 O = 0.32 g. h. H -1 O = 16 H= 20 g O = 320 i. j.
3x10 = 30

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3.   On heating 50g of mercury oxide 46 g  of mercury was obtained. Derive the formula of the mercury oxide.

4.  On heating 1.6g of copper 1.8g of a copper oxide was formed. Find the formula of the oxide.

5.  By reducing 2.39 mg of lead 2.07 mg of lead was obtained. Find the empirical formula of the oxide.

6.A compound has the following percentage composiion. Derive the empirical formula.

Carbon =  40%       Hydrogen = 6.7 %    Oxygen= 53.3 %

8x4 =32

7. These are the 3 common isotopes of Hydrogen.

Although the water forming from them would be chemically the same mass ratios will not be the same.

Calculate the Oxygen percentage in the 3 types of water.

 1H2O 2H2O 3H2O සමස්කඵානික ස්කන්ධය 18 20 v. ඔක්සිජන් ස්කන්ධය .  16 .16 16 ඔක්සිජන්  ප්‍රතිශතය w. x. y.

සැලකිිය යුතුයි:- සාමාන්‍ය හයිඩ්‍රජන් සැම්පලයක H-1 සමස්ඵානික 99.99% ක්ම තිබ‍ෙන බැව් සැලකිිය යුතුය.

2x4= 8 Marks

Answers:1. HgO,    2. H1O1, (There is no compound like this. This compound really is H2O2,)    3. Cu2O,   4.  Ca O2H2  (This really is Ca(OH)2 ,    5.  PbO.

 Q.No. 1.1 1.2 2.1 2.2 2.3 3.1 4.1 4.2 5.1 5.2 Answer 86.61% 13.38% 40 % 12% 48 * 27.27 % 40.51 % 40.51 g 35.8 47.6

3. Steps for the calculation.

Mass ratio of  Hg:O  is  ---:-----

Atom ratio is  ----/201 : -----/16  =  ----- : ------

Get the approximate ratio leaving for experimental errors.

Therefore the formula must be ---------