Key to Symbol, Formula, Valency and Equation
Elements and Compounds
• Elements and Compounds are pure substances.
• An element is a pure substance that cannot be decomposed into simpler substances.
• Elements are listed in the Periodic Table of the Elements.
• Each element has a 1 or 2 letter symbol.
• A compound is a pure substance that can be decomposed into simpler substances.
• Compounds are made up of two or more elements.
Elements
An element is a pure substance that cannot be decomposed (broken down) into simpler substances.
There are 92 naturally occurring elements.
Each element has been given a 1 or 2 letter symbol:
• the first letter of the symbol is always a capital letter
eg, H for hydrogen, C for carbon, N for nitrogen
• if there is a second letter in the symbol it is a lower case letter
eg, He for helium, Ca for calcium, Ne for neon
Elements can be present in nature as solids, liquids or gases.
Liquid Elements Gaseous Elements Solid elements
2 elements exist in nature as liquids: 11 elements exist in nature as gases: all other elements exist in nature as solids, eg:
mercury (Hg)
bromine (Br) hydrogen (H)
helium (He)
nitrogen (N)
oxygen (O)
fluorine (F)
neon (Ne)
chlorine (Cl)
argon (Ar)
krypton (Kr)
xenon (Xe)
radon (Rn) lithium (Li)
carbon (C)
sodium (Na)
magnesium (Mg)
aluminium (Al)
silicon (Si)
phosphorus (P)
sulfur (S)
potassium (K)
calcium (Ca)
zinc (Zn)
The atmosphere is mostly made up of the elements nitrogen (~78%) and oxygen (~21%).
Common elements found in the earth's crust are:
• oxygen (O)
• silicon (Si)
• aluminium (Al)
• iron (Fe)
• calcium (Ca)
• sodium (Na)
• potassium (K)
• magnesium (Mg)
• hydrogen (H)
The most common elements found in living things are:
• carbon (C)
• hydrogen (H)
• oxygen (O)
• nitrogen (N)
• phosphorus (P)
• sulfur (S)
The most common elements found in the universe are:
• hydrogen (H)
• helium (He)
• oxygen (O)
• carbon (C) Some Elements
Name Symbol
hydrogen H
helium He
lithium Li
beryllium Be
boron B
carbon C
nitrogen N
oxygen O
fluorine F
neon Ne
sodium Na
magnesium Mg
aluminium Al
silicon Si
phosphorus P
sulfur S
chlorine Cl
argon Ar
potassium K
calcium Ca
scandium Sc
titanium Ti
vanadium V
chromium Cr
manganese Mn
iron Fe
cobalt Co
nickel Ni
copper Cu
zinc Zn
gallium Ga
germanium Ge
arsenic As
selenium Se
bromine Br
krypton Kr
rubidium Rb
strontium Sr
zirconium Zr
silver Ag
tin Sn
barium Ba
tungsten W
platinum Pt
gold Au
mercury Hg
lead Pb
uranium U
Compounds
Compounds are pure substances made up of 2 or more elements.
Each compound has a formula showing which elements are present in the compound
Examples of some common compounds are shown below.
compound name compound formula elements present
water H2O hydrogen (H) and oxygen (O)
ammonia NH3 nitrogen (N) and hydrogen (H)
carbon monoxide CO carbon (C) and oxygen (O)
carbon dioxide CO2 carbon (C) and oxygen (O)
sodium chloride NaCl sodium (Na) and chlorine (Cl)
sodium hydroxide NaOH sodium (Na), oxygen (O) and hydrogen (H)
calcium chloride CaCl2 calcium (Ca) and chlorine (Cl)
calcium carbonate CaCO3 calcium (Ca), carbon (C) and oxygen (O)
calcium nitrate Ca(NO3)2 calcium (Ca), nitrogen (N) and oxygen (O)
calcium phosphate Ca3(PO4)2 calcium (Ca), phosphorus (P) and oxygen (O)
calcium sulfate CaSO4 calcium (Ca), sulfur (S) and oxygen (O)
methane CH4 carbon (C) and hyrogen (H)
ethanol C2H5OH carbon (C), hydrogen (H) and oxygen (O)
A compound can be decomposed into simpler pure substances.
For example, an electric current can be passed through water to form the elements hydrogen and oxygen.
Water can be decomposed into hydrogen and oxygen.
Water is a compound made up of hydrogen and oxygen.
Shortest expression of atom is symbol.
Formula is written with symbol and valence of different atom.
How can you learn symbol and valence with atomic number?
Ser. Symbol Val ence Ser. Symbol Val ence Ser. Symbol Val ence Ser. Symbol Val ence
1. H 1
Adding 8 with Sr. No.
9. F 1
17. Cl 1 2. He 0
Adding 8 with Sr. No.
10. Ne 0
18. Ar 0 3. Li 1
Adding 8 with Sr. No
11. Na 1
19 K 1 4. Be 2
Adding 8 with Sr. No.
12. Mg 2
20. Ca 2
5. B 3
Adding 8 with Sr. No
13. Al 3
21. Sc 3 6. C 2, 4
Adding 8 with Sr. No
14. Si 2,4 7. N 3,5
Adding 8 with Sr. No
15. P 3,5 8. O 2
Adding 8 with Sr. No
16. O 2,4,6
Molecular Mass (Molecular Weight)
• In theory, the relative molecular mass or molecular weight of a compound is the mass of a molecule of the compound relative to the mass of a carbon atom taken as exactly 12.
• In practice, the molecular mass, MM, (molecular weight, MW) of a compound is the sum of the atomic masses (atomic weights) of the atomic species as given in the molecular formula.
• In theory we can only refer to the Molecular Mass or Molecular Weight of a covalent compound since only covalent compounds are composed of molecules.
Formula Mass (Formula Weight)
• The relative formula mass, FM, (formula weight, FW) of a compound is the sum of the atomic masses (atomic weights) of the atomic species as given in the formula of the compound.
• Formula Mass (Formula Weight) is a more general term that can be applied to compounds that are not composed of molecules, such as ionic compounds.
• In practice, the terms, molecular mass, molecular weight, formula mass and formula weight are used interchangeably by Chemists.
Molecular Mass Calculations
Calculate the Molecular Mass (MM) of the compound carbon monoxide, CO
• The formula for carbon monoxide is composed of one atom of carbon and one atom of oxygen
• Atomic mass carbon = 12.01 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
• Molecular Mass (MM) for cabon monoxide
= atomic mass carbon + atomic mass oxygen
Molecular mass (MM) = 12.01 + 16.00
= 28.01 g/mole
Calculate the Molecular Mass (MM) of the compound carbon dioxide, CO2
• The formula for carbon dioxide is composed of one atom of carbon and two atoms of oxygen
• Atomic mass carbon = 12.01 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
• Molecular Mass (MM) for carbon dioxide
= atomic mass carbon + (2 x atomic mass oxygen)
Molecular Mass (MM) = 12.01 + (2 x 16.00)
= 12.01 + 32.00
= 44.01g/mole
Calculate the Molecular Mass (MM) of the compound water, H2O
• The formula for water is composed of two hydrogen atoms and one oxygen atom
• Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
• Molecular Mass (MM) for water
= (2 x atomic mass hydrogen) + atomic mass oxygen
Molecular Mass (MM) = (2 x 1.008) + 16.00
= 2.016 + 16.00
= 18.016g/mole
Calculate the Molecular Mass (MM) of the compound calcium hydroxide, Ca(OH)2
• The formula for calcium hydroxide is composed of one calcium "atom" (actually an ion) and two hydroxide ions. Each hydroxide ion is composed of one hydrogen "atom" (actually an ion) and one oxygen "atom" (also an ion)
• Atomic mass calcium = 40.08 (from the Periodic Table)
Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
• Molecular Mass (MM) for calcium hydroxide
= atomic mass calcium + (2 x atomic mass oxygen) + (2 x atomic mass hydrogen)
Molecular Mass (MM) = 40.08 + (2 x 16.00) + (2 x 1.008)
= 40.08 + 32.00 + 2.016
= 74.096g/mole
Alternatively, Moelcular Mass = atomic mass of calcium + (2 x molecular mass of hydroxide ions)
Molecular Mass (MM) = 40.08 + [2 x (16.00 + 1.008)]
= 40.08 + [2 x 17.008]
= 40.08 + 34.016
= 74.096g/mole
Calculate the Molecular Mass (MM) of the compound ammonium sulfate, (NH4)2SO4
• The formula of ammonium sulfate is composed of two atoms of nitrogen, eight atoms of hydrogen, one atom of sulfur and four atoms of oxygen
• Atomic mass nitrogen = 14.01 (from the Periodic Table)
Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of sulfur = 32.06 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
• Molecular Mass (MM) for ammonium sulfate
= (2 x atomic mass nitrogen) + (8 x atomic mass hydrogen) + atomic mass sulfur + (4 x atomic mass oxygen)
Molecular Mass (MM) = (2 x 14.01) + (8 x 1.008) + 32.06 + (4 x 16.00)
= 28.02 + 8.064 + 32.06 + 64.00
= 132.144g/mole
Alternatively, Molecular Mass (MM) = [2 x molecular mass ammonium ions (NH4)] + atomic mass sulfur + (4 x atomic mass oxygen)
Molecular Mass (MM) = {2 x [14.01 + (4 x 1.008)]} + 32.06 + (4 x 16.00)
= {2 x [14.01 + 4.032]} + 32.06 + 64.00
= {2 x 18.042} + 32.06 + 64.00
= 36.084 + 32.06 + 64.00
= 132.144g/mol
Calculate the Molecular Mass (MM) of the compound ethanoic acid (acetic acid), CH3COOH
• The formula of ethanoic acid is composed of two carbon atoms, four hydrogen atoms and two oxygen atoms
• Atomic mass carbon = 12.01 (from the Periodic Table)
Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
• Molecular Mass (MM) for ethanoic acid
= (2 x atomic mass carbon) + (4 x atomic mass hydrogen) + (2 x atomic mass oxygen)
Molecular Mass (MM) = (2 x 12.01) + (4 x 1.008) + (2 x 16.00)
= 24.02 + 4.032 + 32.00
= 60.052g/mol
Percent Composition (Percentage Composition)
The percent composition (percentage composition) of a compound is a relative measure of the mass of each different element present in the compound.
To calculate the percent composition (percentage composition) of a compound
• Calculate the molecular mass (molecular weight, formula mass, formula weight), MM, of the compound
• Calculate the total mass of each element present in the formula of the compound
• Calculate the percent compositon (percentage composition): % by weight (mass) of element
= (total mass of element present ÷ molecular mass) x 100
Examples
Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)
• Calculate the molecular mass (MM):
MM = 22.99 + 35.45 = 58.44
• Calculate the total mass of Na present:
1 Na is present in the formula, mass = 22.99
• Calculate the percent by weight of Na in NaCl:
%Na = (mass Na ÷ MM) x 100 = (22.99 ÷ 58.44) x 100 = 39.34%
• Calculate the total mass of Cl present:
1 Cl is present in the formula, mass = 35.45
• Calculate the percent by weight of Cl in NaCl:
%Cl = (mass Cl ÷ MM) x 100 = (35.45 ÷ 58.44) x 100 = 60.66%
The answers above are probably correct if %Na + %Cl = 100, that is,
39.34 + 60.66 = 100.
Calculate the percent by weight of each element present in sodium sulfate (Na2SO4).
• Calculate the molecular mass (MM):
MM = (2 x 22.99) + 32.06 + (4 x 16.00) = 142.04
• Calculate the total mass of Na present:
2 Na are present in the formula, mass = 2 x 22.99 = 45.98
• Calculate the percent by weight of Na in Na2SO4:
%Na = (mass Na ÷ MM) x 100 = (45.98 ÷ 142.04) x 100 = 32.37%
• Calculate the total mass of S present in Na2SO4:
1 S is present in the formula, mass = 32.06
• Calculate the percent by weight of S present:
%S = (mass S ÷ MM) x 100 = (32.06 ÷ 142.04) x 100 = 22.57%
• Calculate the total mass of O present in Na2SO4:
4 O are present in the formula, mass = 4 x 16.00 = 64.00
• Calculate the percent by weight of O in Na2SO4:
%O = (mass O ÷ MM) x 100 = (64.00 ÷ 142.04) x 100 = 45.06%
The answers above are probably correct if %Na + %S + %O = 100, that is,
32.37 + 22.57 + 45.06 = 100
Calculate the percent by weight of each element present in ammonium phosphate [(NH4)3PO4]
• Calculate the molecular mass (MM) of (NH4)3PO4:
MM = 3x[14.01 + (4 x 1.008)] + 30.97 + (4 x 16.00) = 3 x [14.01 + 4.032] + 30.97 + 64.00 = (3 x 18.042) + 30.97 + 64.00 = 54.126 + 30.97 + 64.00 = 149.096
• Calculate the total mass of N present:
3 N are present, mass = 3 x 14.01 = 42.03
• Calculate the percent by mass of N present in (NH4)3PO4:
%N = (mass N ÷ MM) x 100 = (42.03 ÷ 149.096) x 100 = 28.19%
• Calculate the total mass of H present:
12 H are present in the formula, mass = 12 x 1.008 = 12.096
• Calculate the percent by mass of H present in (NH4)3PO4:
%H = (mass H ÷ MM) x 100 = (12.096 ÷ 149.096) x 100 = 8.11%
• Calculate the total mass of P present:
1 P is present in the formula, mass = 30.97
• Calculate the percent by mass P in (NH4)3PO4:
%P = (mass P ÷ MM) x 100 = (30.97 ÷ 149.096) x 100 = 20.77%
• Calculate the total mass of O present:
4 O are present in the formula, mass = 4 x 16.00 = 64.00
• Calculate the percent by mass of O in (NH4)3PO4:
%O = (mass O ÷ MM) x 100 = (64.00 ÷ 149.096) x 100 = 42.93%
The answers above are probably correct if %N + %H + %P + %O =100, that is,
28.19 + 8.11 + 20.77 + 42.93 = 100
Here the Serial number is the atomic number of respective element.
During writing Formula the positive elements or the electropositive radical are placed at left side.
Mg /NH4+---nonmetal or negative radical.
Non metal or negative radicals are placed at right side.
Number of valence of each side transfer one to another.
Symbol Valence Symbol Valence
Ca - 2 P - 3
Valence transfer
Ca P
Formula – Ca3P2
Nomenclature of Bi elemental Compounds:
Compounds which are formed by only two type of elements are called bi elemental compound.like HCl, NaCl. H2S, H2O.
Nomenclature : Full Name of left element + half name of right element + ide
For example: Na2S – Sodium + sulph + ide = sodium sulphide.
Elemental ion: When an atom becomes electrically charged is called elemental ion or ion.
Ion are of two type: positive and Negative.
Nomenclature:
Positive Ion : Same as the name of atom.
Negative ion : Half of its name + ide
For example : Cl- - Chlor + ide = Chloride.
Radical : The group of different atoms which contain some charge and act like a single in chemical reaction.
Radicals are of two type: Positive and negative.
Nomenclature of Positive radical: Half of left part + nium
NH3H+-- ammo + nium =ammonium
PH4+ -- - Phosph + nium = Phosphonium
H3O+-- Hydro + nium = hydronium.
Nomenclature of Negative radical: negative radical are two type, oxradical or oxygen containing radicals and oxygenless radicals ( CN-, CNS-)
Oxygenless radical : after those name “ide” comes.
CN- - Cyan + ide = cyanide
CNS- -- Thiocyan + ide = thiocyanide
Oxoradicals or oxygen containing radical:
Generally after these radical “ate” added
In special cases “ite” added.
Rule: Half name of left element + ate / ite
(Generally ate is added if different radicals are formed with same elements then the radical which less amount of oxgen after those name “ite” is added)
SO42- = sulph + ate = Sulphate
SO32- = Sulph + ite = sulphite
PO43- = phosph + ate = phosphate
PO33- = phosph + ite = Phosphite
NO3- = Nitr + ate = Nitrate
NO2- = Nitr + ite = Nitrite
MnO4- = Mangan + ate = manganate
Same as carbonate, chlorate etc.
Nomenclature of ploy elemental compound:
poly elemental compounds are form with more than two type cof elements where minimum one compound radical remain present.
Because in organic compounds have two part
Like, Radical + element NH4Cl
Or Element + Radical Na2SO4
Or Radical + radical (NH4) 2SO4
If a student can identify the radical or ion in compound, he or she can write the name of these types of compounds easily.
Ca3(PO4)2 : the left part is Calcium and the right sided part is phosphate radical, So the name of the compound is Calcium Phosphate.
Valence of Radical: Radical must have charge; the number of charge on the radical is the valence.
Bi – radical: When the number of charge of the radical is reduced by 1 with a hydrogen or if after adding one hydrogen the radical remain charged then it is called bi- radical.
CO32- = Carbonate HCO3- = Bi-Carbonate
Di valent or tri valent radicals form bi radicals but mono valent radical do not form thiese type of radical rather those form compound.
NO3- = Nitrate adding H with NO3- = HNO3 is a compound because it has no charge.
Structural Formula: How does structural formula become easy?
Structural formula of some covalent compound is determined easily by calculating valency of constituent elements.
a. The element of highest valency are kept in centre
b. The second highest valency holder elements are placed of the surroundings of central atoms.
c. The third lowest valency holder are linked with the second highest elements
d. And so on link third to second to first drawing lines.
e. Now check the second highest valency holders so that they occupy all of their valencies.
Example: Structural Formula of H2SO4.
Step1. Here
valency Hydrogen is 1 third element
valency Oxygen is 2 Second element
Since sulphur is single atom and paled in centre, its valecncy may be more than 2.
Step -2. Now Keeping S in centre Four oxygen are to be placed at surroundings of S.
Step 3. Now third elements are placed to link with second elements
Step 4. Now link elements by single line one to another to reach the central elements.
Step 5. Since left and right sided O having two hand of each,
The upper O and the lower O must be filled with two hands.
Now see that the central S already filled with six hands. So the valency is 6.
By following these steps other some structural formula may be determined.
For self assessment write the structural formula of (i) HNO3 (ii) H2SO4 (iii) H2O2 (iv) HN3 (v) H3PO4 (vi) C2H6 (vii) C2H2 (viii) HClO4.