Chemistry

CLASS 12 - CHEMISTRY
FORMULA BOOK
THE SOLID STATE  Simple d=a
 Calculation of number of particles per unit cell 3
 Contribution of each atom present on the  Body-centred d= a = 0.866 a
2
1 a
corner =  Face-centred d= = 0.707 a
8 2
 Contribution of each atom present on the
d
1  Relation between atomic radius r (= for
face = 2
2 pure elements) and edge (a) of cubic unit
 Contribution of each atom present on the cell
1 Simple Body-centred Face-centred
edge centre =
4
 Contribution of each atom present at the a 3 a
r= r = a r =
body centre = 1 2 4 2 2
 Relation between radius (r) of a void and = 0.433 a = 0.3535 a
the radius (R) of the spheres in the close
 Calculation of density of a cubic crystal from
packing
its edge
 Radius (r) of the tetrahedral void = 0.225R
Z M
 Radius (r) of the octahedral void = 0.414R  = a 3  N  1030 g/cm3
 Radius ratio rules 0

 For elements:
Radius ratio Coordination Structural
Z = no. of atoms/unit cell (1 for simple, 2
(r+ /r ) Number arrangement
for bcc and 4 for fcc)
0.155  0.225 3 Planar M = atomic mass of the element
triangular a = edge of the unit cell in pm
0.225  0.414 4 Tetrahedral N0 = Avogadro’s number.
0.414  0.732 6 Octahedral  For ionic compounds:
0.732  1 8 Body-centred Z = no. of formula units in one unit cell
cubic e.g. 4 for NaCl and ZnS, 1 for CsCl, etc.
M = formula mass (molecular mass) of
 Relation between no. of voids and spheres
the compound
in the packing +
a = edge = 2 × distance between Na and
 No. of octahedral voids = No. of atoms (ions) -
Cl in case of NaCl.
present in the close packing.
 Ionic and covalent radii :
 No. of tetrahedral voids = 2 × No. of atoms + –
 Interionic distance in A B = rA+ + rB–
(ions) in the close packing  Bond length of a covalent molecule
= 2 × No. of octahedral voids. (A–A) = 2 × rA
 Relation between nearest neighbour  Bond length of a covalent molecule
distance (d) and edge (a) of cubic unit cell (A–B) = rA + rB

, 2+
 No. of cation vacancies: Each ion like Sr
wB
+ –
when introduced into Na Cl creates one  Mass fraction of component B xB
cation vacancy because for electrical wA wB
2+ +
neutrality, one Sr replaces two Na ions. x A + xB = 1
SOLUTIONS  Parts per million (ppm) of substance A
Mass of A Vol. of A
wt. of the solute in g = 106 or 106
 % by wt. =  100 Mass of solution vol. of solution
wt. of the solution in g  Normality equation (for dilution of a solution
wt. of solute in g or for a complete reaction between two
 % by wt./vol. =  100
Vol. of solution in cc solutions)
Vol. of solute in cc N1V1 = N2V2.
 % by volume =  100
Vol. of solution in cc  Molarity equation (for dilution of a solution)
wt. of the solution in g M1V1 = M2V2.
 Strength of a solution =  If two non-reacting solutions of different
Vol. of solution in litres
Moles of the solute normalities are mixed, the normality of the
 Molarity =  1000 final solution can be calculated using
Vol. of solution in cc
relations :
Mass of the solute in g
where moles = N1V1 + N2V2 = N3V3.
Molecular mass of the solute
 According to Henry’s law : pA = kHxA
g. Eq. of the solute
 Normality = 1000 where kH = Henry’s constant
Vol. of solution in cc
 According to Raoult’s law, for a solution
Mass of the solute in g
where g eq. = containing volatile components A and B,
Eq. mass of the solute o
pA = xA pA and pB = xB pB .
o

o o
Mol. mass PTotal = pA + pB = xApA + xBpB
 Normality of a solution = Molarity × o o o o o
Eq. mass = (1 – xB)pA + xBpB = (pB – pA ) xB + pA .
 Normality of an acid = Molarity × Basicity Mole fraction of A in the vapour phase
pA
 Normality of a base = Molarity × Acidity =
pA  pB
Moles of solute  Raoult’s law for non-volatile solutes :
 Molality =  1000
Mass of the solvent in g p o  ps n2 n
  2
 Mole fraction of solute in solution p o
n1  n2 n1
n2 w2 M 2 (if solution is dilute i.e. < 5%)
x2
n1 n2 w1 M 1 w2 M 2 w2 M 2 w2 M 2
= w M w M
 Mole fraction of solvent in solution 1 1 2 2 w1 M 1
n1 w1 M 1 (if solution is dilute)
x1
n1 n2 w1 M 1 w2 M 2 n w
 Osmotic pressure, RT where n
where w1, M1 are mass and molecular mass of V M
solvent and w2, M2 for the solute. x1 + x2 = 1.  For isotonic solutions, 1 = 2 which means
at the same temp. C1 = C2.
 In general, for a solution containing many
components (A, B, C.....), mole fraction of A.  Elevation in boiling point, Tb = Kbm where
Kb = molal elevation constant and m = molality
nA
xA of the solution.
n A nB nC .... and so on.
xA + xB + ... = 1 K
 Units of Kb = Tb/m = deg./molality = mol kg 1
wA
 Mass fraction of component A x A
–1
w A wB = K kg mol .