Equivalent Weight and Atomic Weight

Equivalent Weight:

-It is defined as number of parts by weight of that element it is combined or displaced by $1.008$ part by weight of hydrogen or $8$ parts by weight of oxygen or $35.5$ parts by weight of Chlorine.

-In general;

$\frac{Eq . wt o f X}{Eq . wt o f Y} = \frac{wt . o f X}{wt . o f Y}$

i) $Eq . wt o f e l e m e n t = \frac{A t . wt}{v a l e n cy}$

ii) $Eq . wt o f a c i d = \frac{M o l . wt}{N o . o f H ^{+} d i s pl a ce d p er m o l ec u l e} = \frac{M o l . wt}{B a s i c i t y o f a c i d}$ (Basicity=No. of $O H^{-}$ ion reqd. by one molecule of acid for complete neutralization or simply no. of $H^{+}$ displaced.)

iii) $Eq . wt o f ba se = \frac{M o l . wt}{A c i d i t y o f ba se}$ (Acidity=No. of $H^{+}$ combined per molecule per molecule of base)

iv) $Eq . wt o f s a lt = \frac{M o l . wt}{N o . o f + v e or - v e c ha r g e}$

v) $Eq . wt o f r a d i c a l = \frac{M a ss o f r a d i c a l}{N o . o f c ha r g e in r a d i c a l}$

vi) $Eq . wt o f o x i d i z in g an d re d u c in g a g e n t = \frac{M o l . wt}{C ha r g e ( in cre a se / d ecre a se in O . N . )}$

Note:

a) Equivalent weight of $K M n O_{4}$ in: ( $M o l . wt M n = 158$ )

-Basic Medium= $e q . wt = \frac{M}{1}$

-Neutral Medium= $e q . wt = \frac{M}{3}$

-Acidic Medium= $e q . wt = \frac{M}{5}$ (Trick to remember: BAN- $135$ )

b) Equivalent weight of $K_{2} C r_{2} O_{7}$ in any medium is $e q . wt = \frac{M}{6}$ ( $M o l . wt C r = 52$ )

Determination of equivalent weight by hydrogen displacement method:

-Metals that lies above hydrogen in electro-chemical series can displace hydrogen from dilute acid.

$M + 2 H^{+} ⟶ M^{n +} + H_{2} ↑$

$Eq . wt = \frac{wt . o f m e t a l}{wt . o f h y d ro g e n e v o l v e d} \times 1.008$

$wt . o f H_{2} d i s pl a ce d = (V \times 0.000089) g m$

$∴ Eq . wt = \frac{wt . o f m e t a l}{V _{NTP} \times 0.000089} \times 1.008 g m$

Determination of Equivalent Weight by Oxide Formation Method:

$Eq . ma ss o f m e t a l = \frac{wt . o f m e t a l in o x i d e}{wt . o f o x y g e n in o x i d e} \times 8$

Determination of Equivalent Weight Chlorine formation method:

$Eq . wt o f m e t a l = \frac{wt . o f m e t a l}{wt . o f c h l or in e} \times 35.5$

Atomic Weight:

-It is the ratio of the average mass of a chemical element ’s atoms to some standard.

-It is the relative weight of elements expressed using $\frac{1}{12} t h o f C - 12$ atom as reference element.

To Determine Atomic Weight/Mass:

$A t o mi c w e i g h t o f g a seo u s e l e m e n t = \frac{m o l ec u l a r w e i g h t}{a t o mi c i t y}$

Atomicity is defined as the total number of atoms present in a molecule. For example, each molecule of oxygen $(O_{2})$ is composed of two oxygen atoms. Therefore, atomicity of oxygen is $2$ .

$A t o mi c ma ss = e q u i v a l e n t ma ss \times v a l e n cy$
Dulong-Petit law:

- $A t o mi c w e i g h t \times s p ec i f i c h e a t = 6.4$ Or, $A t o mi c h e a t = 6.4$

-This law is only applicable to solids except Boron, Carbon and Silicon.

Molecular Mass\Weight:

It is the relative weight of substance and is algebraic sum of atomic weight of all atoms present in the molecule.
Expressed in terms of grams is gram molecular weight.
Molecular weight is number of times a molecule of substance is heavier than $\frac{1}{12} t h o f C - 12$ atom.
$1 am u = \frac{1}{12} \times w e i g h t o f C - 12 i so t o p e = 1.66 \times 1 0^{- 24} g m = 1.66 \times 1 0^{- 27} k g$
$N o . o f m o l es (a t o m) = \frac{w e i g h t g i v e n in g m}{g r am a t o mi c w e i g h t}$
$N o . o f m o l es (m o l ec u l e) = \frac{w e i g h t g i v e n in g m}{g r am m o l ec u l a r w e i g h t}$
$N o . o f m o l es (g a s) = \frac{g i v e n v o l u m e ( L i t res ) in NTP}{22.4 L i t res}$
$1 m o l e = 6.023 \times 1 0^{23}$ (ions/atoms/molecules) chemical units
$1 m o l e (a t o m) = a t o mi c w e i g h t in g r am \to 1 g r am a t o m = 1 m o l e (a t o m)$
$1 m o l e (m o l ec u l e) = m o l ec u l a r w e i g h t in g r am \to 1 g r am m o l ec u l e = 1 m o l e (m o l ec u l e)$

To Determine Molecular Weight:

For gases only : $M o l ec u l a r w e i g h t = 2 \times Va p or De n s i t y = 2 \times [\frac{wt . ( in g m ) o f V m l o f v o l a t i l e s u b s t an ce}{V \times 0.000089}] g m$
Victor Meyer's method ( for volatile liquid & solids ):

$M o l ec u l a r W e i g h t = \frac{m RT}{P V}$ where,

- $R = 0.0821 l i t er a t m m o l^{- 1} K^{- 1}$

- $P$ = Pressure in atm, $T$ = Temperature in Kelvin, $V$ = Vol. In litres

Diffusion Method ( for gases only ):

- $\frac{r _{1}}{r _{2}} = \frac{M _{2}}{M _{1}}$ where,

$r_{1} \to$ Rate of diffusion of gas $1$

$r_{2} \to$ Rate of diffusion of gas $2$