# Stoichiometry

Stoichiometry Definition

The quantitative relationship existing between the quantities of the reactants and the products in a chemical reaction is known as stoichiometry .

In order to carry out stoichiometric calculations , one should be able to to write a balanced chemical reaction . Chemical reactions are represented in a concise way by chemical equations using the symbols and the formulae of the reactants and the products of the reaction .

Example : When water vapour is passed over heated carbon ( 1000oC) results in formation of carbon monoxide and hydrogen gas. This reaction is represents by the equation .

C(s)   +   H2O (g)    `|->` CO (g)   +  H2 (g)

On counting the number of atoms of carbon , hydrogen and oxygen on both sides of the equation , it ca be seen that , they are present in equal numbers before and after the reaction . A chemical equation must have equal number of atoms of each element on either side of arrow mark in the reaction . The equation where the equality of the number of atoms of elements is signified as described above is called a balanced chemical equation .

## Stoichiometry :Molar Volume

Definition : One gram molecular weight of any gaseous substance at standard temperature and pressure ( 273 K and 1 atm ) occupies 22.414 litres volume . This volume is known as molar volume . This is a constant for all ideal gases . This value is taken as 22.4 litres .

Example :       2 H2 (g)     +   O2 (g)     `|->`    2 H2O (g)

The ratio of the combining volumes as shown by the stoichiometric equation if one gram molecular weight of any gas at STP occupies one unit volume is

H2 (g)    :    O2 (g)      :    H2O   (g)

2 vol       :    1 vol        :    2 vol

Hence under STP conditions    2*22.4 litres of hydrogen gas combines with 1*22.4 litres of oxygen gas and forms 2*22.4 litres of water vapour . Basing on this , stoichiometric calculations can be carried out as illustrated by the following problem .

Problem : 1 litre of oxygen and 3 litres of CO measured at STP are reacted to get CO2 . Calculate the volumes of the gases after the reaction and also the weight of CO2 formed in the reaction .

Stoichiometric equation for the above reaction is

2CO (g)   +   O2 (g)    `|->` 2 CO2 (g)

Solution : As per the stoichiometric equation 2 vol. of carbon monoxide reacts with 1 vol. of oxygen to give 2 vol. of carbon dioxide .

Vol. of O2 before the reaction (STP)      =  1 litre

Vol. of CO before the reaction (STP)     =  3 litres

Vol. of CO2 before the reaction (STP)   =  0 litres

`:.` Volumes of gases after the reaction (STP)

Volume of O2    =    ( 1 - 1 )   =    0 litres

Volume of CO   =   ( 3 - 2 )    =    1 litre

Volume of CO2                       =    2 litres

Weight of 22.4 litres of CO2 at STP    =   44 gms

Weight of 2 litres of CO2 at STP     =     `(44gm*2lit)/(22.4lit)`    =   3.928 gm

## Stoichiometry :Mole concept

Introduction : A number of methods are available for measuring the amount of a given sample of a substance . Generally the amounts are expressed either in the units of volume or in the units of weight . In order to provide a scientific relationship between masses and volumes of substances and the number of particles present in them , chemists have found the mole concept more useful .

Gram atom :

One gram atomic weight of a substance is termed as gram atom .

Ex : Atomic weight of O2    =   16 amu

One gram atom of O2    =    16 gm of O2

Gram atomic weight of O2     =    16 gm

Gram molecule :

One gram molecular weight of a substances termed as one gram molecule .

Ex : Molecular weight of oxygen   =   32 amu

One gram molecule of oxygen   =   32 gm of oxygen

Gram molecule weight of oxygen    =   32 gm   .

By employing various experimental methods , scientists determined the number of atoms present in 12 gm of carbon - 12 isotope ( one gram atomic weight of carbon 12 isotope ) and found it be equal to 6.022*1023 . Later this value was confirmed . They also found that gram molecular weight or gram atomic weight of a substance contained the same number of chemical units . This is  a constant number known as Avogadro number , denoted by the letter N .

Thus 12 gm of C is the basic unit of the amount of substance in chemistry . The is a definite and is given a specified name mole .

Molar Mass : The mass of 1 mole of a substance expressed in grams is known as molar mass .

The molar mass of a substance is numerically equal to the formula weight of the substance in atomic units .

Formula weight : The formula weight of a substance is the total mass of all atoms present in the chemical formula of the substance .