Percentage composition and Molecular formula
Category : NEET
Percentage Composition and Molecular Formula
(1) Percentage composition of a compound : Percentage composition of the compound is the relative mass of each of the constituent element in 100 parts of it. If the molecular mass of a compound is M and B is the mass of an element in the molecule, then
For example, the empirical formula of glucose is \[C{{H}_{2\,}}O\]which shows that C, H and O are present in the ratio 1 : 2 : 1 in a molecule of glucose.
Empirical formula mass of a compound is equal to the sum of atomic masses of all atoms present in the empirical formula of the compound.
Calculation of the empirical formula : The empirical formula of a chemical compound can be deduced by knowledge of the, (i) Percentage composition of different elements. (ii) Atomic masses of the elements.
The following steps are involved in the calculation of the empirical formula,
Step I. Calculate the relative number of atoms or atomic ratio.
\[\text{Atomic ratio}=\frac{\text{Percentage of an element}}{\text{Atomic mass of the same element}}\]
Step II. Calculate the simplest atomic ratio.
Step III. Calculate the simple whole number ratio.
Step IV. Write the empirical formula.
(3) Molecular formula : The chemical formula that gives the actual number of atoms of various elements present in one molecule of the compound. For example, the molecular formula of glucose is \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\].
Relation between empirical and molecular formula : The molecular formula of a compound is a simple whole number multiple of its empirical formula.
Molecular formula =\[n\,\,\times \,\text{Empirical formula}\]; Where n is any integer such as 1, 2, 3….etc.
The value of ‘n’ can be obtained from the following relation: \[n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}\].
The molecular mass of a volatile compound can be determined by Victor Meyer’s method or by employing the relation, \[\text{Molecular mass}=\text{ 2}\times \text{Vapour density}\].
Calculation of the molecular formula : The molecular formula of a compound can be deduced from its :
(i) Empirical formula, (ii) Molecular mass
The determination of molecular formula involves the following steps:
Setp I. Calculation of empirical formula from the percentage composition.
Setp II. Calculation of empirical formula mass.
Setp III. Calculation of the value of ‘n’.
Setp IV. Calculation of molecular formula by multiplying the empirical formula of the compound by ‘n’.
Examples bad on Percentage composition & Molecular formula
Example:12 The oxide of a metal contains 40% by mass of oxygen. The percentage of chlorine in the chloride of the metal is [BIT Ranchi 1997]
(a) 84.7 (b) 74.7 (c) 64.7 (d) 44.7
Solution: (b) % of oxygen = \[\frac{8}{m\,+\,8}\,\,\times \,\,100\,\,=\,\,40\] or, \[40\,(m\,+\,8)\,=\,800\] or, \[m\,+\,8\,=\,20\] or, \[m\,=\,12\]
\[\therefore \] % of chlorine = \[\frac{35.5}{m\,+\,35.5}\,\times \,100\]= \[\frac{35.5}{12\,+\,35.5}\,\times \,100\] = 74.7 (Where m is the atomic mass of metal)
Example: 13 The empirical formula of an organic compound containing carbon and hydrogen is \[C{{H}_{2}}\]. The mass of one litre of this organic gas is exactly equal to that of one litre of \[{{N}_{2}}\]. Therefore, the molecular formula of the organic gas is [EAMCET 1985]
(a) \[{{C}_{2}}{{H}_{4}}\] (b) \[{{C}_{3}}{{H}_{6}}\] (c) \[{{C}_{6}}{{H}_{12}}\] (d) \[{{C}_{4}}{{H}_{8}}\]
Solution: (a) Molar mass of 1L of gas = mass of 1L \[{{N}_{2}}\]
\[\therefore \] Molecular masses will be equal i.e., molecular mass of the gas = 28, hence formula is \[{{C}_{2}}{{H}_{4}}\]
Example: 14 A sample of pure compound is found to have Na = 0.0887 mole, O = 0.132 mole, C = \[2.65\,\,\times {{10}^{22}}\]atoms.
The empirical formula of the compound is [CPMT 1997]
(a) \[N{{a}_{2\,}}C{{O}_{3}}\] (b) \[N{{a}_{3}}\,{{O}_{2}}\,{{C}_{5}}\] (c) \[N{{a}_{0.0887}}\,{{0}_{0.132}}\,{{C}_{2.65\,\times \,{{10}^{22}}}}\] (d) \[NaCO\]
Solution: (a) \[\because \] \[6.02\times {{10}^{23}}\] atoms of C = 1 mole of C
\[\therefore \] \[2.65\times {{10}^{22}}\] atoms of C = \[\frac{1\,\times 2.65\times {{10}^{22}}}{6.02\,\,\times {{10}^{23}}}\] mole = \[\frac{2.65}{6.02\times 10}=0.044\] mole
Now,
|
Element |
Relative number of moles |
Simplest ratio of moles |
|
\[Na\] |
0.0887 |
\[\frac{0.0887}{0.044}\,=\,2\] |
|
\[O\] |
0.132 |
\[\frac{0.132}{0.044}\,=\,3\] |
|
\[C\] |
0.044 |
\[\frac{0.044}{0.044}\,=\,1\] |
Thus, the empirical formula of the compound is\[N{{a}_{2}}C{{O}_{3}}\].
Example: 15 An organic compound containing C, H and N gave the following on analysis: C = 40%, H = 13.3% and
N = 46.67%. Its empirical formula would be [CBSE PMT 1999, 2002]
(a) \[CHN\] (b) \[{{C}_{2}}{{H}_{2}}N\] (c) \[C{{H}_{4}}N\] (d) \[{{C}_{2}}{{H}_{7}}\,N\]
Solution: (c) Calculation of empirical formula
| Element |
Symbol |
Percentage of element |
At. mass of elements |
Relative number of atoms = \[\frac{\text{Percentage}}{\text{At}\text{. mass}}\] |
Simplest atomic ratio |
Simplest whole number atomic ratio |
|
Carbon |
C |
40 |
12 |
\[\frac{40}{12}=3.33\] |
\[\frac{3.33}{3.33}=1\] |
1 |
|
Hydrogen |
H |
13.3 |
1 |
\[\frac{13.3}{1}=13.3\] |
\[\frac{13.3}{3.33}=4\] |
4 |
|
Nitrogen |
N |
46.67 |
14 |
\[\frac{46.67}{14}=3.33\] |
\[\frac{3.33}{3.3}=1\] |
1 |
Thus, the empirical formula is \[C{{H}_{4}}N\].
Example: 16 An organic substance containing C, H and O gave the following percentage composition :
C = 40.687%, H = 5.085% and O = 54.228%. The vapour density of the compound is 59. The molecular formula of the compound is
(a) \[{{C}_{4}}{{H}_{6}}{{O}_{4}}\] (b) \[{{C}_{4}}{{H}_{6}}{{O}_{2}}\] (c) \[{{C}_{4}}{{H}_{4}}{{O}_{2}}\] (d) None of these
Solution: (a)
|
Element |
Symbol |
Percentage of element |
Atomic mass of element |
Relative number of atoms \[=\frac{Percentage}{At.\,mass}\] |
Simplest atomic ratio |
Simplest whole number atomic ratio |
|
Carbon |
C |
40.687 |
12 |
\[\frac{40.687}{12}=3.390\] |
\[\frac{3.390}{3.389}=1\] |
2 |
|
Hydrogen |
H |
5.085 |
1 |
\[\frac{5.085}{1}=5.085\] |
\[\frac{5.085}{3.389}=1.5\] |
3 |
|
Oxygen |
O |
54.228 |
16 |
\[\frac{54.228}{16}=3.389\] |
\[\frac{3.389}{3.389}=1\] |
|
\[\therefore \ \] Empirical formula is \[{{C}_{2}}{{H}_{3}}{{O}_{2}}\]
\[\therefore \] Empirical formula mass of \[{{C}_{2}}{{H}_{3}}{{O}_{2}}\]= 59
Also, Molecular mass = \[2\times \text{Vapour density}\] = \[2\times 59=118\]
\[\therefore \] \[n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}=\frac{118}{59}=2\]
Now, Molecular formula =\[n\times \left( \text{Empirical formula} \right)\] =\[2\times \left( {{C}_{2}}{{H}_{3}}{{O}_{2}} \right)={{C}_{4}}{{H}_{6}}{{O}_{4}}\]
\[\therefore \] Molecular formula is \[{{C}_{4}}{{H}_{6}}{{O}_{4}}\].
Chemical equations and its balancing.
(i) It must represent an actual chemical reaction.
(ii) It must be balanced i.e., the total number of atoms of various substances involved on both sides of the equation must be equal.
(iii) It should be molecular. The elementary gases like hydrogen, oxygen etc. must be written in the molecular form as H2, O2 etc.
(i) Qualitative information : Qualitatively a chemical equation conveys the names of the reactants and products taking part in the reaction.
(ii) Quantitative information : Quantitatively a chemical equation conveys the following information :
(a) It conveys the actual number of reactants and product species (atoms or molecules) taking part in the reaction.
(b) It tells the relative masses of the reactants and products participating in the reaction.
(c) It conveys the relative number of reactant and product moles.
(d) It also conveys the volumes of the gaseous reactants and products if present.
Example : Reaction between CaCO3 and aqueous HCl.
\[\underset{\begin{smallmatrix}
\\
100\,gm.
\\
1\,mole
\end{smallmatrix}}{\mathop{CaC{{O}_{3}}}}\,\,\,\,+\,\,\,\,\,\underset{\begin{smallmatrix}
\\
73\,gm.
\\
2\,mole
\end{smallmatrix}}{\mathop{2HC{{l}_{{}}}}}\,\,\,\,\xrightarrow{\,\,\,}\,\,\underset{\begin{smallmatrix}
\\
111\,gm.
\\
1\,mole
\end{smallmatrix}}{\mathop{CaC{{l}_{2}}}}\,\,\,+\,\,\underset{\begin{smallmatrix}
\\
\,18\,gm.
\\
1\,mole
\end{smallmatrix}}{\mathop{{{H}_{2}}O}}\,\,\,+\underset{\begin{matrix}
\\
44\,gm. \\
1\,mole \\
22.4\,\text{litres} \\
\,\text{at}\,\text{STP} \\
\end{matrix}}{\mathop{C{{O}_{2}}}}\,\]
Note : q All chemical equations are written under STP conditions provided no other conditions are mentioned.
Limitations of a chemical equation and their removal : Although a chemical equation conveys a number of informations, it suffers from certain limitations or drawbacks. The major limitations and the steps taken for their removal are given below:
(i) Physical states of the reactants and products : The chemical equation fails to convey the physical states of the reactants and products. These are specified by the use of letters ‘s’(for solids), ‘l’(for liquids), ‘g’(for gases) and ‘aq’(for aqueous solutions).
Example : \[CaC{{O}_{3}}_{(s)}\,\,+\,\,2HC{{l}_{(aq)}}\,\,\xrightarrow{{}}\,\,CaC{{l}_{2}}_{(s)}\,\,+\,\,{{H}_{2}}{{O}_{(l)}}\,\,+\,\,C{{O}_{2}}_{(g)}\]
(ii) Conditions of temperature, pressure and catalyst : These conditions are normally not mentioned in the equation. These can be expressed on the arrow head which separates the reactants from the products.
Example : \[{{N}_{2}}_{(g)}\,\,+\,\,\,3\,{{H}_{2}}_{(g)}\underset{600atm}{\mathop{\xrightarrow{Fe,723K}}}\,\,\,2\,N{{H}_{3}}_{(g)}\]
(iii) Speed of reaction : The speed of a particular reaction whether slow or fast can be mentioned by writing the word slow or fast on the arrow head.
Example : \[N{{O}_{2}}_{(g)}\,\,\,+\,\,\,{{F}_{2}}_{(g)}\,\xrightarrow{\text{slow}}\,\,N{{O}_{2}}\,{{F}_{(g)}}\,\,\,+\,\,{{F}_{(g)}}\]
\[N{{O}_{2}}_{(g)}\,\,\,+\,\,\,{{F}_{(g)}}\,\,\xrightarrow{\text{fast}}\,\,N{{O}_{2}}\,{{F}_{(g)}}\]
(iv) Heat change accompanying the reaction : The heat evolved or absorbed in a chemical reaction can be written on the product sides. The S.I. unit of heat is kJ.
Example : \[C{{H}_{4(g)}}+2{{O}_{2(g)}}\xrightarrow{{}}C{{O}_{2}}_{(g)}\,\,\,+\,\,\,2\,{{H}_{2}}{{O}_{(l)}}+\ \underset{\text{(Heat is evolved)}}{\mathop{393.5\ KJ}}\,\]
\[{{H}_{2}}_{(g)}\,\,+\,\,{{I}_{2}}_{(g)}\,\,\xrightarrow{{}}\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(Heat is absorbed)}}{\mathop{2\,H{{I}_{(g)}}\,\,-\,\,53.9\,KJ}}\,\]
(v) Reversible nature of a reaction : Certain chemical reactions proceed both in the forward and backward directions. The reversible nature of the reaction can be indicated by two arrows pointing in the opposite direction (?).
Example :\[{{H}_{2}}_{(g)}\,\,+\,\,C{{l}_{2}}_{(g)}\,\,\underset{\text{Backward}}{\overset{\text{Forward}}{\mathop{\ \ \ \ \ \ \ \ \ \ \ \ }}}\,\,2\,HC{{l}_{(g)}}\]
(vi) Formation of precipitate and evolution of a gas : Formation of a precipitate in the chemical reaction can be indicated by writing the word ppt. or by an arrow pointing downwards.
\[Ag\,N{{O}_{3}}_{(aq)}\,\,\,+\,\,\,NaC{{l}_{(aq)}}\,\,\xrightarrow{{}}\,\,\underset{\text{(ppt)}}{\mathop{AgCl}}\,\,\downarrow \,\,+\,\,NaN{{O}_{3}}_{(aq)}\]
The evolution of a gas is expressed by an arrow which points upwards.
\[M{{g}_{(s)}}\,\,\,+\,\,\,2\,HC{{l}_{(aq)}}\,\,\xrightarrow{{}}\,\,Mg\,C{{l}_{2}}_{(aq)}\,\,+\,\,{{H}_{2}}_{(g)}\,\,\uparrow \]
Let us consider a chemical reaction which occurs due to passing of steam over red hot iron forming iron oxide and hydrogen gas. It can be represented as:
Skeleton equation : \[F{{e}_{\,(s)}}\,\,\,+\,\,\,{{H}_{2}}{{O}_{\,(v)}}\,\,\xrightarrow{{}}\,\,F{{e}_{3}}{{O}_{4}}_{(s)}\,\,\,+\,\,\,{{H}_{2}}_{(g)}\]
Balanced equation : \[3\,F{{e}_{(s)}}\,\,\,+\,\,4{{H}_{2}}{{O}_{(v)}}\,\,\xrightarrow{{}}\,\,F{{e}_{3}}{{O}_{4}}_{(s)}\,\,+\,\,4{{H}_{2}}_{(g)}\]
The balancing of equations is done by the following methods:
(i) Hit and Trial method, (ii) Partial Equation method
(iii) Oxidation-Number method, (iv) Ion-Electron method
The first two methods are discussed here, while the remaining two methods will be taken up for discussion in redox reactions.
(i) Hit and Trial method : This method involves the following steps:
(a) Write the symbols and formulae of the reactants and products in the form of skeleton equation.
(b) If an elementary gas like H2, O2 or N2 etc. appears on either side of the equation, write the same in the atomic form.
(c) Select the formula containing maximum number of atoms and start the process of balancing.
(d) In case the above method is not convenient, then start balancing the atoms which appear minimum number of times.
(e) Balance the atoms of elementary gases in the last.
(f) When the balancing is complete, convert the equation into molecular form.
Let us balance the skeleton equation \[M{{g}_{3}}{{N}_{2}}\,\,\,+\,\,\,{{H}_{2}}O\,\xrightarrow{{}}\,\,Mg\,{{(OH)}_{2}}\,\,\,+\,\,N{{H}_{3}}\]
The balancing is done in the following steps
Step I. Balance the Mg atoms \[M{{g}_{3}}{{N}_{2}}\,\,+\,\,{{H}_{2}}O\,\xrightarrow{{}}\,\,3\,Mg\,{{(OH)}_{2}}\,\,+\,\,N{{H}_{3}}\]
Step II. Balance the N atoms \[M{{g}_{3}}{{N}_{2}}\,\,+\,\,{{H}_{2}}O\xrightarrow{{}}3\,Mg{{(OH)}_{2}}\,\,+\,\,2\,N{{H}_{3}}\]
Step III. Balance the O atoms \[M{{g}_{3}}{{N}_{2}}\,+\,6\,{{H}_{2}}O\xrightarrow{{}}3\,Mg\,{{(OH)}_{2}}\,\,+\,\,2\,N{{H}_{3}}\]
The hydrogen atoms are already balanced. Hence, final balanced equation is
\[M{{g}_{3}}{{N}_{2}}\,\,+\,\,\,6{{H}_{2}}O\,\xrightarrow{{}}\,\,3\,Mg\,{{(OH)}_{2}}\,\,+\,\,2\,N{{H}_{3}}\]
(ii) Partial equation method : Chemical equations which involve a large number of reactants and products can not be balanced easily by the hit and trial method. In partial equation method, the overall reaction is assumed to take place into two or more simpler reactions known as partial equations. The balancing of the equation involves the following steps:
(a) Split the chemical equation into two or more simpler equations or partial equations.
(b) Balance each partial equation separately by hit and trial method.
(c) Multiply the partial equations with suitable coefficient if necessary so as to cancel out the final substances which do not appear in the final equation.
(d) Finally, add up the partial equations to get the final equation.
Let us balance the skeleton equation \[NaOH\,+\,\,C{{l}_{2}}\,\xrightarrow{{}}\,\,NaCl\,\,+\,\,NaCl{{O}_{3}}\,\,+\,\,{{H}_{2}}O\]
This reaction is supposed to take place in the following steps:
The probable partial equations for the above reaction are:
\[Na\,OH\,+\,\,C{{l}_{2}}\,\xrightarrow{{}}\,\,Na\,Cl\,\,+\,\,Na\,ClO\,\,+\,\,{{H}_{2}}O\] and \[Na\,Cl\,O\,\xrightarrow{{}}\,\,Na\,Cl\,{{O}_{3}}\,\,+\,\,NaCl\]
Balance the partial chemical equations separately by hit and trial method as
\[2\,Na\,OH\,\,+\,\,C{{l}_{2}}\,\,\xrightarrow{{}}\,\,NaCl\,\,+\,\,Na\,ClO\,\,+\,\,{{H}_{2}}O\] and \[3\,Na\,Cl{{O}_{3}}\,\,\xrightarrow{{}}\,\,Na\,Cl{{O}_{3}}\,\,+\,\,2\,NaCl\]
Multiply the first partial equation by 3 in order to cancel out \[NaClO\]which does not appear in the final equation. Finally add the two partial equations to get the final equation.
\[\frac{\begin{align}
& 2\,NaOH\,+\,C{{l}_{2}}\,\xrightarrow{{}}\,NaCl\,+\,NaClO\,+\,{{H}_{2}}O\,]\,\,\times \,\,3 \\
& \,\,\,\,\,\,\,\,\,\,3\,NaClO\,\,\xrightarrow{{}}\,\,NaCl{{O}_{3}}\,+\,2\,NaCl \\
\end{align}}{6\,NaOH\,\,+\,\,3\,C{{l}_{2}}\,\xrightarrow{{}}\,NaCl{{O}_{3}}\,\,+\,\,5\,NaCl\,+\,3\,{{H}_{2}}O}\]
You need to login to perform this action.
You will be redirected in
3 sec
