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Basic Concepts of Chemistry

Conservation of Mass
According to this law during any chemical and physical change the total mass of product is equal to the total mass of reactants.

Examples

  • 9 g of KClO3 heated produce 1.92 g of O2 and the residue KCl left behind weight 2.96 g show that the data illustrate the law of conservation of mass.
    Solution:
    \(\begin{matrix}
    KClO_3& \xrightarrow{\Delta}& KCl& +& O_2& \\
    4.9\;g& &2.96\;g&+&1.92\;g& \\
    &&&=&\;4.88\;g&\\
    \end{matrix}\)

Law of Definite Proportion
According to this law, a pure chemical compound always contains same elements combined together in the same element combined together in the same definite proportion in the weight.

Examples

  • 16 g of Cu when treated with NO3 followed by ignition of the nitrate give two 2.90 g of CuO. In another experiment 1.15 of CuO upon reduction with H2 give 0.932 g of Cu.
    Solution:
    In 1st case \(\frac{2.16}{2.70}\) \(\times\) 100 = 80%
    % of Cu in CuO = 80 %
    % of O in CuO = 20 %
    \(\frac{0.92}{1.5}\) \(\times\) 100 = 20%

Law of multiple proportion
According to this law when two elements combine to form two or more than two compounds, then the masses of one of the elements which combine with a fixed mass of other are in a simple whole no. ratio.

Examples

  • Carbon is found to form oxides which contains 42.9 % and 27.3 % of a respectively shows the multiple proportion.
    Solution:
    = 100 - 42.9 = 57.1
    = 100 – 27.3 = 72.7
    1 part of O = 42.9 part of C
    1 part of O = \(\frac{42.9}{37.1}\) = 0.75
    In IInd case
    72.7 part of O = 27.3 part of C
    1 part of O = \(\frac{27.3}{72.7}\) = 0.37
    = \(\frac{0.75}{0.37}\):\(\frac{0.37}{0.37}\)
    = 2 : 1

Gay Lussac’s Law
It states that whenever gases reacts with each other or combine with each other or combine with each other to form gases out product they always have a simple whole ratio by volume under similar condition of temperature and pressure.
\(\begin{matrix}
2H_2& +&O_2& \longrightarrow& 2H_2O&\\
2\;Vol&&1\;Vol& \longrightarrow& 2\;Vol&\\
2\;ml&&1\;ml& \longrightarrow& 2\;ml&\\
2\;l&&1\;l& \longrightarrow& 2\;l&\\
\end{matrix}\)

Avogadro’s law
It states that equal no. of all gases contains equal no. of molecules under similar condition of temperature and pressure.
\(\begin{matrix}
2H_2& +&Cl_2& \longrightarrow& 2HCl&\\
1\;Vol&&1\;Vol& \longrightarrow& 2\;Vol&\\
n\;molecules&&n\;molecules& \longrightarrow & 2n\;molecules&\\
1\;molecule&&1\;molecule& \longrightarrow & 1\;molecule&\\
\frac{1}{2}\;molecule&&\frac{1}{2}\;molecule& \longrightarrow & \frac{1}{2}\; molecule&\\
\end{matrix}\)
Molecules mass = 2 \(\times\) U.D
U.D. = \(\frac{density\;of\;gass}{Density\;of\;H_2}\)
Molar mass = mass of 22.4l of gas at STP condition
T = 273 kelvin and P = 1 bar

Examples

  • Molar mass of 44 g = 1 mol
    22400 ml at STP
    1 mol = 22400 ml
    5 mol = 22400 \(\times\) 1.5

Atomic Mass
The atomic mass of an element is the no. of times on atom of that elements is heavier than 1 atom of C12.
1 C atom = 12 amu
1mu   = \(\frac{1}{12}\)\(\times\) mass of 1 C – atom
Atomic mass = \(\frac{mass\;of\;1\;H - atom}{\frac{1}{12}{\times\;mass\;of\;1\;C - atom}}\)
= \(\frac{mass\;of\;1\;H - atom}{\frac{1}{12}{\times\;mass\;of\;1\;C - atom}}\)
= \(\frac{1.67{\times}10^{-24}}{1.66{\times}10_{-24}}\)
= 1.008 u
\(\frac{1}{12}\) \(\times\) 1.99 \(\times\) 10-23
= 1.66 \(\times\) 10-24 g
14 = 1 atom (H)
1 g = NA atom (H)
Atomic mass of an elements is the average relative mass of its atom as compared to the mass of C12 atom

Examples

  • Calculate the no. of atom in 52 u of He.
    Solution:
    52 u of He
    = 1 u = 4
    = 13 \(\times\) NA atoms

Gram Atomic Mass
The atomic mass of an elements is express in g is known as gram atomic mass.
O2 = 32 g
N2 = 28 g
Cl3 = 106.5
Al3 = 71

Examples

  • 1 mol AlCl3 = 1 mol Al3+
    = 27g AL3+
    = NA \(\times\) Al3 ion
    = 3 mol of Al-
    = 106.5 g al-
    = 3 NA al- ions
    1 mol AlCl3 = total no. of ions = 4
    1 mol AlCl3 = 4 NA

Examples

  • How many moles of CH4 are required to produced 22 g CO2 after combustion.
    Solution:
    By formula
    \(\begin{matrix}
    CH_4&+&2H_2O&\rightarrow&CO_2&+&H_2O&\\
    &1\;Mol&&&&2\;Mol&&\\
    \\
    CH_4&\rightarrow&CO_4&\\
    22\;g&&144\;g&\\
    \end{matrix}\)
    1 g CO2 = \(\frac{22}{44}\) = \(\frac{1}{2}\)
    1 CO2 = 1 CH4
    \(\frac{1}{2}\)CO2 = \(\frac{1}{2}\)CH4 = 0.5 mole of CH4
    22 g = \(\frac{1}{44}\) \(\times\) 22 = \(\frac{1}{2}\) CH4 = 0.5 mole of CH4
    Statement
    \(\begin{matrix}
    &CH_4& +& 2O_2& \longrightarrow& 2H_2O& +& CO_2& \\
    &1\;Mol&&&&&&1\;Mol&&\\
    &16\;g&&&&&&44\;g&&\\
    &44\;g\;\;CO_2&&& \longrightarrow&&& 1\;Mol\;\;CH_4&\\
    \end{matrix}\)
    1 g = \(\frac{1}{44}\)
    22 g = \(\frac{1}{44}\) \(\times\) 22 = \(\frac{1}{2}\)

Limiting Reagent
Which completely consume in the reaction.

Excess Reagent
Which do not completely consume in the reaction.

Example

  • 1 L of O2 at STP is made to react 56.66 kg with 3 L of CO. Calculate the mass of each substance found after the reaction and also tell which is limiting reagent?
    Solution:
    \(\begin{matrix}
    2CO&+&O_2&\longrightarrow& 2CO_2&\\
    2\;Vol&&1\;Vol&&2\;Vol&\\
    \end{matrix}\)
    2 L \(\rightarrow\) 2 L
    \(\begin{bmatrix}
    2\;L\;\;O_2& \longrightarrow& O_2&\\
    1\;g& \longrightarrow& \frac{1}{2}&\\
    3\;L& \longrightarrow& \frac{1}{2}\;\times\;3\;=\;\frac{3}{2}&\\
    \end{bmatrix}\)
    Limiting Reagent O2
    2CO + O2 \(\rightarrow\) 2CO2
    4 l \(\rightarrow\) 28 g
    1 l \(\rightarrow\) \(\frac{28}{22.4}\)
    3l \(\rightarrow\) \(\frac{28}{22.4}\) \(\times\) 30 = \(\frac{840}{22.4}\) = 1.25 g
    22.4 l \(\rightarrow\) 44 g CO2
    1 l \(\rightarrow\) \(\frac{44}{22.4}\) = 1.964
    2l \(\rightarrow\) 1.964 \(\times\) 2 = 3.929

Solution
It is a homogenous mixture of two or more compound present in any ratio.

Concentration
Dilute
The concentration can be expressed as follows

  • Mass % (w/w %)
    It is the mass of solute in grams present in 100 g of the solution.
    % of component 1 = \(\frac{w_1}{w_1 + w_2}\) \(\times\) 100
    % of component 2 = \(\frac{w_2}{w_1 + w_2}\) \(\times\) 100
    10% NaOH solution (mass/mass) means that 10 g NaOH is dissolved in 100 g of the solution (90 g H2O)
  • Volume present (v/v %)
    It is the vol. of salute in ml present in 100 ml of solution.
    10 % ethyl alcohol solution means 10 ml ethyl alcohol is present in 100 ml of the salutation.
  • Mass/volume (w/v %)
    It is the mass salute in grams present in 100 ml of the solution. 10 percent NaOH solution (w/v) means that 10 g of NaOH is present in 100 ml of the solution.
  • Strength (g/l)
    It is the amount of solute in g present in 1 l of the solution.
    1 g \(\rightarrow\) 100 ml of solution
    1 g \(\rightarrow\) 100 ml of solution

Example

  • If we have 100 ml of the solution contain 5 g solute what will be its strength.
    5 g \(\rightarrow\) 100 ml
    50 g \(\rightarrow\) 1000 ml
    2 l \(\rightarrow\) 10 g
    1 l \(\rightarrow\) 5 g

Molarity (Mol/L)
It is defined as the no. of moles of solute dissolved per line of the solution. It is represented by capital (M)
Mathematically, M = \(\frac{No.\;of\;moles\;of\;solute}{vol.\;of \;solution(L)}\)
= \(\frac{given\;mass}{molar\;mass}\) \(\times\) \(\frac{1000}{V}\)

Examples

  • Calculate the maturity of NaOH in the solution prepared by dissolving its four 4 g in enough water to form 250 ml of the solution.
    Solution:
    M = \(\frac{4}{40}\) \(\times\) \(\frac{100}{250}\) = 0.4 M

Relation between strength and maturity
As we know
M = \(\frac{given\;mass}{v(L)}\)  \(\times\) \(\frac{1}{molar\;mass}\)
M = strength \(\times\) \(\frac{1}{molar\;mass}\)
Strength = Molarity \(\times\) Molar mass

Molality (m)
mol/kg
It is the no. of moles of solute present in 1 kg of the solute. It is represented by small(m)
m = \(\frac{no.\;of\;moles\;of\;solute}{mass\;of\;solvent}\)
It is also equal to given mass \(\times\) 1000. (W2 = mass of solute); m = molality, W2 =mass of solvent
m = \(\frac{w_2\;(g)}{m_2}\) \(\times\) \(\frac{1000}{w_1\;(g)}\)
m2 = molar mass solute

Example

  • Commercial available concentrated Hal contain 38% by mass. What is the molality of this solution.
    Solventing 1.19 (Density)
    38% HCl by mass means that 38 ml of Hal is present in 100 g of the solution.
    M = \(\frac{given\;mass}{molar\;mass}\) \(\times\) \(\frac{1000}{v}\)
    Mass of solute = 38 g
    Mass of solvent = 100 g
    = \(\frac{38}{36.5}\) \(\times\) \(\frac{1000}{84.03}\) = 12.38

Molarity Equation
m1v1 = m2v2

  • How many (m) grams BaCl2 are needed to prepare 100cm3 of 0.250 m. BaCl2 (Ba = 173 g).
    \(\frac{250}{100}\) \(\times\) \(\frac{1000}{1000}\) \(\times\) 20 m = 5.20 g
    = \(\frac{0.250}{544{\times}1000}\) \(\times\) \(\frac{1000}{100}\) = 5.20 g

Normality (N)
g-equivalent/L    
It is the no. of g - equivalent of the solute dissolved per liter of the solution.
Mathematically = \(\frac{g - equivalent}{vol.\;of\;solution\;in(L)}\)
N = \(\frac{given\;mass}{equivalent\;mass}\)
Units = g-equivalent/L
Equivalent mass = \(\frac{molar\;mass}{n-factor}\), where n = 1, 2, 3, ….
For  acid

\(\begin{bmatrix}
HCl,&Basicity\;=\;1\;=\;n&\\
H_2SO_4,&Basicity\;=\;2\;=\;n&\\
H_3PO_4,&Basicity\;=\;3\;=\;n&\\
\end{bmatrix}\)
Equivalent mass = \(\frac{36.5}{1}\) = 36.5
= \(\frac{98}{2}\) = 49
= \(\frac{98}{3}\) = 32.6
For Base
NaOH, acidity = 1, Acidity = 1, eq = \(\frac{40}{1}\) = 40
Ca(OH)2, acidity = \(\frac{74}{2}\) = 37
For salts
Na+ al- = n = Change on any ion = 1
Or no. of e-s transferred = 1
MgCl2 = \(\frac{58.5}{1}\) = 58.5

Examples

  • MgCl2 = n = 2
    mass = \(\frac{24+2(35.5}{2}\) = 47.5
    AlCl2 = n= 3
    Eq. mass = \(\frac{27 + 3(35.5)}{3}\) = 36.5

Relation between molality and normality
N = \(\frac{givem\;mass}{eq.\;mass}\)  \(\times\) \(\frac{1000}{u (ml)}\)
N = n-factor \(\times\) \(\frac{given\;mass}{mol.\;mass}\) \(\times\) \(\frac{1000}{v(ml)}\)
N= n \(\times\) m
1 N HCl \(\rightarrow\) 1m HCl, n = 1
1 m H2SO4 \(\rightarrow\) 2NH2SO4, n = 2
2 N Ca(OH)2 \(\rightarrow\) 1 m Ca(OH)2

Relation between normality and strength
N = \(\frac{given\;mass}{equ.\;mass}\) \(\times\) \(\frac{1}{v(L)}\)
Normality = \(\frac{strenth}{eq.\;mass}\)
Strength =  N \(\times\) eq. mass
= m \(\times\) molar mass

Normality Equation
N1v1 = N2v2
N1v1 + N2v2 = N3v3

Examples

  • H2 gas is prepared in the lab by reacting dal. Hal with 2n following reaction takes place. Calculate the vol of H2 gas liberated at Zn + 2HCl \(\rightarrow\) 2ZnCl2 + H2
    When 32.65 g of Zn react with Hal, 1mole of gas occupies 22.7 l at STP.
    Zn + 2HCl \(\rightarrow\) 2ZnCl2 + H2
    65
    65.3 g \(\rightarrow\) 22.7 l
    1 g \(\rightarrow\) \(\frac{22.7}{65.3}\) L
    32.65 g \(\rightarrow\) \(\frac{22.7}{65.3}\) \(\times\) 32.65 = 11.35

PPM (Parts per Million (106))
It is no. of parts of solute by mass present in million parts by mass of the solution.
(PPM)A = \(\frac{Mass\;of\;Solute}{Mass\;of\;Solution}\) \(\times\) 106

Examples

  • Calculate the amount of C formed.
    2 A \(\rightarrow\) 4 B
    1 A \(\rightarrow\) \(\frac{4}{2}\) \(\times\) 5
    4 D \(\rightarrow\) 3 C
    1 D \(\rightarrow\) \(\frac{3}{4}\) \(\times\) 6 = 4.5 C
  • Calculate the average atomic mass using the isotope

     

    % data

    Molar Mass

    H

    99.985

    1

    2H

    0.015

    2

    Solution:
    = \(\frac{1 \times 99.985}{100}\) = 0.99985
    = \(\frac{2 \times 0.015}{100}\) = 0.0003
    In the reaction A + B \(\rightarrow\) AB2 identify the limiting reagent if any in the following mixture. (1) 300 atoms of A + 200 molecules of B (2) 2 mole of A + 3 mole of B (3) 100 atoms of A + 100 molecules of B (4) 5 mole of A + 2.5 mole of B (5) 2.5 mole of A + 5 mole of B.
    300 atoms of A \(\rightarrow\) 200 molecules of B
    \(\rightarrow\) \(\frac{200}{300}\)= 0.6
    Limiting regent = B
    2 Mole of A \(\rightarrow\) 3 mole of B
    Limiting reagent = A

Limiting regent Nitrogen
28 g N \(\rightarrow\) 34 g NH3
1 g \(\rightarrow\) \(\frac{34}{28}\)
2 \(\times\) 103 \(\rightarrow\) \(\frac{34}{28}\) \(\times\) 2 \(\times\) 1023 

Example

  • What will be the mass of C12
    Mass of 6.022 \(\times\) 1023 atoms of carbon = 12 g
    1 atom of carbon = \(\frac{12}{6.022{\times}10^{23}}\)
    = 1.99 \(\times\) 1023

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