Some basic concepts of chemistry

Law of chemical combination• Law of conservation of mass• This law states that mass can neither be created nor be destroyed in a

chemical reaction.• Law of definite proportions• This law states that mass is always present in a definite proportions by weight.• Law of multiple proportions• If two elements can combine to form more than one compound the masses one

element that combine with a fixed mass of other element are in a small whole number ratio.

• Law of gaseous volumes or gay lussac's law:• When gases combine or produced in a chemical reaction they do so by

combining in a small ratio by mass provided the temperature and pressure remain constant.

• Avogadro's law:• Equal volumes of gases at the same temperature should have equal number of

volumes.

Dalton’s Atomic theory• Matter consists of indivisible particles called

atoms.• All atoms of a given element have identical

properties including identical mass. Atoms of different elements have different mass.

• Compounds are formed when atoms of different elements combine in a fixed ratio.

• Chemical reactions involve reorganization of atoms. These can neither be created nor destroyed in a chemical reaction.

Atomic mass•Atomic mass is the mass of one atom due

to relative charges of the other atom.1 atomic mass unit (amu)=1.606656x10-24g1 atomic mass can actually be defined as

the mass of 1/12 of carbon isotope.Atomic mass can be also known as unified

mass.

Molecular mass• Molecular mass can be defined as the sum of atomic masses present

together as the sum of attractive forces in a molecule.

• Example:• Calculate molecular mass of glucose(C6H12O6) molecule.

• SolutionMolecular mass of glucose (C6H12O6)

= 6(12.011 u) + 12(1.008 u) + 6(16.00 u)= (72.066 u) + (12.096 u) + (96.00 u)=180.162 u

Mole concept•One mole can be defined as the number of

particles exactly equal to 6.022 x 1023 particles

•These particles can be atoms, ions or molecules

•The mass of one mole is said to be its molar mass.

• ‘6.022 x 1023 ‘ is said to be known as Avogadro's constant

Percentage composition• Mass % of an element = x100

• Example mass of H2o.• Molar mass of water = 18.02 g• Mass % of hydrogen = • 2×1.008 ×100• 18.02• = 11.18• Mass % of oxygen = • 16.00 ×100• 18.02• = 88.79

Empirical formula for molecular formula.

•An empirical formula represents the simplest•whole number ratio of various atoms present• in a compound whereas the molecular• formula shows the exact number of different• types of atoms present in a molecule of a•compound.

A compound contains 4.07 % hydrogen,24.27 % carbon and 71.65 % chlorine.Its molar mass is 98.96 g. What are itsempirical and molecular formulas ?

Solution: molar mass of chloromethyl is 12.01+2 x 1.008+35.453=49.48

Therefore, molar mass = 98.96 empirical formula mass 49.48 = n=2Therefore when we multiply the formula by 2 we get

C2H4Cl2

PRACTICAL PROBLEMS ON REACTION OF SOLUTIONS• Problem 1.3Calculate the amount of water (g)produced by the combustion of 16 g ofmethane.SolutionThe balanced equation for combustion of methane isCH4(g)+202(g) CO2(g)+2H2O(g)

(i)16 g of CH4 corresponds to one mole.(ii) From the above equation, 1 mol of CH4 (g) gives 2 mol of H2O (g).Therefore 2 mol of H2O=2X(2+16)=36g

Therefore 1 mol of H2O 18g of H2O=18g of H2O

1 mol of H2O = 18g

Therefore 2 mole of H2O = 2 x 18g of H2O=36g of H2O.

SOME MORE PROBLEMS•Problem 1.4How many moles of methane are requiredto produce 22 g CO2(g) after combustion?SOL: According to the chemical equation,

CH4+202 CO2+2H2OTHERFORE 44G OF CO2THERFORE 22G OF CO2 will be 22g x 1 mol of CO2 44g of CO2 =

½mole of CO2Therefore 0.5 mole of CO2 corresponds to 1 mole of CH4 after

combustion.

Limiting reagent• In a chemical reaction when one reactant gets consumed the

other reactant consuming it is called limiting reagent, in short-reactant which gets consumed limits the amount of product formed and is therefore called limiting reagent.

• Problem 1.5• 50.0 kg of N2(g) and 10.0 kg of H2 (g) are mixed to produce

NH3(g). Calculate the NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.

SOLUTION: The chemical reaction can be written as N2+H2 NH3

(SOLUTION CONTINUED IN THE NEXT SLIDE)

• Now 50.0 Kg of N2 x 1000g of N2 x 1 mol N2 1 kg of N2 28.0 g N2 = 17.86x 102 mol moles of

H2Now similarly , 10.00 kg H2 x 1000g H2 x 1 mol of H2 1 kg of H2 2.016g of H2 =4.96x103 mol

according to the above equation, 1mol N2 (g) requires 3 mol H2 (g), for the reaction.

Hence, for 17.86 mol of N2 the mole of H2 (g) required would be

17.86x 102 moles of N2 x 3 moles of H2(g) 1 mole of N2 (g) = 5.36x 102 mole of N2

But we have only 4.96×103 mol H2 . Hence, dihydrogen is the limiting reagent in this case. So NH3(g) would be formed only from that amount of available dihydrogen i.e., 4.96x103 mol

(cont of this ex on next slide)

• Now since 3 mol of H2 gives 2 moles of NH3(g)4.96x103 of H2 x 2 moles of NH3 3 mol H2 =3.30x103 mol of NH3Now they are to be converted to grams,Now 1 mole of ammonia= 17.1 grams of ammoniaTherefore 3.30 x103 mol x 17.01 g of NH3 1 mole of ammonia = 3.30x103 x

17g of ammonia

= 56.1x103 g of NH3= 56.1 kg NH3

CONCENTRATION OF SOLUTIONS

•To find out concentration in a solution we express it in 4 ways

• molality

mass percent mole fraction molality

Mass per cent• Mass percent = mass of solute x100 mass of solution Problem 1.6A solution is prepared by adding 2 g of a substance A to 18 g

of water. Calculate the mass per cent of the solute.Mass percent of A= mass of A x100 mass of solution = 2g of A x100 = 2g x 100 = 10% 2g of A +18g of H2O 20g

Mole fraction• It is the ratio of number of moles of a particularcomponent to the total number of moles of thesolution. If a substance ‘A’ dissolves insubstance ‘B’ and their number of moles are NA

and NB respectively; then the mole fractions

of A and B are given as• Mole fraction of A = no of moles of A = na

no of moles in the solution. na + nb

Mole fraction of B is just the vice versa

Two main concentrations of solutionsmolality molarity

• It is the most widely used unit and is denoted by M. It is defined as the number of moles in solute upon volume of solution in litres(L).

• Molality= no of moles in solute/volume of solution in litres.

• It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by (m).

• Molarity= no of moles in solute/ mass of solvent in kg.

Some numerical’s on molality• Q.1 : Suppose we have 1 M solution of a substance, say

NaOH and we want to prepare a 0.2 M solution from it.• Sol: if 1 mole is present in 1000 litres then 0.2 mole =

1000mL = 0.2 mol = 200 mL. 1 molNow we know from general calculations thatM1 X v1 = M2 x v2

Therefore 0.2M x 1000mL = 1.0M x v2

v2 = 0.2M x 1000 mL = 200mL. 1.0M

SOME EXAMPLES.ON MOLALITY• Calculate the molarity of NaOH in the solution

prepared by dissolving its 4 g in enough water to form 250 mL of the solution.

• Solution:• Since molarity (M) = No of moles of solute volume of solution in litres. = mass of NaOH/molar mass of NaOH 0.250 L= 4g/40g = 0.1 mol 0.250L 0.250 L = 0.4M.

EXAMPLES ON MOLARITY.• The density of 3 M solution of NaCl is 1.25 g mL-1

Calculate molality of the solution.• Solution:M=3mL-1

Therefore Mass of NaCl in 1litre solution =3 x 58.5 =175.5Therefore mass of 1L solution = 1000x1.25 (since density =

1.25)Mass of water in solution= 1250-175.5 = 1074.5Therefore on applying molality formula=

3mol/1074.5=2.79m