CLASS :- 11
SUBJECT :- CHEMISTRY
CHAPTER :- 1
TOPIC
Limiting Reactant
The react and which is present in laser amount get consume after sometime and after that no further reaction take places even do the other reactant It is present the react and which get completely consume in in a reaction is called limiting reactant
The concentration of the limiting reactant limit the amount of product form the other reactant present in the quantities greater than those need to to react with the quantity of the the limiting reactant present would be left unreacted it is also called excess reagent
example formation of ammonia from nitrogen and hydrogen
One mole nitrogen reacts with 3 Mol hydrogen and forms 2 mole ammonia. if in reaction Reaction vessels content 2 mole nitrogen and 3 mole hydrogen then definitely One More Nitogen to combined with 3 Mol hydrogen and form 2 mole ammonia In the reaction vessels hydrogen completely consume and nitrogen Ramain 1 mole in reaction vessels hydrogen hear limiting reactant and nitrogen excess reactant In a reaction
Identify the limiting reactant if any in the following reaction mixture
1 300 atom of a 200 molecules of B
2 two mole of a a three mole of B
3 100 atom of a 100 molecules of B
4 5 mole of a a 2.5 mole of B
5 2.5 Mol of a 5 mole of B
Answer:
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
.•. 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be
left unreacted. Hence, B is the limiting reagent while A is the excess reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
.•. 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant.
(iii) No limiting reagent.
(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.
50 kg of Nitrogen and 10 kg Hydrogen gas are mix to produce Ammonia gas .calculate the ammonia form
Identify the limiting reactant In this reaction if any
TOPIC
Concentration
The amount of solute present in a given quantity of solvent our solution is expressed in term of concentration
1 mass percentage
the mass of the component per hundred gram of the solution is called mass percentage
Mass of solute
Mass % = ……………………….. ⤮ 100
Mass of solution
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute
2 volume percentage
the volume of the component per hundred parts of the volume of solution
mass to volume percentage the mass of the component per hundred part of the volume of the solution
Volume of solute
Volume % = ……………………….. ⤮ 100
Volume of solution in ml
3 Molarity of a solution
the number of mole of the solute dissolve per litre of the solution represent as M
Example :- 1 m sodium carbonate solution mans 106 gram of solute present per litre of the solution
mole of solute
M = ……………………….. ⤮ 1000
Volume of solution in ml
However it has one disadvantage It's changed with temperature because of expansion are are contraction of the liquid with temperature
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Molality
the number of mole of the solute dissolve per 1 kg of the solvent
it is denoted by m
mole of solute
M = ……………………….. ⤮ 1000
Mass of solvent (g)
Molality does not change with temperature because mass of the solvent does not change with change in temperature
The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate the molality of the solution.
Mole fraction
it is the ratio of of number of mole of one component to the total number of mole present in the solution
it is denoted by X
let's suppose that a solution of content and na mole of solute and nb mole of the solvent then
Mole fraction of solute x A = na / na+nb
mole fraction of solvent x B = nb/na+nb
the sum of mole fraction of all the component in solution is always equal to 1
mole fraction is independent of temperature because all term use in mass
Normality
number of Gram equivalent of the solute dissolved per litre of the solution it is denoted by N
number of Gram equivalent of solute
N =…………………………………… X 1000
volume of solution in ml
Mass Percentage Formula :-
Mass percent is the mass of the element or solute divided by the mass of the compound or solute. The result is multiplied by 100 to give a percent.
The formula for the amount of an element in a compound is:
mass percent = (mass of element in 1 mole of compound / mass of 1 mole of compound) x 100
The formula for a solution is:
mass percent = (grams of solute / grams of solute plus solvent) x 100
or
mass percent = (grams of solute / grams of solution) x 100
The final answer is given as %.
Mass Percentage Examples
Question :- Calculate the mass percent of different elements present in sodium sulphate (Na2 SO4).
How much copper can be obtained from 100 g of copper sulphate (CuSO4 )? (Atomic mass of Cu= 63.5 amu)
Answer: 1 mole of CuS04 contains 1 mole (1 g atom) of Cu
Molar mass of CuS04= 63.5 + 32 + 4 x 16 = 159.5 g mol-1
Thus, Cu that can be obtained from 159.5 g of CuS04 = 63.5 g
Question :- Calculate the mass percentage composition of glucose (C6H12O6)
Solution :- the formula of glucose C6H12O6
Molar Mass of glucose = 6X12+ 12X1+ 6X16 = 180
In molecules of glucose contain 6 carbon , 12 hydrogen and 6 oxygen
Mass of carbon = 6 X 12 = 72
Mass % of carbon = 72 X100 /180 = 40.0%
Mass of hydrogen = 1 X 12 = 12
Mass % of hydrogen =12X100 /180 =6.67%
Mass of oxygen = 6X 16 = 96
Mass % of oxygen = 96X100 /180 = 53.33%
Molecular formula and Empirical formula
The empirical formula of a compound gives the simplest ratio of the number of different atoms present in a compound ,
whereas
The molecular formula gives the actual number of each different atom present in a molecule.
If the formula is simplified then it is an empirical formula. The molecular formula is commonly used and is a multiple of the empirical formula.
The general statement relating molecular formula and the empirical formula is
Molecular Formula = n × Empirical Formula
Molecular mass
----------------------------- = n
Empirical mass
Molecular formula
The molecular formula is the formula derived from molecules and is representative of the total number of individual atoms present in a molecule of a compound.
Molecular formulas are associated with gram molecular masses that are simple whole number multiples of the corresponding empirical formula mass.
Empirical formula
The empirical formula is the simplest formula for a compound which is defined as the ratio of the smallest possible whole number atom of the elements present in the formula. It is also known as the simplest formula.
An empirical formula for a compound is the formula of a substance written with the smallest integer subscript.
The empirical formula gives information about the ratio of numbers of atoms in the compound. The percent composition of a compound directly leads to its empirical formula
Example (Glucose Molecular Formula Vs Glucose Empirical Formula)
Let’s take the example of glucose. The molecular formula of glucose is C6H12O6 and the empirical formula of glucose is CH2O.
We can derive a relation between the Molecular formula and the empirical formula of glucose.
We can derive a general expression as,
Molecular formula = n × empirical formula where n is a whole number
Sometimes, the empirical formula and molecular formula both can be the same.
Molecular mass /empirical mass =n
1) The first step in this problem is to change the % to grams.
2) Next divide all the given masses by their molar mass.
2) Then change the % to grams
3) Then, pick the smallest answer in moles from the previous step and divide all the answers by that. Remember that if you calculate a number that is x0.9 round to the nearest whole number
3) Next, divide all the masses by their respective molar masses
4) Pick the smallest answer of moles and divide all figures by that
5) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula
6) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula
(A) Calculation of Empirical Formula
(B) Molecular formula = (Empirical formula) X n
Question 1 Determine the empirical formula of an oxide of Iron which has 69.9 % iron and 30.1 % dioxygen by mass.
Question :-2 Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.8 g mol-1(Atomic mass: Fe = 55.85, O = 16.00 amu)Calculation of Empirical Formula. See Q3.
Answer: Empirical formula mass of Fe203 = 2 x 55.85 + 3 x 16.00 = 159.7 g mol–1
Hence, the molecular formula is the same as the empirical formula, viz.,Fe203.
Question :- 3. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. Answer:
Question :- 4 A compound on analysis was found to contain C = 34.6%, H = 3.85% and O = 61.55%. Calculate the empirical formula.
Answer: Step I. Calculation of simplest whole number ratios of the elements.
The simplest whole number ratios of the different elements are: C:H:O::3:4:4
Step II. Writing the empirical formula of the compound.
The empirical formula of the compound = C3H4O4.
QUESTION 5 :- An organic compound on analysis gave the following percentage composition C =40.0 % H = 6.67 % and the rest is oxygen . the vapour density of the compound was found to be 90 . Find out the molecular formula of the compound .
(A) Calculation of Empirical Formula
% Of oxygen =100 - ( % of carbon + % of hydrogen)
=100 - (40+ 6.67)
% Of oxygen = 53.33 %
Empirical formula = CH2O
Empirical mass = 12+2+16 = 30
(B) Molecular formula
Molecular mass = 2 x vapour density
Molecular mass = 2x90 = 180
n = molecular mass / empirical mass
n = 180 / 30 = 6
Molecular formula = (empirical formula)xn
Molecular formula = ( CH2O ) x 6
Molecular formula = C6H12O6
TOPIC
Chemical equation
a chemical equation is a statement of a chemical reactions in turn off the symbol and the formula of the species involved in the reactions the chemical reactions may be defined as
a brief representation of chemical change in term of symbol and formula of substance involve in it for example the reactions of silver nitrate with sodium chloride to give silver chloride and sodium nitrate may be represent as
the reactions of of barium chloride with sodium sulphate to give barium sulphate and sodium chloride may be represented as
The substance which react among themselves to bring about the chemical changes are now as a reactant where is the substance which are produced as a result of the chemical change and now as product
Essential of chemical equation
1 it must be consistent with the experimental factor
2 It should be balance
3 it should be molecular the elementary gases like Hydrogen oxygen extractor
information conveyed by a chemical equation
chemical equation convey what qualitative and quantitative information
qualitatively a chemical equation tell the names of the various reactant and the product
quantitatively a chemical equation represent
1 the relative number of reactant and product
2 the number of mole of the reactant and product
3 the relative mass of the reactant and product
4 the relative volume of gas is react and and product
The chemical equation Combustion of Methane gas
Give the following information
one mole of Methane react with 2 mole of Oxygen to give one mole of carbon dioxide and two mole of water gas
Limitation of chemical equation
in a chemical equation as such they do not gives us the following information
1 Physical state
2 Concentration of of reactant and product
3 The conditions as temperature pressure are catalyst
4 The heat exchange
5 The formation of precipitate are evolution of gas during chemical change
6 The speed of reactions
7 Nature of reactions as the chemical reactions is Reversible or Irreversible reactions
Removal of drawback of chemical equation
1 The physical state of reactant and product can specified G for gas l for liquid and s for solid
2 The strength of acid are ways as represent dil for dilute conc for concentration
3 The conditions of the reactions such as Ham pressure pressure catalyst may be written on the arrow sign
4 Hit exchange during chemical change represent in form of enthalpy
5 Precipitation Express byppt i and gases represent over direction arrow
6 Chemical Kinetic Li slow or fast reactions on the arrow head
7 Reversible nature of the reactions represent are indicated by why double headed Arrow sign
TOPIC
Balancing of chemical Equation
According to love conservation of mass The mass Of product must be equal to Mass of reactant
During a chemical change it is a required in a chemical equation which has an equal number of atom of each element in the reactant and the product The chemical equation is called balanced chemical equation
Methods to balance a chemical equation some of these are method are
1 Hit and trial method
2 Partial equation method
3 Oxidation number method
4 Ion electron method
1 Hit and Trial method of trial and error method
this method involves the following steps
write the symbol and formulae of the chemical species in the form of Skeleton equation
if any elementary gas a change in AB form
select the chemical species containing maximum number of atom at start trial and error method
incase the above method fails then start balancing application from minimum number of atom of chemical species
was all the atom of valence once all the atoms are balanced than change the equation into the molecular form
Example let us considered the below chemical equation
2 Partial equation method
according to this concept chemical reactions progress in more than one step each step is called partial equation
each partially equation is balanced separately by hit and trial process
partial equation are multiplied by suitable number if required
Finally , the partial equation are are added to get the final equation
Stoichiometric calculation
1 mole to mole relationship
2 mass to mass relationship
3 mass - volume relationship
4 volume-volume relationship