UNIT 1 ( PAGE 2)
THE MOLE CONCEPT
The word ‘mole’ was introduced around 1896 by Wilhelm Ostwald who derived the term from Latin word moles meaning a ‘heap’ or ‘pile’. A substance may be considered as a heap of atoms and molecules. The unit ‘mole’ in 1967 , was accepted to provide a simple way of reporting a large number – the massive heap of atoms and molecules in a sample.
For a chemist, the unit for counting atoms, molecules, ions, electrons, etc is the mole. A mole represents 6.023 x 1023 particles irrespective of their nature. The number 6.023 x 1023 is called Avogadro’s number(N). Thus 1 mole is equal to 6.023 x 1023 particles.
The mole is also related to the mass of the substance. The quantity of matter containing Avogadro’s number (6.023 x 1023 ) of particles is called one mole of that species. Mass of one mole atoms of any element in grams is equal to atomic mass or one gram atom. Similarly, mass of one mole (6.023 x 1023) molecules of a substance is equal to its gram molecular mass.
According to SI definition, a mole is the amount of substance that contains as many particles(atoms, ions, molecules, etc) as there are atoms in exactly 0.012 kg ( i.e., 12 g) of carbon-12 (12C)
The number of atoms in 12 g of carbon and hence the number of particles in a mole has been found to be 6.023 x 1023.
It is important to note that while using the term mole we must specify the kind of particles because the amount of substance changes with change in the nature of particles. For example , one mole of sulphur atoms contains 6.023 x 1023 atoms of S but a mole of sulphur molecules (S8) contains 6.023 x 1023 of molecules of S8. Therefore, a mole of sulphur molecules contain 8 moles of sulphur atoms.
It must be remembered that :
Mass of one mole of atoms = Gram atomic mass
of the element
Mass of one mole of molecules = Gram molecular
mass of the
substance
Mass of one mole of formula units of ionic compounds
= Gram formula mass of the ionic compound
Mole in terms of Volume
Volume occupied by one gram molecular mass or one mole of a gas under standard conditions (273 K and 760 mm) pressure is called gram molecular volume(G.M.V) . It is 22.4 L for each gas. According to Avogadro’s law, equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. Therefore it follows that 22.4 L of each gas at STP contain the same number of molecules , i.e., Avogadro’s number of molecules (6.023 x 1023). Hence ,
1 mole = gram molecular mass
= 6.023 x 1023 molecules
= 22.4 L at STP
Mole concept in Ionic compounds
Mole concept can be applied even to ionic compounds. One mole of ionic compounds represents one mole(6.023 x 1023 ) formula units of the compound. For example,
One mole of sodium chloride ( NaCl)
= 6.023 x 1023 NaCl units.
= 6.023 x 1023 Na+ ions + 6.023 x 1023 Cl- ions.
Similarly , one mole of Calcium chloride ( CaCl2 )
= 6.023 x 1023 Ca2+ ions + 2 x 6.023 x 1023 Cl- ions.
Mass of one mole of an ionic compound is equal to its gram formula mass. For example,
Mass of one mole of NaCl
= gram formula mass of NaCl = 58.5 g
Conclusion
A mole of substance represents:
¨ 6.023 x 1023 particles
¨ 22.4 L of gas at STP
¨ 1 gram atom of an element
¨ 1 gram molecular mass of a substance
¨ 1 gram formula mass of an ionic substance
Applications of mole concept
This concept is used to calculate:
¨ Absolute mass of an atom or a molecule
¨ Number of atoms or molecules present in a given mass of a substance
¨ Mass of a given number of atoms or molecules
Calculation of Number of Moles
Calculation of mass of an atom of an element.
Calculation of mass of a molecule of an element or compound.
Kilomoles and Millimoles.
Similarly, for small quantities ( in mg ) we may express the amount in millimoles.
EQUIVALENT WEIGHTS OF ELEMENTS
Equivalent weight of an element is defined as the number of parts by weight of the element which combines with or displaces 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen or 35.5 parts by weight of chlorine or one equivalent of any other element.
The fundamental principle is that one equivalent of an element will always combine with or displace one equivalent of any other element.
Illustration
The following examples illustrate the definition of equivalent weight.
1. Chlorine combines with hydrogen to form hydrogen chloride, HCl as :
H2 + Cl2 ® 2 HCl
2 71
Hence 2 parts by weight of hydrogen combine with 71 parts by weight of chlorine, or 1 part by weight of hydrogen will combine with 35.5 parts by weight of chlorine. Hence equivalent weight of chlorine is 35.5.
2. Magnesium burns in oxygen gas to form magnesim oxide.
2 Mg + O2 ® 2 MgO
(2 x 24) (32)
32 parts by weight of oxygen combine with 48 parts by weight of magnesium. 8 parts by wt. of oxygen will combine with 12 parts by wt. of magnesium.
Hence the equivalent weight of magnesium is 12.
3. Sodium combines with chlorine to form sodium chloride.
2 Na + Cl2 ® 2 NaCl
(2 x 23) 71
Here 71 parts by wt. of chlorine is equivalent to 46 parts by wt. of sodium.
Thus, 35.5 parts by weight of chlorine is equivalent to 23 parts by wt. of sodium.
Hence, equivalent weight of sodium = 23.
4. Zinc metal reacts with dil. H2SO4 to liberate hydrogen
gas.
Zn + H2SO4 ® Zn SO4 + H2
65 98 2
Zinc metal reacts with dil. H2SO4 to liberate hydrogen gas.
2 parts by weight of hydrogen is equivalent to 65 parts by wt. of zinc.
1 part by wt. of hydrogen is equivalent to 32.5 parts by weight of zinc.
Hence equivalent weight of zinc = 32.5
Similarly,
2 parts by weight of hydrogen is equivalent to 98 parts by weight of H2SO4
1 part by weight of hydrogen is equivalent to 49 parts by wt. of sulphuric acid..
Hence equivalent wt. of H2SO4 = 49
Equivalent weights can also be calculated from the formulae of the compounds. For example:
5. AlCl3
3 x 35.5 parts by wt. of chlorine combine with 27 parts by weight of Al. 35.5 parts by wt. of chlorine will combine with 9 parts by weight of Al.
Equivalent weight of Al = 9
6. FeO
16 parts by weight of oxygen equivalent to 56 parts by wt. of Fe . 8 parts by wt. of oxygen is equivalent to 28 parts by wt. of Fe.
Equivalent weight of Fe = 28
Gram Equivalent weight
The weight of a substance in grams which is numerically equal to its equivalent weight is called gram equivalent weight or one gram equivalent weight. It is simply the equivalent weight of the substance expressed in grams. For example, 1 gram equivalent of oxygen is 8 g .
Gram equivalent weight = 1 gm-equivalent.
Relation between Equivalent weight and valency
Equivalent weight of Acids
Equivalent weight of an acid is the number of parts by weight of it which can give one part by weight of H+ ions in its aqueous solutions or one part by weight of replaceable hydrogen.
The number of H+ ions that one molecule of an acid can give in its aqueous solution is known as its basicity.
Equivalent weight of a Base
Equivalent weight of a base is the number of parts by weight of it which can neutralise one equivalent of an acid or it is the number of parts by weight of it which can furnish one equivalent or 17 parts by weight of OH- ions in aqueous solution.
The number of OH- ions that one molecule of a base can give in its aqueous solution is called its acidity. Therefore,
Equivalent weight of a Salt
The equivalent weight of a salt is that weight of it which contains one equivalent of the metal.
Total valency of a metal is equal to the valency of the metal ion multiplied by the number of atoms present in the formula of the salt.
Equivalent weight of oxidising and reducing agents.
A given substance may have different equivalent weights in various reactions. In a simple way, the equivalent weight of an oxidising agent is calculated from the number of oxygen atoms it produces and equivalent weight of a reducing agent is calculated from the number of oxygen with which it combines. For example,
i) Equivalent weight of KmnO4 in the reaction :
2 KMnO4 +3 H2SO4 ® K2SO4 + 2 MnSO4 + 3 H2O+ 5[O )
2 x 158 g 5 x 16 g
5 x 16 parts of oxygen are liberated from KMnO4 = 316 parts
8 parts of oxygen are liberated from KMnO4 = (316/ 10)
= 31.6
Equivalent weight of KMnO4 = 31.6
ii) Equivalent weight of K2Cr2O7 in the reaction :
K2Cr2O7 + 4 H2SO4 ® K2SO4 + Cr2 ( SO4 )3 +4 H2O +3(O )
294 g 3 x 16 g
48 parts of oxygen are liberated from K2Cr2O7 = 294
8 parts of oxygen are liberated from K2Cr2O7
= ( 294 x 8 ) / 48 = 49
Equivalent weight of K2Cr2O7 = 49
iii) Equivalent weight of FeSO4 in the reaction :
2 FeSO4 + H2SO4 + O ® Fe2 ( SO4 )3+ H2O
2 x 152 16
16 parts of oxygen react with FeSO4 = 304 parts
8 parts of oxygen react with FeSO4 =
= ( 304 x 8 ) / 16 = 152
Equivalent weight of FeSO4 = 152
Problems
01. Calculate the amount of carbon dioxide that could be produced when :
(i) 1 mole carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g dioxygen.
02. In three moles of ethane (C2H6) , calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number molecules of ethane.
03. How are 0.5 mol Na2CO3 and 0.5 M Na2CO3 different ?
04. How many atoms of oxygen are present in 0.5 mole of oxygen atoms ?
05. How many moles of sulphur are there in 80 g of sulphur if the molecular formula is S8 ?
06. Calculate the number of methane molecules and number of hydrogen and carbon atoms in 25 g of methane.
07. Calculate the approximate volume of 1 molecule of water.
08. Calculate the difference in the number of atoms in 1 g of C-14 ( atomic mass = 14.01 g / mol ) and 1 g of C-12 isotope of carbon.
09. How many years it would take to spend Avogadro number of rupees at a rate of 10 lakh rupees per sec.
10. Calculate the total number of electrons present in 1.6 g of methane.
11. Calculate the total number of moles present in 12 g ozone and 32 g of sulphur dioxide in a gaseous mixture.
12. Calculate the number of molecules in a drop of water weighing 0.05 g.
13. Calculate the mass of a silver atom.
14. How many molecules of water are there in 1 litre of water ?
15. How many : (a) moles of P atoms (b) moles of P4 molecules (c) molecules of P4 are contained in 92.91 g of phosphorus ? ( Atomic mass of P = 30.974 )
16. Which of the following weighs most ?
i) 1 mole of Na2CO3 ii) 10 g atom of Helium
iii) 22.4 L of CO2 at STP iv) 6.023 x 1024 atoms
of hydrogen.
17. Calculate the mass of one molecule of oxygen.
18. How many (a) atoms and (b) gram-atoms are contained in 20 g of calcium.
19. Calculate the number of atoms in each of the following :
i) 52 moles of He ii) 52 u of He iii) 52 g of He
20. Calculate the mass of the following :
i) one atom of Ca ii) one molecule of sulphur dioxide.
21. Calculate the mass of sodium which contains same number of atoms as are present in 4 g of Ca.
22. A complex of magnesium contains 2.68 % of magnesium by mass. Calculate the number of magnesium atoms in 2 g of this complex.
23. Calculate the number of atoms of each type in 3.42 g of sucrose .
24. Find out the volume of the following at STP :
i) 14 g of nitrogen
ii) 6.023 x 1022 molecules of ammonia
iii) 0.1 mole of sulphur dioxide.
25. Calculate the volume occupied at STP by :
i) 16 g oxygen
ii) 50 moles of oxygen
iii) 6.023 x 1024 molecules of oxygen
EMPIRICAL FORMULA AND MOLECULAR FORMULA
A formula which gives the simplest whole number ratio between atoms of various elements present in one molecule of the substance is known as the Empirical formula. This does not convey any information about the actual number of atoms in the molecule. The empirical formula of glucose is CH2O i.e., the ratio of various atoms is C : H : O = 1 : 2 : 1. Actually glucose contains 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms.
The empirical formula can be calculated from the Percentage composition of the compound by the following steps.
i) Divide the percentage composition of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.
ii) Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.
iii) Multiply the figures obtained by suitable integer
iv) Finally write down the symbols of the various elements side by side and put the above numbers as subscripts to the lower right hand corner of each symbol. This will represent the empirical formula of the compound.
MOLECULAR FORMULA
Molecular formula of a compound may be defined as the formula which gives the actual number of atoms of various elements present in the molecule of the compound.
Molecular formula of the compound is a simple whole number multiple of its empirical formula . The two are related by the following relation.
Molecular formula = n ( Empirical formula )
where ‘n’ is a simple whole number and may have values 1,2,3.... and it is the ratio between the molecular mass of the compound and its empirical formula weight.
The steps involved in the calculation of molecular formula of a substance are as under :
i) Calculation of empirical formula.
ii) Calculation of molecular mass of the compound.
iii) Calculation of the value of ‘n’ by dividing the molecular mass by empirical formula weight.
iv) Calculation of molecular formula by multiplying empirical formula with ‘n’.
Problems
26. Write the empirical formula of the compounds having the following molecular formula.
i) C6H6 ii) C6H12 iii) H2O2
iv) H2O v) Na2CO3 vi) B2H6 vii) N2O4
27. A substance on analysis gave the following percentage composition. Na = 43.4 % ; C = 11.3 %; O = 45.3 % . Calculate the empirical formula of the substance.
28. Naphthalene (moth balls) contains 93.71% carbon and 6.29% hydrogen. Its molecular mass is 128 g mol-1 , calculate its molecular formula.
29. (a) Butyric acid contains only c, H and O . A 4.24 mg
sample of butyric acid is completely burned, it gives 8.45 mg of CO2 and 3.46 mg of water. What is the mass percentage of each element in butyric acid ?
(b) The elemental composition of butyric acid is found to be 54.2 % C, 9.2 % H and 36.6 % O. Determine the empirical formula.
(c) The molecular weight of butyric acid was determined to be 88 amu. What is the molecular formula ?
30. A crystalline salt being rendered anhydrous, loses 45.6 % of its weight. The percentage composition of the anhydrous salt : Al = 10.5 % ; K = 15.1 % ; S = 24.8 % O= 49.6 %. Find the simplest formula of the anhydrous and crystalline salts.
31. A substance is found to contain :
C u =25.45 % ; S=12.82 % ; O = 57.7 % and H = 4.01 % Calculate the formula of the substance on the assumption that all hydrogen in the compound is present in combination with oxygen as water of crystallization.
32. Calculate the molecular formula of the following two compounds from the given data :
I Compound : Na = 28.55 % ; S = 40 % ; O = 31.45 % ; and
vapour density = 79.
II Compound : C = 19.5 % ; Fe = 15.2 % ; N = 22.8 % ;
K = 42.5 % ; Its molecular mass = 368.
Writing formulae
We use symbols, formulae of radicals and valencies to write the formulae of compounds.The formulae and valence of some common ions are given in the following Table 1.8.
Table 1.8
Radical | Formula | valency |
Ammonium | NH4+ | 1 |
Hydoxide | OH- | 1 |
Nitrate | NO3- | 1 |
Nitrite | NO2- | 1 |
Bicarbonate | HCO3- | 1 |
Bisulphate | HSO4- | 1 |
Bisulphite | HSO3- | 1 |
Permanganate | MnO4- | 1 |
Chlorate | ClO3- | 1 |
Perchlorate | ClO4- | 1 |
sulphate | SO42- | 2 |
Sulphite | SO32- | 2 |
Carbonate | CO32- | 2 |
Manganate | MnO42- | 2 |
Chromate | CrO42- | 2 |
Dichromate | Cr2O72- | 2 |
Silicate | SiO32- | 2 |
Zincate | ZnO22- | 2 |
Thiosulphate | S2O32- | 2 |
Aluminate | AlO33- | 3 |
Phosphate | PO43- | 3 |
Suppose the formulae of barium chloride and potassium sulphate are to be written. The following rules are used :
1. Write the symbols for elements and radicals.
Ba Cl K SO42-
2. Write the valencies at the right hand top corner, which gives the formulae of the compound.
Ba2 Cl1 K1 SO42
3. Reverse the valencies and write at the right hand bottom corner, which gives the formulae of the compound.
Problem
90. Write the molecular formula of the following compounds.
i) Ammonium dichomate v) Sodium carbonate
ii) Silver nitrate vi) Aluminium phosphate
iii) Calcium bromide vii) Barium oxide
iv) Auric chloride viii) potassium chlorate
CHEMICAL EQUATIONS
Representation of chemical change in terms of symbols and formulae of the reactants and products is called a Chemical Equation.
A chemical equation must fulfill the following conditions:
i) It should represent a true chemical change, ie., if a reaction is not possible between certain substances, it cannot be represented by a chemical equation.
ii) It should be balanced.
iii) It should be molecular, ie., all the species should be represented in their molecular form. For example, elementary gases like hydrogen, oxygen etc., should be represented as H2 and O2.
Information conveyed by a Chemical Equation
A chemical equation has both qualitative as well as quantitative significance.
i) Qualitatively, a chemical equation tells us the names of the various reactants.
ii) Quantitatively, it expresses :
a) The relative number of molecules of reactants and products taking part in the reaction.
b) The relative number of moles of reactants and products.
c) The relative mass of reactants and products.
d) The relative volumes of reactants and products.
Illustration
As an illustration consider the following example:
Zn + 2 HCl ® ZnCl2 + H2
(65 g) ( 2 x 36.5 ) (65 + 2 x 35.5 = 136 g) (2g)
Qualitatively, it tells that zinc reacts with hydrogen chloride to produce zinc chloride and hydrogen.
Quantitatively
i) One atom of zinc reacts with two molecules of hydrogen chloride to produce one molecule of zinc chloride and one molecule of hydrogen.
ii) 65 parts by weight of zinc reacts with 73 parts by weight of hydrogen chloride to produce 136 parts by weight of zinc chloride and 2 parts by weight of hydrogen.
iii) 22.4 L of hydrogen is produced at STP when one mole of zinc atoms reacts with 2 moles of hydrogen chloride molecules.
Limitations of Chemical Equations
Chemical equations are useful mode of representing the chemical changes. However, they have a number of drawbacks. The equation do not tell us the following facts.
i) The physical state of reactants and products.
ii) The concentrations of reactants and products.
iii) The conditions of the reaction.
iv) The speed of the reaction.
v) The heat changes accompanying the reaction.
vi) Precipitation or evolution of gas.
Removal of Drawbacks of Chemical Reactions.
1. The physical states of the reactants and products can be specified by the use of certain letters. For example, the letters s, l and g are used to denote solid, liquid and gaseous state of substance respectively. The word ‘aq’ is used to represent that the substance is in the form of its solution in water. These letters are written after the symbol or formula of the substance. For example,
CaCO3 (s) + H2SO4(aq) ® CaSO4(s) + H2O (l) + CO2 (g)
ii) Actual concentration of reactants and products can not be mentioned. However, in the case of an acid or base the words con. for concentrated and dil. for dilute are used before the formula of the acid or base. For example,
Zn(s) + dil 2 HCl ® ZnCl2 (aq) + H2 (g)
iii) The conditions of temperature, pressure, catalyst etc. can be mentioned on the arrow separating reactants and products. For example,
The above equation indicates that this reaction between
hydrogen and nitrogen has been carried out in the presence of Fe/Mo (catalyst) at 273 K and 600 atm. pressure.
iv) The speed or rate of the reaction cannot be mentioned
in the equation.
v) Heat changes can be represented by writing +q on the product side in case heat is produced during the reaction. The absorption of heat can be represented by writing -q kJ on the product side. The letter q can be replaced by actual numerical value also . For example,
C(s) + O2 (g) ® C O2 (g) + 394 kJ
( heat is evolved)
N2 (g) + O 2 (g) ® 2 NO(g) -180.5 kJ
(heat is absorbed )
Balancing Chemical Equations
Following methods are generally used for balancing chemical equations.
1. Hit and Trial Method
There is no definite rule for balancing equation by this method and it requires a good deal of skill and practice. A few guidelines to balance an equation are :
a. Select the biggest formula (containing largest number ofatoms), no matter on what side of the equation it occurs.
b. Balance the atoms combined in this formula on both
sides.
c. If an elementary gas ( like H2, N2, F2,Cl2,Br2, I2) is present in the equation , balance the equation by keeping such a gas in the atomic state . Multiply the atomic equation by 2 to get the molecular equation.
For example , to balance the equation :
Al + HCl ® AlCl3 + H ( skeleton equation )
We select the biggest formula AlCl3 . In order to balance three chlorine atoms in it, we must multiply HCl on the left by 3. Then we balance the 3 hydrogen atoms of hydrochloric acid by multiplying hydrogen on the right by 3.
Al + 3 HCl ® AlCl3 + 3 H ( atomic equation)
Finally , we make equation molecular by multiplying whole
equation by 2.
2 Al + 6 HCl ® 2 AlCl3 + 3 H2 (molecular equation)
2. Partial Equation Method
This method is mostly used for balancing complicated equations. The method involves the following steps.
Step I
Split the chemical reaction in two or more stages.
Step II
Write equations for different stages and balance by hit and trial method. These reactions are called Partial Equations .
Step III
If necessary, the partial equations are multiplied by suitable numbers to cancel those intermediate products which do not occur as products in the resultant equation.
Step IV
Finally, the partial equations are added to get balanced equation.
The method is illustrated by the following example. Action of chlorine on sodium hydroxide to form sodium chloride , sodium chlorate and water.
NaOH + Cl2 ® NaCl + NaClO3 +H2O ( unbalanced )
Balancing it by partial equation method :
Cl2 + H2O ® HCl + HClO ......... (1)
NaOH + HCl ® NaCl + H2O .......... (2)
NaOH + HClO ® NaClO + H2O .......... (3)
3 NaClO ® 2 NaCl + NaClO3 ........ (4)
In order to cancel the intermediate product NaClO, we multiply the partial equation (3) by 3. In order to cancel the 3 HClO, we multiply equation (1) by 3. Finally, in order to cancel 3 HCl we multiply equation (2) by 3, so we get :
3 Cl2 + 3 H2O ® 3 HCl + 3 HClO
3 NaOH + 3 HCl ® 3 NaCl + 3 H2O
3 NaOH + 3 HClO ® 3 NaClO + 3 H2O
3 NaClO ® 2 NaCl + NaClO3
_______________________________________
Adding : 6 NaOH + 3 Cl2 ® 5 NaCl + NaClO3 + 3 H2O
( Balanced equation )
Problems
91. Balance the following equations :-
i) CaCO3 + HCl ® CaCl2 + CO2 + H2O
ii) Cu + HNO3 ® Cu(NO3)2 + NO + H2O
iii) Fe2O3 + CO ® Fe + CO2
iv) Fe + H2O ® Fe3O4 + H2
v) P + NaOH ® NaH2PO2+ PH3
vi) KmnO4 + FeSO4 + H2SO4 ®
K2SO4 + MnSO4 + Fe2 (SO4)3 + H2O
STIOCHIOMETRY OF CHEMICAL REACTIONS
In chemical reactions, the reactants and products are in the ratio of their moles. This forms the basis of all chemical calculations. These calculations are helpful in quantitative chemical calculations. The calculation of relative amounts of substances in chemical reaction is called stoichiometry ( derived from Greek words Stoicheion = element and metron = measure)
Following are the important steps involved in the chemical calculations.
i) Write the balanced chemical equation for the chemical reaction involved and write the number of moles under each reactant or product.
ii) Convert the given quantity of substance into moles.
iii) From the ratio of the moles, calculate the number of moles of the desired substance or substances involved in the reaction.
iv) Convert the moles back into grams of the substance asked for.
v) For reactions involving gases, apply Avogadro’s hypothesis as well as the relation :
1 mole of a gas = 22.4 litres at S.T.P
vi) For reactions taking place in solution, calculations are made with molarity ( number of moles per litre) or normality ( number of gram equivalents per litre).
Limiting Reagents
In many situations, one of the reagents is present in excess, therefore, some of the reactant is left over on completion of the reaction. For example consider the combustion of hydrogen,
2 H2(g) + O2(g) ® 2 H2O(g)
Suppose that 2 moles of H2 and 2 moles of O2 are available for the reaction. It follows that only 1 mole of O2 is required for complete combustion of 2 moles of H2 ; 1 mole of O2 will therefore left over on completion of the reaction. The amount of product obtained is determined by the amount of the reactant that is completely consumed in the reaction. The reactant is called Limiting Reagent. Thus the limiting reagent may be defined as the reactant which is completely consumed during the reaction. In the above reaction, H2 is the limiting reagent.
Problems
92. Calculate the amount of water produced by combustion of 16 g of methane.
93. How many moles of methane are required to produce 22 g of
CO2(g) after combustion ?
94. 50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g) . Calculate NH3(g) formed. Identify the limiting reagent in the production of NH3 in this situation.
95. Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO3). Decomposition of potassium chlorate gives potassium chloride and oxygen. If 2.4 mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed ?
96. If 20.0 g CaCO3 is treated with 20.0 g of HCl, how many grams of CO2 can be generated according to the equation
CaCO3(s) + 2 HCl(aq) ® CaCl2(aq) + H2O(ℓ) + CO2(g)
97. What weight of iron will be converted into its oxide(Fe3O4) by the action of 18 g of steam ?
98. How much weight of potassium chlorate (KClO3) will be required to decompose to give 0.96 g of oxygen ?
99. One gram of a sample of magnesium containing its own oxide as impurity , on treatment with excess of dilute sulphuric acid gave out 896 ml of H2 at S.T.P. Calculate the percentage purity of the sample.
STIOCHIOMETRIC CALCULATIONS INVOLVING SOLUTIONS
A solution is a single phase homogeneous system, containing two or more substances. The component of a solution forming the larger proportion is referred to as solvent ; while the other component which is present in minor proportion in solution , is called solute. The composition of the solution may vary within wide limits. Examples of solutions are common salt in water, alcohol in water, sugar in water etc.
UNITS OF CONCENTRATIONS OF SOLUTIONS
The concentration of a solution may be defined as the amount of solute present in the given quantity of the solution. It is usually expressed in any one of the following ways .
1. Mass Percentage
The mass percentage of a component in a given solution is the mass of the component in 100 g of the solution.
If mA and mB are the masses of the two components A and B respectively, in a binary solution, then
2. Volume Percentage
Volume percentage is defined as the volume of the component per 100 parts by volume of solution. If VA ml is the volume of one component A and VB ml is the volume of the second component B, then :
3. Parts per million
When a solute is present in very minute amounts, the concentration is expressed in parts per million abbreviated as ppm. It is the parts of a component per million parts of a solution. It is expressed as :
This mode is generally used to express very low concentration such as hardness of water or concentration of chlorine in public supply of potable water.
4. Molarity (M)
It is the number of moles of the solute dissolved per litre of the solution. If nB moles of solute are present in V litres of solution , then :
* A solution having molarity one is called molar solution. Such a solution contains one mole of solute per litre of solution.
* Molarity is expressed in moles per dm3.
* Molarity of a solution changes with change in temperature.
5. Molality (m)
It is the number of moles of the solute dissolved per 1000 g (= 1 kg) of the solvent.
If nB is the number of moles of solute and WA is the weight of the solvent in grams, then molality of the solution is :
· A solution containing one mole of solute per 1000 g of solvent has molality equal to one and is called a molal solution.
· Molality is expressed in units of moles per kilogram
( mol / kg).
Note : Molality is considered better for expressing the concentration as compared to molarity because the molarity changes with temperature because of expansion or contraction of the liquid with temperature. However, molality does not change with temperature because mass of the solvent does not change with change in temperature.
6. Normality(N)
It is the number of gram equivalents of the solute dissolved per litre of solution.
· A solution having normality equal to one is called a normal solution. Such a solution contains one gram equivalent of solute per litre of solution.
· A decinormal solution contains 1/10 g equivalents of solute per litre of solution.
· A seminormal solution contains ½ g equivalents per litre.
· A centinormal solution contains 1/100 g equivalents per litre.
Relationship between Normality and Molarity of solution
The molarity and normality of the solution are related as :
* For acids, Normality = Molarity x Basicity of acid
* Basicity is the number of H+ ions furnished by each molecule of acid in aqueous solutions.
* For bases, Normality = Molarity x Acidity of a base.
* Acidity is the number of OH- ions furnished by each molecule of base in solutions.
7. Mole fraction (X)
Mole fraction of any component in a solution is the ratio of the number of moles of that component to total number of moles of solute plus solvent in solution.
Let us suppose that a solution contains nA moles of solvent and nB moles of solute
The sum of the mole fractions must be equal to 1 i.e.,
Thus, if the mole fraction of one component is known, that of the other can be calculated. For example ,
XA = 1 - XB or XB = 1 - XA
8. Formality (F)
Formality of a solution may be defined as the number of gram formula masses of ionic solute dissolved per litre of the solution.
· Formality is used to express the concentration of the ionic solids which do not exist as molecules but exist as net work of ions.
· A solution containing one gram formula mass of the solute per litre of solution has formality equal to one and is called formal solution.
· Formality of a solution changes with change in temperature.
VOLUMETRIC CALCULATIONS
Equivalent weight of acids and bases and salts are commonly determined by volumetric analysis, i.e., analysis involving measurement of volumes. In volumetric analysis , a known volume of the given solution is taken in a conical flask (titration flask) . The solution taken in the titration flask is called titant and the solution which is made to react with titrant is called titre. The process of completely reacting a known volume of the solution of one substance with the solution of another substance is known as titration. The point at which the chemical reaction involved in the titration is just complete is called the end point and the substance used to indicate the end point in the titration is called the indicator.
Normality Equation
The expression which relates the normalities and volumes of the two reacting solutions in a titration is called normality equation. It is expressed as :
N1 V1 = N2 V2
where N1 and V1 represent normality and volume of one solution and N2 and V2 represent normality and volume of the second solution.
Molarity equation or dilution formula
Acids like HCl, H2SO4, CH3COOH etc ., are available as concentrated aqueous solutions. The solutions required in the laboratory are prepared by diluting the concentrated solutions with water. When a concentrated solution is diluted by adding more solvent, the number of moles of solute in the solution remains unchanged.
Number of moles of solute before dilution = Number of moles of solute after dilution.
Number of moles of the solute = M x Volume of solution in litres.
M1 V1 = M2 V2
where, M1 = Initial molarity ; M2 = molarity after
dilution
V1 = Initial volume ; V2 = volume after
dilution.
Molarity of a mixture of two solutions of same substance
If V1 ml of a solution of molarity M1 is mixed with V2 ml of the solution of the same substance with molarity M2. Then the molarity (M) of the resulting solution is given by :
M (V1 + V2) = M1 V1 + M2 V2
Problems
100. A solution is prepared by adding 2 g of a substance A to 18 g
of water. Calculate the mass percentage of the solute.
101. Calculate the molarity of NaOH in a solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
102. Density of 3 M solution of NaCl is 1.25 g mL-1. Calculate molality of the solution.
103. A solution is prepared by dissolving 18.25 g of sodium hydroxide in distilled water to give 200 mL of solution. Calculate the molarity of sodium hydroxide solution.
104. How many moles and how many grams of sodium chloride are present in 250 mL of 0.50 M NaCl solution ?
105. Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density 1.84 g mL-1. What volume of concentrated sulphuric acid is required to make 5.0 L of 0.50 M H2SO4 solution ?
106. 500 mL of 0.250 M Na2SO4 solution is added to an aqueous solution of 15.00 g of BaCl2 resulting in the formation of the white precipitate of insoluble BaSO4 . How many moles and how many grams of BaSO4 are formed ?
107. Reaction 2 Br- (aq) + Cl2(aq) ® 2 Cl- (aq) + Br2(aq) is used for commercial preparation of bromine from its salts. Suppose we have 50.0 mL of a 0.060 M solution of NaBr. What volume of a 0.050 M solution of Cl2 is needed to react completely with Br- ?
108. A sample of NaNO3 weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck. What is the molarity of the solution ?
109. In a reaction vessel 0.184 g of NaOH is required to be added for completing the reaction. How many millilitre of 0.15 M NaOH solution should be added for this requirement ?
110. Zinc and hydrochloric acid react according to the reaction :
Zn (s) + 2 HCl(aq) ® ZnCl2(aq) + H2(g).
If 0.30 mol of Zn are added to hydrochloric acid containing
0.52 mol of HCl, how many moles of H2 are produced ?
111. Calculate the molarity of a solution containing 6.84 g of a substance in 250 cm3 of a solution ( molecular mass of the substance = 342 )
112. How much of urea should be added to 200 cm3 of water to get 0.1 molar solution ( molecular mass = 60 ).
113. (a) A sample of NaOH weighing 0.38 g is dissolved in water and the solution is made to 50 cm3 in a volumetric flask. What is the molarity of the resulting solution ?
(b) How many moles of NaOH are contained in 27 cm3 of 0.15 M NaOH ?
114. A solution of NaOH contains 2 g of NaOH per litre of solution. Calculate the molarity of the solution.
115. 2.8 g of KOH dissolved in water to give 200 cm3 of solution. Calculate the molarity of KOH solution.
116. Calculate the number of moles and amount in grams of NaOH in 250 cm3 of a 0.100 M NaOH solution.
117. How much volume of 10 M HCl should be diluted with water to prepare 2 L of 5 M HCl ?
118. What volume of 6 M HCl and 2 M HCl to should be mixed to one litre of 3 M HCl ?
119. Commercially available concentrated hydrochloric acid contains 38 % HCl by mass. (a) What is the molarity of the solution ? (b) What volume of concentrated HCl is required to make 1 L of 10 M HCl . Density of HCl = 1.19 g cm-3.
120. How many grams of oxygen are required to completely react with 0.2 g of hydrogen to yield water. Also calculate the amount of water formed.
121. 1 g of Mg is burned in a closed vessel, which contains 0.5 g of oxygen. What reactant is left in excess ?
122. 100 g of CaCO3 is treated with 500 cm3 of M/2 solution of HCl . Find out the volume of CO2 evolved at STP. What substance is acting as limiting reagent ?
123. A welding fuel gas contains carbon and hydrogen only. Burning 1 g of a 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 STP) of this welding gas is found to weigh 11.6 g . Calculate (i) empirical formula (ii) molecular mass of the gas , and (iii) molecular formula.
124. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3(s) + 2 HCl(aq) ® CaCl2(aq) + CO2(g) + H2O(ℓ)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl ?
125. In the reaction A + B2 ® AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol of A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol of A + 2.5 mol of B
(v) 2.5 mol A + 5 mol B
QUESTIONS
1. What is a unit
2. What is meant by fundamental quantities ?
3. What are the fundamental quantities (or units) in SI system ?
4. What is derived unit ? Give some examples.
5. Deduce the SI units of :
i) Area ii) Force iii) Volume iv) Work v) Density vi) Energy vii) Speed viii) Concentration ix) Acceleration
6. What is meant by significant figures ? Illustrate giving examples.
7. Explain the term uncertainty.
8. Differentiate between accuracy and precision by giving suitable examples.
9. What are the rules observed in counting significant figures, when result of an experiment is expressed ?
10. What is matter ? Give few examples.
11. In how many states do matter exist ? Give examples of each state.
12. Give the points of difference between solids, liquids and gases.
13. Define the term element.
14. What is meant by a compound ? Give examples.
15. What do you understand by the terms
(i) atom (ii) molecule ?
16. Define mixture and give some examples.
17. What are various types of mixtures ?
18. Give three differences between compounds and mixtures.
19. What is a pure substance ? What are its characteristics ?
20. What is atomic mass of an element ?
21. Why is atomic mass called relative atomic mass ?
22. Why atomic mass is not expressed in absolute weight ?
23. What is meant by molecular mass of a compound ?
24. State and explain :
i) The law of conservation of mass.
ii) The law of definite proportions.
iii) Law of reciprocal proportions
iv) Gay- Lussac's law of Gaseous volumes.
25. State and illustrate Avogadro's hy[pothesis.
26. What is the importance of Avogasdro's law ?
27. What is meant by Avogadro's number ?
28. Explain the terms :
i) Gram atomic mass.
ii) Gram molecular mass.
29. What is meant by molecular formula of the compound ? How is it calculated ?
30. What is meant by chemical equation ? What are its limitations ?
31. Explain the significance of chemical equation. What are its essentials ?
32. What is meant by molarity of a solution ?
33. What is meant by the term 'mole' ?
34. Describe the methods of balancing a chemical equation ?
35. What do you understand by the term 'limiting reagent’ ?
36. Define different methods used for expressing the concentration of a solution.
37. When do zeros present in a number become significant ?
38. Is velocity a basic or derived quantity according to SI system ?
39. Why we regard the gaseous state of water as vapours while that of ammonia as a gas ?
40. Why is distilled water a compound while tap water a mixture ?
41. Give one limitation of law of conservation of masas.
42. Which law correlates the mass and volume of a gas ?
43. An element has fractional atomic mass. What does it indicate ?
44. What is the difference between the mass of a molecule and molecular mass ?
45. Where do we use the words mole and mol ?
46. Does one gram mole of a gas occupy 22.4 L under all conditions of temperature and pressure ?
47. Why is it necessary to balance a chemical equation ?
48. In the combustion of methane, why methane is regarded as the limiting reactant ?
49. What is the difference between molarity and molality ?
50. What is a limiting reagent ? What is its significance in stoichiometric calculations ?