# Chapter 3 Atoms and Molecules

Atom (Parmanu) is the basic unit for the chemicals or it better to say that existence of everything around us. If we look in our surrounding we see a lot of thing each and every part is made of atom of different element. We will understand how atom of an element exist all about it mass and properties and what what make them to form molecule and compound.

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### Exercise 1

•  Q1 In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium observations are in agreement with the law of conservation of mass. sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water Ans: Law of conservation of mass: Mass can neither be created nor be destroyed in a chemical reaction. Given reaction: sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water Total mass of reactants = 5.3 + 6 = 11.3 g. Total mass of products = 8.2 + 2.2 + 0.9 = 11.3 g  Total mass of reactants = Total mass of products Hence, these observation are in agreement with the law of conservation of mass. Q2 Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas? Ans: Given, Ratio of hydrogen and oxygen by mass = 1:8  Mass of oxygen gas that completely reacts with 1g of hydrogen gas is equal to 8 g i.e Mass of oxygen gas that completely reacts with 3 g of hydrogen gas = 8 x 3 = 24g.  Hence, 24 g mass of oxygen gas is required to react completely with 3 g of hydrogen gas. Q3 Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? Ans: Postulate of Dalton's atomic theory: Atoms are indivisible particles which cannot be created or destroyed in a chemical reaction. It is the result of the law of conservation of mass. Q4 Which postulate of Dalton’s atomic theory can explain the law of definite proportions? Ans: The postulate of Dalton's theory, "The relative numbers and finds of atoms are content in a given compound” can explain the law of definite proportion.

### Exercise 2

•  Q1 Define the atomic mass unit. Ans: Atomic mass unit: Mass unit equals to exactly 1/12 th the mass of one atom of carbon-12 is known as an atomic mass unit. Note: Carbon-12: Standard reference for measuring atomic Masses.  1 amu (atomic mass unit) = 1.66x10-24 g. Q2 Why is it not possible to see an atom with naked eyes? Ans: The size of an atom is very very small that we do not see with our naked eyes. Atoms of mait elemente cannot exist independently.

### Exercise 3

•  Q1 Write down the formulae of (i) sodium oxide (ii) aluminium chloride (iii) sodium suphide (iv) magnesium hydroxide Ans: (i)  Na2O (Sodium oxide) (ii)  AlCl2 (Aluminium chloride) (iii)  Na2S (Sodium sulphide) (iv)  Mg(OH)2 (Magnesium hyelsonide) Q2 Write down the names of following formulae: (i) Al2(SO4)3 (ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3. Ans: (i)  Aluminium sulphate (ii)  Calcium chloride (iii)  Potassium sulphate (iv)  Potassium Nitrate (v)  Calcium Carbonate Q3 What is meant by the term chemical formula? Ans: It is the symbolic representation of the compound's composition. For example, the chemical formula of Calcium Carbonate is CaCo3. Q4 How many atoms are present in a (i) H2S molecule and (ii) PO43- ion? Ans: (i)  3 atoms are present in the H2S molecule. There are 2 atoms of hydrogen and one is sulphur. (ii)  5 atoms are present in PO43- ion. There are 4 atoms of oxygen and one in phosphorus.

### Exercise 4

•  Q1 Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH. Ans: Atomic mass includes the masses of three subatomic particles which make up an atom i.e. proton, neutron and electron. Unit of atomic mass is 'u' ‘u’ = unified atomic mass unit & 1/12 th mass of single carbon-12 atom  ‘u’ = 1.66 x 10-24 g (Just for understanding - Atomic mass approximately equal to twice of atomic no.) Molecular mass of a substance is the sum of atomic masses of all the atoms in a molecule of a subitance. Unit of molecular mass = 'u’ Molecular mass of H2 = 2 x 1 = 2u (Atomic mass of H=1) Molecular mass of O2 = 16 x 2 = 32 u (Atomic mass of O=16) Molecular Mass of Cl2 = 35.2 x 2 = 71 u (Atomic mass of Cl = 35.5) Molecular Mass of CO2 = 12 x 1 + 16 x 2                 = 12 + 32 = 44 u ( Atomic Mass of C = 12) Molecular Mass of CH4 = 12 x 1 + 4 x 1                = 12 + 4 = 16 u Molecular Mass of C2H6 = 12 x 2 + 6 x 1                = 24 + 6 = 30 u  Molecular Mass of C2H4 = 12 x 2 + 4 x 1                = 24 + 4 = 28 u Molecular Mass of  NH3 = 14 X 1 + 3 x 1                = 14 x 3 = 17 u  (Atomic Mass of N = 14) Molecular Mass of CH3OH = 12 x 1 + 4 x 1 + 16 x 1                = 12 + 4 + 16 = 32 u Q2 Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn=65 u, Na=23 u, K=39 U, C=12 u and O=16 u. Ans: Formula unit mass of ZnO = 1 x Atomic mass of Zn + 1 x Atomic mass of ‘O’ atom  = 1 x 65 + 1 x 16 = 65 + 16 = 81 u   Formula unit mass of Na2o   = 2 x Atomic mass of Na + 1 x Atomic mass of ‘O’ atom = 2 x 23 + 1 x 16 = 46 + 16 = 62 u   Formula unit mass of K2CO3 = 2 x Atomic mass of K + 1 x Atomic mass of C + 3 x Atomic mass of ‘O’ = 2 x 39 + 1 x 12 + 3 x 16 = 78 + 12 + 48 = 138 u

### Exercise 5

•  Q1 If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon? Ans: 1 mole of carbon atoms = 6.022 x 1023 1 mole of carbon atoms weighs = 12 gram Mass of 6.022 x 1023 atoms of carbon = 12 gram  1 atom of carbon = 12 / 6.022 x 1023     = 1.99 x 10-23 gram Hence, Mass of 1 atom of carbon    = 1.99 x 10-23 gram Q2 Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)? Ans: Atomic mass of sodium = 23 u Gram atomic mass of sodium = 23 g  1 mole or 23 g of sodium has = 6.022 x 1023 atom 100 g of sodium has = 6.022x1023 / 23 x 100        = 2.618 x 1024 atom Atomic mass of Ison (fe) = 56 u Gram atomic mass of fe = 56 g 1 mole of fe has = 6.022 x 1023 atom 56 g of fe = 6.022 x 1023 atom 100 g of fe = 6.022 x 1023 / 56 x 100       = 1.075 x 1024 atom Hence, 100 gram of Na has more number of atoms than 100 gram of fe.

### Exercise 6

•  Q1 A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight. Ans: Mass of Boron  =  0.096 g Маss of Oxygen = 0.144 g Mass of sample of compound of oxygen and Boron = 0.24 g  % of Boron by weight in the compound = 0.096 x 100 / 0.24 = 40% % of Oxygen by weight in the compound = 0.144 x 100 / 0.24 = 60% Q2 When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer? Ans: 1 mole of C = 12 g 1 mole of O2 = 32 g C + O2 → CO2 12 g  +  32 g  →  44 g  4           4            4 [3g + 8g - 11g] → constant 8 g of O2 will be consumed from 50g of O2 and 42 g of O2 will be unreactive. It shows that 12 g of carbon burns in 32 g oxygеn from 44 g of CO2 bесоuse 3 g of carbon reacts with 8 g of oxygen to form 11 g of CO2. The same is given to us. Consequently 11.0 g of CO2 will be formed when 3.0 g of carbon is burnt in 50 g of oxygen consuming 8 g of O2, leaving behind 42 g of O2. The answer governs the Law of constant Proportion. Q3 What are polyatomic ions? Give examples. Ans: Polyatomic Ions are that ions which have more than one atom of same kind of different kind and behave as a single unit is known as polyatomic ion. E.g. : OH- , SO42- , PO43- Q4 Write the chemical formulae of the following. (a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate. Ans: (a)     MgCl2 (b)     Cao (c)     Cu(NO3)2 (d)     AlCl3 (е)     CаCо3 Q5 Give the names of the elements present in the following compounds. (a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate. Ans: (a) Calcium oxide (Cao), There are two type of elements are present first in calcium and the other is oxygen  (b) Chemical formula of Hydrogen Bromide is HBr. There are two types of element present, the first one in hydrogen and the other in Bromine. (c) Chemical formula of Baking powder is NaHCO3 (Sodium Hydrogen Carbonate). There are 4 types of element sodium, hydrogen, carbon and oxygen. (d) Chemical formula of Potassium sulphate is K2SO4. There are three types of element: Potassium, sulphur and oxygen. Q6 Calculate the molar mass of the following substances. (a) Ethyne, C2H2 (b) Sulphur molecule, S8 (c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31) (d) Hydrochloric acid, HCl (e) Nitric acid, HNO3 Ans: Molar mass: The mass of the mole of substance is known as molar mass. Unit = gm/mol. (a) C2H2 = H - C = C - H      = 2 x Atomic mass of C + 2 x Atomic mass of H      = 2 x 12 + 2 x 1      = 24 + 2 = 26 g (b) S8 =  8 x atomic mass of sulphur.       = 8 x 32 = 256 g   (c) P4 = 4 x atomic mass of phosphorus      = 4 x 31 = 124 g (d) HCl = 1 x atomic mass of H + 1 x atomic mass of Cl       = 1 x 1 + 1 x 35.5       = 1 + 35.5 = 36.5 g (e) HNO3 = 1 x atomic mass of H + 1 x atomic mass of N + 3 x atomic mass of O       = 1 x 1 + 1 x 14 + 3 x 16       = 1 + 14 + 48 = 63 g Q7 What is the mass of— (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)? (c) 10 moles of sodium sulphite (Na2SO3)? Ans: 1 mole of any particle = Relative mass of that particle in grams 1 mole of any molecule = molecular mass in grams (a) 1 mole of nitrogen atom = 14 grams (b) 1 mole of aluminium atom = 27 grams      4 mole of aluminium atom = 27 x 4 = 108 grams (c) Na2SO3 = 2 x atomic mass of Na + 1 x atomic mass of sulphur 3 x atomic mass of O      = 2 x 23 + 1 x 32 + 3 x 16      = 46 + 32 + 48 = 126 g 1 mole of Na2SO3 = 126 g  10 moles of Na2SO3 = 126 x 10 = 1260 grams Q8 Convert into mole. (a) 12 g of oxygen gas (b) 20 g of water (c) 22 g of carbon dioxide. Ans: Mass (in grams) = no. of moles x molecular mass (g/mol) (a) Molar mass of O2 = 2 x 16 = 32 g/mol       Mass = no. of mole x molar mass       12 = n x 32       n = 12/3 = 3/8       n = 0.375 mole  (b) Given mass = 20 g       Molar mass of H2O = 2 x 1 + 1 x 16                                       = 2 + 16 = 18 g/mol       Mass = no. of mole x molar mass       20 = n x 18       n = 20/18 = 10/9       n = 1.11 mole (c) Given mass = 22g       Molar mass of CO2 = 1 x 12 + 2 X 16                                       = 12 + 32 = 44 g/mol       Mass = no. of moles x molar mass       22 = n x 44       n = 22/44 = ½       n = 0.5. mole Q9 What is the mass of: (a) 0.2 mole of oxygen atoms? (b) 0.5 mole of water molecules? Ans: (a) No. of moles = 0.2      Molar mass of oxygen = 16 g/mol      Mass = no. of moles of molar mass      Mass = 0.2 x 16 = 3.2 g (b) No. of moles = 0.5      Molar mass of H2O = 2 x 1 + 16                                      = 2 x 16 = 18 g/mole      Mast = no. of moles x molar mass      Mass = 0.5 x 18 = 9.0 g Q10 Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur. Ans: Molar mass of S8 = 8 x 32 = 256 g            256 g = 6.022 x 1023 no. of molecules of S8 No. of molecules present in 16 g of solid sulphur = 6.022x1023 / 256 x 16 = 0.376 x 1023           = 3.76x1022 Q11 Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u) Ans: Molar Mass of Al2O3 = 2 x Atomic mass of Al + 3 x Atomic mass of O     = 2 x 27 + 3 x 16     = 54 + 48 = 102 g  102 g of Al2O3 contain = 2 x 6.022 x 1023 aluminium ion    = 0.051 g of Al2O3 = 2 x 6.022 x 1023 / 102 x 0.051    = 6.022 x 1023 x 10-3 = 6.022 x 1020