4.1 ATOMIC WEIGHT (ATOMIC MASS) ‘The atomic weight of an clement is the average relative weight of its atoms as compared to the weight of carbon atom taken as 12 amu. The atomic weight of an element expressed in. gram is called its gram atomic weight. 4.2. DIFFERENT METHODS TO FIND ATOMIC WEIGHT OF ELEMENTS ‘These are given below : I. Average Method. In this method, the average of atomic weight of all the isotopes of an element gives the atomic weight of the element. Exaneue 1. (2) Different naturally occurring isotopes of oxygen contain 0.204% "90 with mass 17.9991 amu, 0.037% 0 with mass 16.9991 amuand 99,759% "0 with mass 15.9949 ‘amu. Find the value ofchemieal atomic weight of natural oxygen. (6) Which isotope of carbon is used to express relative atomic ‘mass ofan element ? (HP. Board, 2007) ‘SoLurion. (@) Since data is given in percentage, let total number of oxygen atoms be 100. So, the average atomic weight of oxygen will be: (0.204 x 17.9991) + (0.037 « 16.9991) +(99.759 15.9949) 00 3.672 + 0.629 +1595.635 00 5.9994 Ans. (8) The isotope of carbon, "%C is used to express relative atomic mass of an element. Th ‘mass of an element is the number of times one atom of the element is heavier than = of « 8C atom, Exaupte 2. Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic ‘weight of natural boron is 10.81. Calculate the percentage ofeach Weight Cc 4 Men Atomic CHAPTER. ~ —_ — ‘stop in natural boron. axr.1978) Bouton Let 1 = *ageofizotopehavingat wt 1001 100-x = Sbageotisoope havingat wt 1101 . __ 110014 (100-1) «11.01 Averageatwt. = 00 001 +101-11.01% or 19g, = 100+ MOL=11018 tote 1001y = 1101-1081 x= 20;100-» Isotope with mass 10.01 = 20%; Isotope with mass 11.01 = 80% Ans. Exawpve3. Naturally occurring copper consists of isotopes of Cuand “Cu in the ratio 8 3. What will be the mean atomic weight of copper? SotuTioN. Total number of atoms are 8 atoms of ©'Cu and 3 atoms of Cu. 8x 634365 8+3=11 84311. There Mean at. wt. of C 504+195 _ 699 STS SP = 63.54 Ans, Il, Equivalent weight and valency method. Equivalent weight, valency and atomic weightare related as: At wt. = Eq. wt. x valency. Eq, wt. of an element can be determined by any one of the methods described in chapter 3 Valency of an element isa whole number. Valency of an element is equal to the number of H-atoms with which one atom of an element combines. It is also equal to the number of electrons transferred or shared during combination by an atom of the element.36 NUMERICAL CHEMISTRY FOR COMPETITIONS. 4.3 DETERMINATION OF VALENCY AND EXACT ATOMIC WEIGHT 1. Dulong and Petit’s law. (For metals only) According to this law: Approximate at wt. of an element 64 = Specific heat of element _ Approximate at. wt. Valency = qa The valency found above is changed to nearest whole number. For example, if valency comes out to be 2.99, itis changed to 3. Exams 4. A sample of a metal chloride weighing 0.22¢ required 0.51 g of silver nitrate to precipitate the chloride completely. The specific hent of the metal is 0.057. Find the molecular formula of the chloride if the symbol of the metal isM. (1. 1976) SotuTion. Let eq. wt. of metal = E, wt. of metal chloride = 0.22 g; wt of silver nitrate = 0.51 g. We know that: Wt. of metal chloride _ Eq. wt. of metal chloride Wt. of AgNO, Eq. wE. of AgNO, 022g E4355 051g ~ T0414 +Ox16) _ E4355, ="T0* 9224170 955 = £=37.83 _ 64 64 Approx.at. wt = > eag=pagy = 1228 __ approx at.wt._ 112.28 Valency = "Eg. wt 3783 =296=3 Molecular formula of metalchloride=MCl, Ans. Exavere6. (a) A bivalent metal M has the equivalent weight of 12. Find the at. wt. of M; the forma ofits chloride and the mol. rot. ofits oxide. (®) The chloride of a metal M having the formula MCI, contains 65.55% of chlorine, Find out its a. cv (©) What isthe valency of the element of which the eq wt is 12 and specific heat is 0.252 (7. 1973) SouuTion. (@) At-wt.= Eq. wt. x valenc) 24 Ans. Formula of chloride of bivalent metal M= MCI, Ans. Mol. wt. of bivalent metal oxide = MO = 24 + 16 = 40 Ans. © Wt of Cl, = 65.55 g; Wt of metal 100 ~ 65.55 = 34.45 g Wt.ofmetal ___Eq.wtof metal WE OfCl, ~ Eq-wt of CL@ 355)" 3445 _E_ 6555 ~ 355 = AB 355 _ p= ee = 18.66 From MCL, valency of metal, M=3 Exact at. wt. = Eq. wt. x Valency = 18.66 3=55.98 Ans. 64 _ 6a (0 Approv at wt = 45 = 8 225.6 Approx at wt. Vateney = APBONALE 25.6.0 ag Exauete 6.2.0 g of a metal (specific heat = 0.057 cal g-) on treatment with H,SO, gives 4.51g of metal sulphate. Calculate the valency and atomic weight of the metal. (LL-T. 1978) SOLUTION. 64 64 © Approx.at. w= Sar = gggy = 1228 (j ——Weofmetal "We of metal sulphate ___Fa.wt.of metal = Eq. wt. of metal sulphate 20g __E Tig ~ E+a8* 2E +96 =4.51E; ~ 96. E= Pe =3024 Approx. at. wt. 112.28 Gi) Valeney = aga Enact at. wt. of metal = Eq. wt. x Valency =98.24x3=11472 Ans. Exaurte 7. 32.1 g ofa metal were dissotved in HCland the hydrogen evotved was burnt to form water which weighed 9.1 g The specific heat of the metal is 0.094, Calculate its at. wot. SowuTion. (i) Wt: of metal = 32.1 g; wt. of H,O= 9.1 g. 18 g H,O contains H, =2g [HO = 2H; 18 = mol. wt. of H,0] 2 9.1 g HO contain H, = 29.1 = 101g. We know: Wt.of metal __Eq.wt-of metal 32.1_E. WtofH, ~ Eq.wtof@ 0)’ 101 1 E =317864 64 =p. heat 0.094 __ Approx. at. wt. Valency = "Eq. wt 22122, | wt. x Valency =3178x2=63.56 Ans. Exawete 8. A metal piece weighing 0.45 to 0.55 g when dissolved in dil. HCI produced 90 ml. of H, at NTP. If exact at wt. of metal is 65, calculate the exact weight of the metal ‘SowuTION. Let wt. of metal = 0.5 g; wt. of H, at NTP. = 90 x 0.00009 g, Exact at. wt. of metal = 65; Eq. wt. of metal = E. We know: Exact at. wt Wt. of metal wt. of metal 4) Wi off, atNIP ~ Eq.wiofHeioey ~ 05 E 350.0005 ~ 7.008 0:5 1,008 = 99x 0.00009 = 9% Atwt _ 65. Valency = Fa wt. 622 Exact eq. wt Let x From relation (1), we have: x 6 390.0005 ~ 7.008 71008 15223 Ans. Exauete 9. Wien 0.25 ampere current was passed through ‘a metal salt solution for 10 minutes, 0.1677 g ofthe metal was deposited. If specific heat ofthe metal is 0.0556 cal g” (degree? {find the at. wt of metal. (1 faraday = 96500 coulomb) ‘SoLUuTION. ( Current passed, C = 0.25 ampere; time (#) in seconds = 10 x 60 = 600 s, Wt. of metal deposited 1677 §-€q. wt. of metal Wet. of metal deposited x 96500 coulomb ‘Quantity of electricity passed (= ¢) 0.1677 5, x 96500 coulomb ‘(WBampere x 600s) i2, 150 coulomb 64 __ 64 Sp-heat ~ 0.0556 0789 g. 15.1 (i) Approx. at. wt. ATOMIC WEIGHT 37. 115.1 10789 Exact at. wt. = Eq. wt. x Valency = 10789 x 1 = 107.89 Ans. 2. Calculation of valency from the molecular weight of the volatile chloride. Different steps to calculate valency by this method are: (0 Find the eq. wt. of the element by usual methods. Let E = eq. wt. of the element. (ii) Let x= valency of element. So, at. wt. = Ex x. (iii) Let M be the one atom of the element having valency, x.So, formula of element chloride =MCI,, (i) Calculate vapour density (V.D,) of the chloride of the element. Then, mol. wt. of the chloride of element =2 x V.D. (©) Mol. wt. of MCI, = Ex +355 x=x(E +355) Mol. wt. of MCI, iz, 2x V.D. = E+355 Exaueie 10. The chloride of a metal contains 79.8% chlorine. The vapour density ofits chloride is 133.4. Ifsp. heat of the metal is 0.224, calculate the equivalent weight, valency, at, wt and molecular formula ofits chloride ‘SowuTION. (Wt of chlorine = 798 g; wt. of metal = 100-798 = 20.2 g; sp-heat = 0.224; \V.D. of the metal chloride = 133.4. We know: Wr. of metal Eq.wt.of metal Wt of chlorine ~ Eq. wt. of chlorine (= 353)" 2028 _ E_ 798g ~ 355 20.2255 p= 202x255 Wg « 4S Sp. heat” 028 ‘Approx. at “rw Eq. wt. x Valency 3227 Ans. Empirical formula of metal chloride = MCI,. But nx BF. Molecular formula, M.B. where n ERwt 1335 “EF. wt. of MCI, at. wt. of M + (3 « 35.5) = 2741065 = 1335