{"id":2296,"date":"2023-10-02T12:02:19","date_gmt":"2023-10-02T12:02:19","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2296"},"modified":"2023-10-02T12:15:14","modified_gmt":"2023-10-02T12:15:14","slug":"week-2-ss1-second-term-chemistry-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss1-second-term-chemistry-notes\/","title":{"rendered":"Week 2 &#8211; SS1 Second Term Chemistry Notes"},"content":{"rendered":"<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>WEEK  2<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<strong>MOLE IN TERMS OF THE RELATIVE ATOMIC MASS OR RELATIVE MOLECULAR MASS OF ASUBSTANCE.<br \/>\n<\/strong>The mole can be expressed in terms of the R.A.M of an element or the R.M.M. of a substance\/molecule\/compound this:<br \/>\n1 mole of any substance = the R.A.M. of the substance or the R.M.M of the substance.  <\/p>\n<p>\u00a0<br \/>\n\u00a0NOTE:\u00a0\u00a0\u00a0\u00a0<br \/>\n1 mole of Na (<sub>g<\/sub>)\u00a0\u00a0\u00a0\u00a0= 23g mole<br \/>\n1 mole of 0<sub>2<\/sub> (aq)\u00a0\u00a0\u00a0\u00a0= 16g mole<br \/>\n1 mole of 0<sub>2<\/sub> (g)\u00a0\u00a0\u00a0\u00a0= 16 x 2 = 32glmol<br \/>\n1 mole of C0<sub>2<\/sub> (g)\u00a0\u00a0\u00a0\u00a0= 12 + 16 x 2 = 44g1mol<br \/>\n1 mole of H<sub>2<\/sub>S0<sub>4<\/sub> (aq)\u00a0\u00a0\u00a0\u00a0= (2 x 1) + 32 +64 \u00a0\u00a0\u00a0\u00a0= 98glmol<\/p>\n<p>\u00a0Example:<br \/>\n1.\u00a0\u00a0\u00a0\u00a0Calculate the number of moles present in 11g of carbon dioxide or carbon (iv) oxide (C0<sub>2<\/sub>) gas\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n<strong>Solution:<br \/>\n<\/strong>1 mole C0<sub>2 <\/sub>(<sub>g<\/sub>)\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0Rmm of C0<sub>2<\/sub> (<sub>g<\/sub>)<br \/>\n1 mole of C0<sub>2 \u00a0\u00a0\u00a0\u00a0<\/sub>=\u00a0\u00a0\u00a0\u00a012 + (16&#215;2) = 44g g\/mol<br \/>\n:. 44g of C0<sub>2\u00a0\u00a0\u00a0\u00a0<\/sub>= \u00a0\u00a0\u00a0\u00a01 mole of C0<sub>2<\/sub> (g)<br \/>\n1 g of C0<sub>2<\/sub>\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a01\/44 mole of C0<sub>2<\/sub> (g)<br \/>\n:. 11g of C0<sub>2<\/sub>\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a01\/44 x 11 mole of C0<sub>2<\/sub> (g)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a00.25 mole of C0<sub>2<\/sub> (g)<\/p>\n<p>\u00a0<br \/>\n\u00a0<\/p>\n<ul>\n<li>\n<div>Determine the number of grammes of substance present in 0.05 of sodium carbonate (sodium trioxocarboante IV) (Na<sub>2<\/sub> C0<sub>3<\/sub>).\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0<strong>Solution:<br \/>\n<\/strong>1 mole Na<sub>2<\/sub> C0<sub>3 <\/sub>= Rmm of Na<sub>2<\/sub> C0<sub>3<\/sub><br \/>\n\t\t1 mole Na<sub>2<\/sub>C0<sub>3<\/sub> = (23 x 2) + 12 + (16 x 3)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 106glmol<br \/>\n:. 0.05 mole Na<sub>2<\/sub> C0<sub>3<\/sub><br \/>\n\t\tOf Na<sub>2<\/sub> C0<sub>3<\/sub><br \/>\n\t\t= 0.055moles<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>MOLES IN TERMS OF NUMBER<br \/>\n<\/strong><br \/>\n\u00a0The term number takes into consideration the number of particles such as atoms, ions, molecules, electrons, protons, and neutrons etc, contained by a certain amount of a substance.<\/p>\n<p>\u00a0NOTE:\u00a0\u00a0\u00a0\u00a0 The number of particles taking part and formed in a chemical reaction can be determined.<\/p>\n<p>\u00a0Avogadro determined the actual number of atoms of carbon in 12.00g of <sup>12<\/sup><sub>6<\/sub>C isotope in various ways.  He found out that 12.00g of <sup>12<\/sup><sub>6<\/sub>C contains 6.02 x 10<sup>23<\/sup> atoms of carbon.  He worked with a large number by elements, compounds and ions and came to the conclusion that:<br \/>\na.\u00a0\u00a0\u00a0\u00a0The gram atomic man of all elements always contains the same number of atoms.<br \/>\nb.\u00a0\u00a0\u00a0\u00a0The gram molar mass of all compounds always contain the same number of molecular.<br \/>\nc.\u00a0\u00a0\u00a0\u00a0The gram formula mass of all ions also contain the same number ions. <\/p>\n<p>\u00a0Avogadro established that the number of particles (ions, atoms, molecules, electrons, protons etc.) present in one gram meformula (atomic, ionic, molecular etc.) mass of a substance is 6.02 x 10<sup>23<\/sup>.<\/p>\n<p>\u00a0This value is called the <strong>Avogadro&#8217;s number<\/strong> or <strong>constant N<sup>A<\/sup><\/strong><\/p>\n<p>\u00a0NOTE: This number of particle is contain and in one mole of any substance<br \/>\n1 mole = 6.02 x 10<sup>23<\/sup> particles<br \/>\nExample:<br \/>\n1.\u00a0\u00a0\u00a0\u00a0Calculate the number of particles<br \/>\ni.\u00a0\u00a0\u00a0\u00a044g of iron (II) sulphide (Fes)<br \/>\nii.\u00a0\u00a0\u00a0\u00a05.5g of manganese (Mn)<br \/>\niii.\u00a0\u00a0\u00a0\u00a08g of oxygen molecule (0<sub>2<\/sub>)<br \/>\niv.\u00a0\u00a0\u00a0\u00a08g of oxygen atom (0)<br \/>\n\u00a0\u00a0\u00a0\u00a0(Mn \u2013 55, 0-16, Fe =56, S=32)<br \/>\n\u00a0\u00a0\u00a0\u00a0Mole = mass of given substance (g)<br \/>\n\t\t(n)\u00a0\u00a0\u00a0\u00a0 Gram atomic\/molar mass<\/p>\n<p>\u00a0 ii.\u00a0\u00a0\u00a0\u00a0Mole of Mn =   5.5g<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a055 g\/mole<br \/>\n\u00a0\u00a0\u00a0\u00a0Number of particles<br \/>\n\u00a0\u00a0\u00a0\u00a0= N<sup>A<\/sup> x n<br \/>\n\u00a0\u00a0\u00a0\u00a0: .1 mole of Mn = 6.023 x 10<sup>23<\/sup> atoms<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\n 0.1mole Mn<br \/>\n\u00a0\u00a0\u00a0\u00a0= 0.1 x 6.023 x 10<sup>23<\/sup> atoms<br \/>\n\u00a0\u00a0\u00a0\u00a0= 6.023 x 10 <sup>22<\/sup> atoms<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\nii.\u00a0\u00a0\u00a0\u00a0Mole of Fe = massing of Fes<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G|.M.M\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0=    44g\u00a0\u00a0\u00a0\u00a0      =   44g<br \/>\n\t\t   (56 +32)           88g1mole<br \/>\n\u00a0\u00a0\u00a0\u00a0= 0.5 mole<\/p>\n<p>\u00a01 mole fess = 6.023 x 10<sup>23<\/sup>molecules<br \/>\n:. 0.5 mole= 6.023 x 10<sup>23<\/sup> x .05<br \/>\n\u00a0\u00a0\u00a0\u00a0= 3.012 x 10<sup>23<\/sup> molecules of Fes<\/p>\n<p>\u00a0iii.\u00a0\u00a0\u00a0\u00a0Mole of 0<sub>2<\/sub> =   massing<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G.M.M<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0Gmm of 0<sub>2<\/sub> = 16 x 2 = 32\/gmole<br \/>\n\u00a0\u00a0\u00a0\u00a0:. Mole = 8 g<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  32\/gmole<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.25 mole<br \/>\n1 mole of 0<sub>2<\/sub> = 6.023 x 10<sup>23<\/sup> mole of 0<sub>2<\/sub><br \/>\n\t\t:. 0.25 moles of 0<sub>2<\/sub><br \/>\n\t\t= 6.023 x 10<sup>23<\/sup> x 0.25 molecules<br \/>\n= 1.506 x 10<sup>23<\/sup>\u00a0\u00a0\u00a0\u00a0molecules of 0<sub>2<\/sub><\/p>\n<p>\u00a0iv.\u00a0\u00a0\u00a0\u00a0G.m.m of oxygen = 16g\/mole<br \/>\n\u00a0\u00a0\u00a0\u00a0Mole of 0 = massing of 0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G.a.m\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=   8g<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16g\/mole<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.5mole<br \/>\n\u00a0\u00a0\u00a0\u00a01 mole of 0 = 6.023 x 10<sup>23<\/sup> atoms of 0<br \/>\n\u00a0\u00a0\u00a0\u00a0:. 0.5 mole of 0<br \/>\n\u00a0\u00a0\u00a0\u00a0= 0.5x 6.023 x 10<sup>23<\/sup> atoms<br \/>\n\u00a0\u00a0\u00a0\u00a0= 3. 012 x10<sup>23<\/sup> atoms of 0<\/p>\n<p>\u00a0<br \/>\n\u00a02.\u00a0\u00a0\u00a0\u00a0A sample of nitric (trioxonitrate (v) acid contains 1.2 x 10<sup>23<\/sup> molecules of the acid.  <\/p>\n<p>\u00a0Calculate<br \/>\na.\u00a0\u00a0\u00a0\u00a0The number of moles<br \/>\nb.\u00a0\u00a0\u00a0\u00a0The mass of the acid (HN0<sub>3<\/sub>) in the sample<br \/>\n\u00a0\u00a0\u00a0\u00a0(N<sup>A<\/sup> = 6.023 x 10<sup>23<\/sup>particles mol-<sup>1<\/sup>, H=1, N=14, 0=16)<\/p>\n<p>\u00a0a.\u00a0\u00a0\u00a0\u00a01 mole of HN0<sub>3<\/sub> acid =<br \/>\n\u00a0\u00a0\u00a0\u00a06.02 x 10<sup>23<\/sup>molecules<br \/>\n\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a01 mole =         1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0mole\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0     6.02 x 10<sup>23 <\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0:. 1.2 x 10<sup>23<\/sup>molecules<br \/>\n\u00a0\u00a0\u00a0\u00a0=           1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x 1.2 x 10<sup>23<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0         6.02 x 10<sup>23<\/sup>\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=    1.2   mole<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    6.02<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.2 mole<br \/>\nb.\u00a0\u00a0\u00a0\u00a0mole = mass of substance on g<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G. m. m of substance<br \/>\n\u00a0\u00a0\u00a0\u00a0mmHN0<sub>3<\/sub><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0= 1x 14 + (16 x 3)<br \/>\n\u00a0\u00a0\u00a0\u00a0= 63g\/mole<br \/>\n\u00a0\u00a0\u00a0\u00a0Mass of HN<sup>0<\/sup>3<br \/>\n\u00a0\u00a0\u00a0\u00a0= mole x G.m.m.<br \/>\n\u00a0\u00a0\u00a0\u00a00.2 mole x 63 g<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Mole<br \/>\n\u00a0\u00a0\u00a0\u00a0= 12.6g\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<strong>THE MOLE CONCEPT IN TERMS OF VOLUME<br \/>\n<\/strong>There three states of matter solid, liquid and gas.  A good example of a substance that can form<br \/>\nthe three states is water.<\/p>\n<p>\u00a0NOTE: Matter can change its state when there is a considerable change in KE.<\/p>\n<p>\u00a0The effect of temperature and pressure in which more pronounced in gaseous sate because f the<br \/>\nVolume it occupies compared to the and solid and liquid states which have a definite<br \/>\nVolume.<\/p>\n<p>\u00a0Experimentally, it has been proved that the gramme molar mass amount of any gaseous substance<br \/>\nwill always occupy a volume of 22.4dm<sup>3<\/sup> at standard temperature and pressure (s.t.p) and 24dm<sup>3<\/sup><br \/>\n\t\tat room temperature and pressure (r.t.p), standard temperature = 0<sup>0C<\/sup>or 273k and standard<br \/>\npressure = 760 mmHg or 1.01&#215;10<sup>5<\/sup> Nm<sup>-2<\/sup><\/p>\n<p>\u00a0<br \/>\n\u00a0NOTE: 1 mole of any gaseous substance = molar volume of a gas at s.t.p. i.e.<\/p>\n<p>\u00a01 mole of 0<sub>2<\/sub> (32g) = 22.4dm<sup>3<\/sup> at s.t.p<br \/>\n1 mole of C0<sub>2<\/sub> (44g) = 22.4dm<sup>3<\/sup> at s.t.p<br \/>\n1 mole of N<sub>2<\/sub> (28g) = 22.4dm<sup>3<\/sup> at s.t.p<br \/>\n1 mole of S0<sub>2<\/sub> (64g) = 22.4dm<sup>3<\/sup> at s.t.p<br \/>\n1 mole of C1<sub>2<\/sub> (71) 22.4dm3 at s.t.p<\/p>\n<p>\u00a0Examples:<br \/>\n1.\u00a0\u00a0\u00a0\u00a0Calculate the volume occupied by 5 moles of carbon dioxide (carbon (iv) oxide) at s.t.p <\/p>\n<p>\u00a0SOLUTION:<br \/>\n1 mole of gas at s.t.p = 22.4dm<sup>3<\/sup><br \/>\n\t\t1 mole of C0<sub>2<\/sub> = 22.4dm3 at s..t.p<br \/>\n5 mole of CO<sub>2<\/sub> =5 x 22.4 at s.t.p<br \/>\n= 112.0dm<sup>3<\/sup> at s.t.p<\/p>\n<p>\u00a0<br \/>\n\u00a02.\u00a0\u00a0\u00a0\u00a0Determine the number of mole present in 11.2dm<sup>3<\/sup> of nitrogen (IV) oxide (nitrogen dioxide) = N0<sub>2<\/sub> (g) at s.t.p<br \/>\n\u00a0\u00a0\u00a0\u00a0Mole = volume<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G.m.v<\/p>\n<p>\u00a01 mole of N0<sub>2<\/sub> (g) = 22.4dm<sup>3<\/sup> at s.t.p<br \/>\n:. 22.4dm of N0 (g) = 1 mole of N0<sub>2<\/sub> at s.t.p<br \/>\n1 dm<sup>3<\/sup> of N0<sub>2<\/sub> (g) =<br \/>\n1 ofN0<sub>2<\/sub> (g) at s.t.p<br \/>\n22.4dm<sup>3<\/sup><br \/>\n\t\t:.11.2dm<sup>3<\/sup> of N0<sub>2<\/sub>(g)<br \/>\n=   1   x 11.2dm3 of N0<sub>2<\/sub> (g) at s.t.p<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\n22.4dm<sup>3<\/sup>   = 11.22m<sup>3<\/sup> of N0<sub>2<\/sub> (g) at s.t.p<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.5mole of N0<sub>2 <\/sub>at s.t.p<\/p>\n<p>\u00a0<\/p>\n<ul>\n<li>\n<div>How many grammes of gas are present in 5600cm<sup>3<\/sup> of chlorine gas at s.t.p? (Cl=35.5)\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0<strong>SOLUTION:<br \/>\n<\/strong>1 mole of Cl<sub>2<\/sub> (g) = Rmm of Cl<sub>2<\/sub> = molar volume of gas at s.t.p.<br \/>\nRmm of Cl<sub>20 <\/sub>= 35.5 x 2 = 71g\/mol<br \/>\n1 mole of Cl<sub>2<\/sub>= 71gl\/mol = 22400cm<sup>3<\/sup> of Cl<sub>2 <\/sub>at s.t.p.= 71   g<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a022400<br \/>\n:. 5600cm3 of Cl<sub>2<\/sub>=   71     x 5600<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0         22400 \u00a0\u00a0\u00a0\u00a0  = 17.56g of Cl<sub>2<\/sub><\/p>\n<p>\u00a0<\/p>\n<ul>\n<li>\n<div>Calculate the number of molecules of hydrogen gas present in 2.24dm<sub>3<\/sub> of the gas at s.t.p\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0<strong>SOLUTION:<br \/>\n<\/strong>1 mole H<sub>2<\/sub> (g) = Avogadro&#8217;s No of molecule =<br \/>\nMolar volume of H<sub>2<\/sub> at s.t.p<br \/>\n1 mole H<sub>2<\/sub>(g) = 6.0<sub>2<\/sub> x 10<sup>23<\/sup> molecules =<br \/>\n: .22.4dm3 of H<sub>2<\/sub> (g) at s.t.p<br \/>\nMolecules of H<sub>2<\/sub> (g) at s.t.p<br \/>\n1dm3 = 6.02 x 10<sup>23<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0   22.4\u00a0\u00a0\u00a0\u00a0<br \/>\n: .2.24dm3 = 6.02 x 10<sup>23<\/sup> x 2.24<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0          22.4\u00a0\u00a0\u00a0\u00a0         1\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<\/p>\n<ul>\n<li>\n<div>6.02 x 10<sup>22<\/sup> molecules of H<sub>2<\/sub>(g)\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a05.\u00a0\u00a0\u00a0\u00a0Calculate the volume at s.t.p which would occupy 2.5.6g of 5<sub>8<\/sub>vapour (S = 32)<\/p>\n<p>\u00a0<strong>SOLUTION<\/strong>Mole of substance = massing<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G.m.m<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0G.m.m. of S<sub>8<\/sub> = 32 x 8<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Mole of S<sub>8<\/sub><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 2.56g<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  256g\/mole<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.01mole<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0Volume (dm<sup>3<\/sup>) = mole x molar volume (22.4dm<sup>3<\/sup>)<br \/>\n1mole of S<sub>8<\/sub>vapour = 22.4dm<sup>3<\/sup> at s.t.p<br \/>\n1mole of S<sub>8<\/sub>vapour<br \/>\n= 0.01 x 22.4dm<sup>3<\/sup><br \/>\n\t\t  = 0.224dm<sup>3<\/sup><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n<strong>SUMMARY<br \/>\n<\/strong>i.\u00a0\u00a0\u00a0\u00a01mole of any Rmm of = \u00a0\u00a0\u00a0\u00a0Na.GMV<br \/>\n\u00a0\u00a0\u00a0\u00a0Substance\u00a0\u00a0\u00a0\u00a0substance\u00a0\u00a0\u00a0\u00a0 of<br \/>\n\u00a0\u00a0\u00a0\u00a0Gaseous\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0gas at s.t.p<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0This summary holds for a gaseous substance only<\/p>\n<p>\u00a0ii.\u00a0\u00a0\u00a0\u00a01 mole of    =     Rmm of =     Avogadro&#8217;s<br \/>\n\u00a0\u00a0\u00a0\u00a0Any substance \u00a0\u00a0\u00a0\u00a0substance     number<br \/>\n\u00a0\u00a0\u00a0\u00a0Solid\/liquid\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(N<sup>A<\/sup>)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0This summary holds for solid and liquid substances<\/p>\n<p>\u00a0<br \/>\n\u00a01.A volume of a gas was found to weigh 5.6g and when corrected to s.t.p measured 4.48dm<sup>3<\/sup>.  Calculate the G.mm. of the gas<\/p>\n<p>\u00a02.\u00a0\u00a0\u00a0\u00a0Calculate the number of:<br \/>\na.\u00a0\u00a0\u00a0\u00a0atoms in 2.5mole of Na (sodium)<br \/>\nb.\u00a0\u00a0\u00a0\u00a0ions present in 0.5 moles of copper (II) ions (Cu<sup>2+<\/sup>)<\/p>\n<p>\u00a03..\u00a0\u00a0\u00a0\u00a0A volume of a gas Z was found to weigh 6.5g and when corrected to s.t.p it measured 4.84dm3.  Calculate the G.m.m of the gas Z<br \/>\n\u00a0\u00a0\u00a0\u00a0(H =1, S=32, 0=16)<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>EMPIRICAL AND MOLECULAR FORMULAE<br \/>\n<\/strong><strong>Empirical formula: <\/strong>The formula which shows the simplest ratio of the atoms of the elements that make up a compound.<br \/>\n<strong>Molecular formula:<\/strong> The formula which shows the actual number of atoms present in one molecule of the element or compound.<br \/>\n<strong>Exercise:<\/strong>Ethanoic Acid<br \/>\nMol. Formula = CH<sub>3<\/sub>COOH or C<sub>2<\/sub> H<sub>4<\/sub> O<sub>2<br \/>\n<\/sub><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1202_Week2SS1Se1.png\" alt=\"\"\/>        2 atoms of carbon, 4 atoms of H<sub>2 <\/sub>and 2 atoms of O<sub>2<br \/>\n<\/sub>Empirical formula = CH<sub>2<\/sub>O Ratio 1: 2: 1<\/p>\n<p>\u00a0<strong>Note: <\/strong> the 3<sup>rd<\/sup> type of formula is structural formula<\/p>\n<p>\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1202_Week2SS1Se2.png\" alt=\"\"\/>                                CH<sub>3<\/sub>COOH     =        \u00a0\u00a0\u00a0\u00a0  H-C-C=O-H<\/p>\n<p>\u00a0OH<\/p>\n<p>\u00a0<strong>CALCULATIONS INVOLVING E.F AND M.F<br \/>\n<\/strong><strong>Exercise 1:  <\/strong>A compound has the following % composition by mass, C= 40%, H= 6.67% and O = 53.3% calculate the E.F of the compound. If its ml. mass is 180, find its mol. Formula (C=12, H= 1, O= 16)<br \/>\n<strong>                                                                    Answer:<\/strong> E.F = CH<sub>2<\/sub>O<br \/>\nM.F = C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<br \/>\n<\/sub><strong>Note: <\/strong>M.F = (E.F)n = Rmm<\/p>\n<ul>\n<li>Rmm = 2 x v.d\n<\/li>\n<\/ul>\n<p>Where mass is given instead of  % composition, it can still be used i.e. mass = % composition.<\/p>\n<p>\u00a0<strong>Exercise 2: <\/strong>the analysis of a compound gave the Hg results:<br \/>\n                     5.2g of the compound contained 1.935g of carbon,<br \/>\n                     3.2g of the compound contained o.46g of hydrogen,<br \/>\n                     1.2g of the compound contained 0.6g of oxygen.<br \/>\n          Calculate the E.F (C = 12, O= 16, H=1) <\/p>\n<p>\u00a0<strong>Solution:<\/strong>                   C                            H                                 O<br \/>\n       % Composition 1.935 x 100%     0.46 x 100%          0.6 x 100%<br \/>\n                                    52                       3.60                        1.2<\/p>\n<p>\u00a0                                    37.2%                   12.8%                   50%<\/p>\n<p>\u00a037.2%12.8%50%<br \/>\n                                         12                          1                        16<br \/>\n<strong>Note: <\/strong> if 5.2g has been for all the components i.e. no 3.6g and 1.2g, then you solve directly.<\/p>\n<p>\u00a0<strong>Exercise 3: <\/strong>6g of metal x reacts completely with 23.66g of chlorine to form 29.66g of the metallic chloride.<\/p>\n<ol>\n<li>Find the E.F of the metallic chloride\n<\/li>\n<li>If the v.d of the compound is 133.5\n<\/li>\n<li>Find its mol. Formula (x = 27, Cl = 35.5)\n<\/li>\n<\/ol>\n<p>\u00a0<strong>Solution:<\/strong>                                         x                              Cl<\/p>\n<ol>\n<li> Mass composition            6                              23.66\n<\/li>\n<\/ol>\n<p><sup>6<\/sup>\/<sub>27<\/sub><sup>23.66<\/sup>\/<sub>35.5<\/sub><br \/>\n\t\t                                                          0.22                        0.67<br \/>\n0.220.67<br \/>\n                                                          0.22                        0.22<br \/>\n                                                              1                             3<br \/>\nE.F = XCl<sub>3<br \/>\n<\/sub><br \/>\n\u00a0<\/p>\n<ol>\n<li> mol. Mass\n<\/li>\n<\/ol>\n<p>v.d = 133.5<br \/>\nmol mass = 133.5 x 2 = 267<\/p>\n<p>\u00a0 (XCl<sub>3<\/sub>)n = 267<br \/>\n27 + [35.5 x 3]n = 267<br \/>\n133.5n = 267<\/p>\n<p>\u00a0n= 267         =2<br \/>\n      133.5 <\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0Mol. Formula = (XCl<sub>3<\/sub>)<sub>2 <\/sub>= X<sub>2<\/sub>Cl<sub>6<br \/>\n<\/sub><br \/>\n\u00a0<strong>Exercise 4: <\/strong>A hydrocarbon on combustion given 0.704 g of CO<sub>2 <\/sub>and 0.216g of H<sub>2<\/sub>O. If the relative mol. Mass of the compound is 54, calculate E.F. and M.F.<br \/>\n<strong>Solution:<\/strong> Hydrocarbon contains carbon and hydrogen only.<br \/>\nRmm of CO<sub>2<\/sub> = 44g, RAM of Carbon = 12<br \/>\nRmm of H<sub>2<\/sub>O = 18g, RAM of Hydrogen = 2<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1202_Week2SS1Se3.png\" alt=\"\"\/>4 4g of CO<sub>2<\/sub> \u2192 12g of C<br \/>\nTherefore 0.704g CO<sub>2<\/sub> \u2192 x<br \/>\nX = <sup>12<\/sup>\/<sub>44<\/sub> x 0.704 = 0.192 of C<\/p>\n<p>\u00a018g of H<sub>2<\/sub>O \u2192 2g of H<br \/>\nTherefore 0.216g \u2192 x<br \/>\nX = 0.216 x 2 = 0.024g of hydrogen<br \/>\n         18<br \/>\n                                        C            H<br \/>\nMass composition       0.192     0.024<br \/>\nRAM                               12           1<br \/>\n0.1920.024<br \/>\n                                         12          1<br \/>\n0.0160.024<br \/>\n                                        0.016      0.016<br \/>\n= 1   :  1.5<br \/>\n    2       3<br \/>\nE.F = C<sub>2<\/sub>H<sub>3<br \/>\n<\/sub>M.F = (E.F)n = mm<br \/>\n           (C<sub>2<\/sub>H<sub>3<\/sub>)n = 54<br \/>\n            27n = 54<br \/>\n              n = 2<br \/>\nTherefore M.F = C<sub>4<\/sub>H<sub>6<br \/>\n<\/sub><strong>Note: <\/strong> Where the V.D and RMM are not given use this formula to get V.D.<br \/>\nv.d = mass of a certain vol. of a gas<br \/>\nmass of an equal vol. of H<sub>2<br \/>\n<\/sub><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>PERCENTAGE OF ELEMENT<br \/>\n<\/strong><\/p>\n<ol>\n<li>Calculate the percentage by mass of nitrogen in HNO<sub>3 <\/sub>(H = 1, N = 14, O = 16)\n<\/li>\n<\/ol>\n<p><strong>Solution :<\/strong>Percentage by mass of N = molar mass of N             x 100%<br \/>\n                                                      Molar mass of HNO<sub>3<br \/>\n<\/sub>=           14<br \/>\n1 + 4 + (16 x 3)<br \/>\n\t\t\t14      x 100%      = 22.2%<br \/>\n\t\t\t  63<\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>Calculate the percentage by mass of all the component elements in NaNO<sub>3 <\/sub>( Na = 23, N= 14, O= 16)\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>Solution:<\/strong><br \/>\n\t\t%by mass of Na = 23 x 100% = 27%<br \/>\n                                 85<br \/>\n% by mass of N = 14 x 100% = 16.5%<br \/>\n                                85<br \/>\n% by mass of O = 3 x 16 x 100% = 56.5%<br \/>\n                                   85<br \/>\n<strong>QUESTIONS<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0A volume of a gas Z was found to weigh 6.5g and when corrected to s.t.p it measured 4.84dm3.  Calculate the G.m.m of the gas Z<br \/>\n\u00a0\u00a0\u00a0\u00a0(H =1, S=32, 0=16)<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Calculate the number of:<br \/>\na.\u00a0\u00a0\u00a0\u00a0atoms in 2.5mole of Na (sodium)<br \/>\nb.\u00a0\u00a0\u00a0\u00a0ions present in 0.5 moles of copper (II) ions (Cu<sup>2+<\/sup>)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a03. A compound contains 40 C, 6.66% H and a certain % of O. Calculate the E.F. If its mol. mass is 180. Calculate M.F.<br \/>\n4.  5.05g of a compound was found to contain 4g of Ca and 0.35g of S and 0.70g of Oxygen calculate its E.F (Ca = 40, S= 32, O=16)<br \/>\n5. A gaseous hydrocarbon contains 92.3% C, and 7.7% hydrogen by mass. 300cm<sup>3<\/sup> of the hydrocarbon weighs 0.301g and under the same conditions of temperature and pressure;<br \/>\n6.  300cm<sup>3<\/sup> of H weighs 0.023g.<\/p>\n<ol>\n<li>Find the E.F of the hydrocarbon\n<\/li>\n<li>Determine the M.F of the hydrocarbon (C = 12, H =1).\n<\/li>\n<\/ol>\n<p>1.The relative atomic mass of calcium atom is 40.This means that<br \/>\n(A).the mass of calcium is 40g<br \/>\n(B).the calcium is 40 times heavier than that of 1 atom of hydrogen<br \/>\n(C).calcium is 40 times that of 1g of hydrogen<br \/>\n(D).calcium is related to hydrogen through 40 digits<br \/>\n2. The relative molecular mass of lead (ii ) trioxonitrate (v) is (Pb=108,N=14,O=16).<br \/>\n(A).170  (B).222   (C).232   (D).132<br \/>\n3. Which is heavier?1 mole of PbCl<sub>2<\/sub> ,1 mole of H<sub>2<\/sub> and 1 mole of pb(NO<sub>3<\/sub>)<sub>2<\/sub>?<br \/>\n(A) .PbCl<sub>2<\/sub>  (B).H<sub>2<\/sub>  (C).None of them  (D).Pb(NO<sub>3<\/sub>)<sub>2<br \/>\n<\/sub>4. How many toms are contained I mole of hydrogen molecule.<br \/>\n(A).18.09&#215;10<sup>23<\/sup>atoms(B).12.06 x10<sup>23<\/sup>atoms<br \/>\n\t\t(C).6.02&#215;10<sup>23<\/sup>atoms   (D) 6.02 x10<sup>23<\/sup> molecules<\/p>\n<p>\u00a06.The percentage of oxygen in sulphur (iv)oxide is (s=32, o=16)<br \/>\n(A).5%   (B).50%   (C).500%    (D).25%<\/p>\n<p>\u00a07. The empirical formula  of C<sub>6<\/sub>H<sub>6 <\/sub>is<br \/>\n(A).CH   (B).C<sub>3<\/sub>H<sub>3 <\/sub> (C).C<sub>6<\/sub>H<sub>6<\/sub> (D).3CH<\/p>\n<p>\u00a07 .If the relative molecular mass of CH<sub>2<\/sub>O IS 60, calculate the empirical formula.(C=12,H=1)<br \/>\n(A) .4  (B).1  (C).2   (D).3<\/p>\n<p>\t\t\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0 \u00a0 \u00a0WEEK 2 \u00a0\u00a0\u00a0\u00a0 MOLE IN TERMS OF THE RELATIVE ATOMIC MASS OR RELATIVE&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,196],"tags":[],"class_list":["post-2296","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2296","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2296"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2296\/revisions"}],"predecessor-version":[{"id":2297,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2296\/revisions\/2297"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2296"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2296"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2296"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}