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In an inorganic compound, the mass % of Na = 14.3, S = 9.97, H = 6.22 and O = 69.5. The hydrogen and oxygen are present as water. The vapour density of the compound is 161, what is its molecular formula? |
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Answer» ANSWER : The Molecular Formula Of Compound is Na2SH20O14 ANSWER : The Empirical Formula Mass Of the Compound = 2*23+1*32+20*1+14*16 Now Calculate the Value Of N = MOLECULAR MASS/EMPIRICAL MASS = 322/322 = 1 ∴ Vapour density of compound is = 161 ∴ Molecular weight = 2 x vapour density = 2 x 161 = 322 g ∴ Number oxygen atoms = \(\cfrac{322\times\frac{69.5}{100}}{16}\) = 13.986 = 14 Oxygen atom Number of hydrogen atoms = \(\cfrac{322\times\frac{6.22}{100}}{1}\) = 20.02 = 20 Hydrogen atoms ∴ Maximum number of water molecule can present = 10 H2O ∴ remaining number of oxygen atoms = 4 Number of S atom = \(\frac{322\times9.97}{32\times100}\) = 1.003 = 1 sulphur atom Number of sodium atoms = \(\cfrac{322\times\frac{14.3}{100}}{23}\) = 2.003 = 2 sodium atoms Therefore, the molecular formula will be Na2SO4.10H2O |
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