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Na=14.31% ,S =9.97% , O=69.50% , H=6.22% . Calculate the molecular formula of compound assuming that the whole of hydrogen in the compound is present as water of crystallization . Molecular mass of compound is 322. (related to empirical formula) .............please tell |
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Answer» ____________________________________________________________ Question: Na=14.31% ,S =9.97% , O=69.50% , H=6.22% . Calculate the molecular formula of compound assuming that the whole of hydrogen in the compound is PRESENT as water of crystallization . Molecular mass of compound is 322. (related to empirical formula) __________________________________________________________ Answer: Take the approximate ratio between the PERCENTAGE composition and the MOLAR mass of the elements given Element mass % Atomic mass mass percent/Atomic mass ratio Na 14.31 23 0.622 2 S 9.97 32 0.311 1 H 6.22 1 6.22 20 O 69.5 16 4.34 14 Empirical formula is therefore Na₂SH₂₀O₁₄
Empirical mass = (2×23) + (32 )+ (20×1) + (14×16) =322g Given: Molecular mass = 322 g ∴ molecular formula = empirical formula = Na₂SH₂₀O₁₄
Given that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation If water of crystallization are nH₂O Then, 2n = 20 or n=10 Crystallized water is 10H₂O
therefore, the molecule is Na₂SO₄. 10H₂O __________________________________________________________ Please mark me brainliest ( ̄︶ ̄)↗ |
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