Organic compound contains 57.8 percent of carbon and 3.6 % of hydrogen and rest is oxygen the vapour density of the compound was found to be 83 you calculate empirical formula and molecular formula
Answers
Answer:
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Answer and explanation:
Percentage of carbon = 57.8 %
Percentage of hydrogen = 3.6 %
Percentage of oxygen = 100 - (Percentage of carbon + Percentage of hydrogen)
= 100 - (57.8 + 3.6)
= 100 - 61.4 = 38.6 %
Vapour density, V.D. = 83
Now, for finding empirical formula we will make a table -
Element Atomic %age Gram Molar ratio Simple whole
Mass atoms number ratio
Carbon (C) 12 57.8 57.8/12 4.82/2.41 2×2 =4
= 4.82 = 2
Hydrogen(H) 1 3.6 3.6/1 3.6/2.41 2×1.5 = 3
= 3.6 = 1.5
Oxygen(O) 16 38.6 38.6/16 2.41/2.41 1×2 = 2
= 2.41 = 1
Hence, the empirical formula for the compound is C₄H₃O₂.
Empirical formula mass = (4×12) + (3×1) + (2×16)
= 48 + 3 + 32 = 83 u
Molecular mass = 2 × V.D. = 2 × 83 u = 166 u
n = Molecular mass/ Empirical formula mass
n = 166 u/83 u = 2
Hence, Molecular formula = n × C₄H₃O₂ = 2 × C₄H₃O₂
= C₈H₆O₄.
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