Q.(a) Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38% of hydrogen and rest oxygen. Its vapour density is 47. OR
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Start your 14-day free trial to unlock the full solution →The mole ratio of C:H:O works out to 6:6:1, giving empirical formula C6H6O; since its formula mass (94) exactly equals the molecular mass found from the vapour density (94), the molecular formula is also C6H6O.
Step 1 — Percentage of oxygen (by difference):
%O = 100 - 76.6 - 6.38 = 17.02%
Step 2 — Convert each percentage to moles (divide by atomic mass), taking 100 g of the compound:
Moles of C = 76.6 / 12 = 6.383
Moles of H = 6.38 / 1 = 6.380
Moles of O = 17.02 / 16 = 1.064
Step 3 — Divide each by the smallest value (1.064) to get the simplest mole ratio:
C: 6.383 / 1.064 = 6.00
H: 6.380 / 1.064 = 6.00
O: 1.064 / 1.064 = 1.00
So the atoms are in the ratio C:H:O = 6:6:1, giving the empirical formula C6H6O.
Step 4 — Empirical formula mass:
= (6 x 12) + (6 x 1) + (1 x 16) = 72 + 6 + 16 = 94
Step 5 — Molecular mass from vapour density: …
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