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Question 51 of 66

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

(b)
(i) Calculate the total number of angular nodes and radial nodes present in 3d and 4f orbitals.
(ii) Explain why the electron affinity of Be and N is almost zero.
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2020Subjective· 5mImportance★★★★★
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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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