Q.(a) A compound on analysis gave Na=14.31%, S=9.97%, H=6.22%, O=69.5%. Calculate the molecular formula of the compound, if all the Hydrogen in the compound is present in combination with Oxygen as Water of Crystallisation. [molecular mass of the compound is 322] OR
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Converting the mass percentages to moles and simplifying gives the empirical formula Na2SO4.10H2O, whose formula mass (322) exactly matches the given molecular mass -- so the molecular formula equals the empirical formula. (This answers part (a) of the OR question.)
Given (per 100 g of compound): Na = 14.31%, S = 9.97%, H = 6.22%, O = 69.5% (all H present as water of crystallisation). Molecular mass = 322.
Step 1 -- Moles of each element (using atomic masses Na=23, S=32, H=1, O=16):
Moles of Na = 14.31/23 = 0.622
Moles of S = 9.97/32 = 0.3116
Moles of H = 6.22/1 = 6.22
Moles of O = 69.5/16 = 4.344
Step 2 -- Divide each by the smallest value (0.3116) to get simplest mole ratio:
Na: 0.622/0.3116 is about 2.0
S: 0.3116/0.3116 = 1.0
H: 6.22/0.3116 is about 19.96, round to 20
O: 4.344/0.3116 is about 13.94, round to 14
Step 3 -- Empirical formula: Na2SH20O14
Empirical formula mass = 2(23) + 32 + 20(1) + 14(16) = 46 + 32 + 20 + 224 = 322
…
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.