CBSE Class 11-science Answered
A compound on analysis was found to have the following compositions,Na-14.13%,S-9.97%,oxygen-69.50%,hydrogen-6.22%..Calculate the molecular formula of the compound assuming that whole of hydrogen in the compound is present as water of crystallization and molar mass of compound is 322g
Asked by Braham | 20 Jun, 2013, 11:45: PM
Expert Answer
moles Simplest Molar Ratio simple ratio
Na 14.13% 23 14.13/23 = 0.614 0.614/0.311 = 1.97 2
S 9.97% 32 9.97/32 = 0.311 0.311 / 0.311 = 1 1
O 69.5% 16 69.5/16 = 4.343 4.34 / .311 = 13.97 14
H 6.22% 1 6.22 / 1 = 6.22 6.22 / .311 = 20 20
Therefore, the emperical formula is = Na2S1H20O14
Therefore, emperical formula mass = 2x23 + 1x32 + 20x1 + 16x14 = 322
Now,
molecular mass
n = emperical formula mass
= 322/ 322 = 1
therefore molecular mass = Na2S1H20O14
Finally, since all Hydrogen is present as water, the molecular formula would be, Na2SO4.10H20
Answered by | 25 Jun, 2013, 01:01: AM
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