Question

A compound has the following percentage composition.
Al=0.2675g.;P=0.3505g.;O=0.682g.
If the molecular weight of the compound is 122 and its original weight which on analysis gave the above results 1.30g. Calculate the molecular formula of the compound.
[Al=27,P=31,O=16]

Answer 1

Answer must be  AlPO₄
GIVEN : Molecular mass of the compound = 122 g
0.26g of Al is present in 1.3g sample of the compound.
So mass of Al pressent in 122 g of the same compound = (0.26/1.3) * 122
                                                                                  = 24.4 ≈ 27
Similarly mass of P present in 122 g of the same compound =( 0.35/1.3) * 122
                                                                                       = 32.84 ≈ 31
And mass of O in it = (  0.682 / 1.3 ) * 122 = 64.0030 ≈ 64

Comparing these values with the molecular masses of the respective compounds , we find that a molecule of the given compound contains 1 Al atom , 1 P atom and 4 O atoms...

Hence the molecular formula of the compound is AlPO₄ .
  

Answer 2

Answer:

Explanation:

GIVEN : Molecular mass of the compound = 122 g

0.26g of Al is present in 1.3g sample of the compound.

So mass of Al pressent in 122 g of the same compound = (0.26/1.3) * 122

                                                                                  = 24.4 ≈ 27

Similarly mass of P present in 122 g of the same compound =( 0.35/1.3) * 122

                                                                                       = 32.84 ≈ 31

And mass of O in it = (  0.682 / 1.3 ) * 122 = 64.0030 ≈ 64

Comparing these values with the molecular masses of the respective compounds , we find that a molecule of the given compound contains 1 Al atom , 1 P atom and 4 O atoms...

Hence the molecular formula of the compound is AlPO₄ .

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