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A non-metal X forms two oxides I and II. The mass percentage of oxygen in I (X4O6) is 43.7% which is same as that of X in the 2nd oxide. What is the formula of 2nd oxide?

Asked by Topperlearning User 3rd March 2015, 1:22 PM
Answered by Expert
Answer:

                         Oxygen                       X

 I oxide              43.7                         56.3
II oxide              56.3                         43.7
 
From the data given above we can say that 43.7 parts of Oxygen corresponds to = 6 oxygen atoms
 
begin mathsize 11px style therefore space 56.3 space parts space of space Oxygen space in space II space correspond space to space equals fraction numerator 6 space cross times space 56.3 over denominator 43.7 end fraction space equals space 7.730 space atoms
end style
Also 56.3 parts of X in I correspond to = 4 X atom

begin mathsize 11px style therefore space 43.7 space parts space of space straight X space in space II space will space correspond space to space equals fraction numerator 4 space cross times space 43.7 over denominator 56.3 end fraction space equals space 3.1 space atoms end style
 
begin mathsize 11px style Now comma space the space ratio space of space straight X space colon space straight O space in space second space oxide space equals space fraction numerator 3.1 over denominator 3.1 end fraction space colon space fraction numerator 7.73 over denominator 3.1 end fraction space or space 1 space colon space 2.5 space or space 2 space colon space 5 end style 
Hence, the formula of second oxide is X2O5
Answered by Expert 3rd March 2015, 3:22 PM
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