There is a series of nitrogen oxides with the general formula NxOy. What is the empirical formula of one
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There is a series of nitrogen oxides with the general formula NxOy. What is the empirical formula of one

[From: ] [author: ] [Date: 11-12-15] [Hit: ]
we have 25.9 g nitrogen in it.Then subtract 25.9 from 100= 74.1 g of oxygen.Now atomic mass of oxygen is 16,......
containing 25.9% nitrogen by mass?

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Answer is N2O5.
Here's why---->
First consider 100 g of that gas. So as per your data, we have 25.9 g nitrogen in it.
Then subtract 25.9 from 100= 74.1 g of oxygen.
Now atomic mass of oxygen is 16, so divide 74.1 by 16, we get 4.63 which is rounded to 5 because only whole natural numbers are acceptable in formulae.
Similarly, divide 25.9 by atomic mass of nitrogen ie by 14, we get, 1.85 which is rounded to 2, now you know why, right?!
so there it is, N2O5, as per your question x=2, y=5, in formula NxOy.
I hope that helps. Thanks!
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