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An oxide of nitrogen is 25.9% N by mass, has a molar mass of 108 g/mol, and contains no nitrogen-nitrogen or oxygen-oxygen bonds. Draw its Lewis structure (including all lone pair electrons) and name it. Include any nonzero formal charges in your structure.

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Answer:

The compound is N2O4

Explanation:

We have certain important pieces of information about the compound;

1) it is an oxide (a binary compound of nitrogen and oxygen)

2) there are no N-N bonds present

3) there are no O-O bonds present

Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen

Relative atomic mass of oxygen = 16

Relative atomic mass of nitrogen = 14

We now deduce the empirical formula

Nitrogen. Oxygen

25.9/14. 74.1/16

1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)

1 2

Empirical formula is NO2

To find the molecular formula

(NO2)n = 108

(14+2(16))n= 108

46n=108

n= 108/46

n= 2

Therefore molecular formula= N2O4

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