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The empirical mass of NO is 14 + 16 = 30

Divide 138 / 30 ~ = 4.5.

The fact that the numbers are not whole number would suggest that the empirical formula is incorrect.

However, if we use say NO2 as the empirical formula , then the empirical mass is 14 + 16 + 16 = 46

divide 138 /46 = 3

So NO2 has a formula of N3O6 . This substance does not exist, So it would suggest that the formula is 3NO2 (3/2)N2O4 .

I would suggest that the original empirical formula is incorrect. and that the substance is either nitrogen dioxide (NO2) or dinitrogen tetroxide N2O4.

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lenpollock

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12mo ago
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12y ago

The formula of NO2 has a molecular weight of 46 g/mol. Your compound has a molecular weight of 92 g/mol. As you can see the molecular weight of the compound is twice that of the empirical formula. Therefore the molecular formula of your compound is:

2 *(NO2) ---> N2O4

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7y ago

Ar of N = 14g/mol Ar of O = 16g/mol

Mr of NO2 = 14+2(16) = 46g/mol

Mr of NxO2x = 138.02g/mol

(Mr of NxO2x / Mr of NO2) = 138.02/46 = 3.000434783 = 3 Therefore x = 3 and the molecular formula is N3O2(3) = N3O6 (but this compound is only hypothetical)

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11y ago

The molecular formula is the same as the empirical formula, NO2. The compound NO2 has a molar mass of 46g/mol, so the empirical and molecular formulas are the same.

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9y ago

The molecular formula is N2o4.

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12y ago

92 g

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11y ago

N3o6

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Q: What is the Molar mass of empirical formula NO2?
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