Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o. what is the molecular formula for this gas? express your answer as a chemical formula.

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Assuming we have 100 g of sample30.45/MW of N 14g = 2.17569.55/MW of O 16g = 4.344.34/2.185 = 2for every 1 mole of N we have 2 moles of Oso the empirical formula would be NO2 without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question

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