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.

Assuming we have 100 g of sample

30.45/MW of N 14g = 2.175

69.55/MW of O 16g = 4.34

4.34/2.185 = 2

for every 1 mole of N we have 2 moles of O

so 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