To find the empirical and molecular formulas for this nitrogen oxide, you can follow these steps:
Step 1: Calculate the number of moles of each element (oxygen and nitrogen) in 100 g of the compound using their respective percentages.
For oxygen:
Number of moles of oxygen = (69.57% / 100%) * (100 g) / (atomic weight of oxygen)
Atomic weight of oxygen (O) ≈ 16 amu
Number of moles of oxygen = (69.57 / 100) * (100 g) / (16 amu) ≈ 43.48 moles of oxygen
For nitrogen:
Number of moles of nitrogen = (30.43% / 100%) * (100 g) / (atomic weight of nitrogen)
Atomic weight of nitrogen (N) ≈ 14 amu
Number of moles of nitrogen = (30.43 / 100) * (100 g) / (14 amu) ≈ 21.73 moles of nitrogen
Step 2: Find the simplest whole number ratio of moles of oxygen to moles of nitrogen.
Divide both moles by the smaller number of moles (21.73 in this case) to get the ratio:
Oxygen to Nitrogen ratio ≈ 43.48 / 21.73 ≈ 2
So, the empirical formula is NO2.
Step 3: Determine the molecular formula using the molecular weight provided (92.02 amu) and the empirical formula.
The empirical formula mass (EFM) of NO2 is:
EFM(NO2) = (1 * atomic weight of N) + (2 * atomic weight of O) = (1 * 14 amu) + (2 * 16 amu) = 46 amu
Now, divide the given molecular weight (92.02 amu) by the empirical formula mass (EFM) to find the integer multiple:
Molecular formula multiple = 92.02 amu / 46 amu = 2
So, the molecular formula is (NO2)2, which can be simplified to N2O4.