a certain substance is analyzed and found to contain the following weight percentages 36.84% nitrogen (N) and 63.16% oxygen (O). determine the empirical formula of this compound. (Atomic Weight N=14,O=16)
empirical formula is the simplest ratio of whole numbers of components in a compound in the given compound lets calculate for 100 g of the compound N O mass 36.84 g 63.16 g number of moles 36.84 g / 14 g/mol 63.16 g / 16 g/mol = 2.63 mol = 3.94 mol divide by least number of moles 2.63/2.63 = 1.00 3.94 / 2.63 = 1.50 multiply by 2 to get numbers that can be rounded off to whole numbers N - 1.00 x 2 = 2.00 O - 1.50 x 2 = 3.00 therefore empirical formula is N₂O₃
