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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)

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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
User Scott Joudry
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