Acompound is 12.7% al, 19.7% n, and 67.6% o. determine its emperical formula.

AlN₃O₉

Explanation:

Assume that you have 100 g of the compound.

Then you have 12.7 g Al, 19.7 g N, and 67.6 g O.

1. Calculate the moles of each atom

Moles of Al = 12.7 × 1/26.98 = 0.4707 mol Al

Moles of N = 19.7 × 1/14.01    =  1.406   mol N

Moles of O = 67.6 × 1/16.00  = 4.225   mol O

2. Calculate the molar ratios.

Al: 0.4707/0.4707 = 1

N: 1.406/0.4707    = 2.987

O: 4.225/0.4707   = 8.976

3. Determine the empirical formula

Round off all numbers to the closest integer.

Al: 1

N: 3

O: 9

The empirical formula is AlN₃O₉.

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