The empirical formula for a substance that is 12.67 % aluminum, 19.73% nitrogen, 67.60% oxygen is Al(NO₃)₃.
given that :
aluminum = 12.67 % = 12.67 g
nitrogen = 19.73 % = 19.73 g
oxygen = 67.60 % = 67.60 g
molar mass of aluminum = 12.67 / 26.98
= 0.46 mol
molar mass of nitrogen = 19.73 / 14
= 1.40 mol
molar mass of oxygen = 67.60 / 16
= 4.2 mol
divided by the smallest one , we get:
aluminum = 0.46 / 0.46 = 1
nitrogen = 1.40 / 0.46 = 3
oxygen = 4.2 / 0.46 = 9
The empirical formula is = Al(NO₃)₃
To learn more about empirical formula here
https://brainly.com/question/14462365
#SPJ4
