calculate the emperical formula of the compound which contains 26.6% of K(potassium),35.5% of Cr(chromium),38.1% of O(oxygen)
atomic weight of K=39
atomic weight of Cr=52
atomic weight of O= 16
irrelevant answers are to be reported
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Answer:
K(potassium)= 26.6/39
= 0.68
= 0.68/0.68=1
Cr(chromium) = 35.5/52
= 0.68
= 0.68/0.68=1
O(oxygen) = 38.1/16
= 2.38
= 2.38/0.68
= 3
emperical formula
= (K Cr O 3)n
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