JEE Main Chemistry question
A compound (60 g) on analysis gave C = 24 g, H = 4 g and O = 32 g. Its empirical formula is:
Answer: (D)
Solution
Moles: C = 24/12 = 2, H = 4/1 = 4, O = 32/16 = 2. Ratio C : H : O = 2 : 4 : 2 = 1 : 2 : 1, giving the empirical formula CH₂O.