A compound on analysis gave the following results.C= 54.54%. H=9.09% O= 36.37%. And vapour density of compound = 88. Find the molecular of formula of the compound.

Answer 1

Firstly, we need to calculate the empirical formula of the compound:
Assume we have 100g of the compound, then:
C = 54.54g
H = 9.09g
O = 36.37g

Now, we need to convert the masses of the elements into moles:
moles of C = 54.54/12 = 4.55
moles of H = 9.09/1 = 9.09
moles of O = 36.37/16 = 2.27

Next, we need to divide the number of moles of each element by the smallest number of moles:
4.55/2.27 = 2
9.09/2.27 = 4
2.27/2.27 = 1

Therefore, the empirical formula of the compound is C2H4O.

Now we need to find the molecular formula by using the vapour density of the compound.

Molecular weight = 88

Empirical formula weight = (2x12) + (4x1) + 16 = 44

Dividing the molecular weight by the empirical formula weight gives: 88/44 = 2

Therefore, the molecular formula of the compound is (C2H4O)2 which simplifies to C4H8O2.