Question

combustion analysis of a compound, Q, of mass 17.0 indicated that Q contained 9.6g of carbon, 1.0 of hydrogen, and the remainder oxygen. The empirical formula of Q is

Answer 1

Solution:- mass of C = 9.6 g

mass of H = 1.0 g

mass of O = 17.0 - (9.6 + 1.0) = 17.0 - 10.6 = 6.4 g

moles of C = 9.6/12.01 = 0.799

moles of H = 1.0/1.01 = 0.990

moles of O = 6.4/16 = 0.400

Now we would alculate the mol ratio and for this we divide all by the least one. here the least one is O. so, we would divide all of them by 0.400.

C = 0.799/0.400 = 2

H = 0.99/0.400 = 2.50

O = 0.400/0.400 = 1.00

2.50 is not a whole number so we need to multiply all of them by 2.

The whole number ratio would be..

C = 4, H = 5 and O = 2

So, the empirical formula of Q is C₄H₅O₂