Ca =4%, h =56%, to=40% vapour density =90,empirical formula =90 and molecular formula=
Answers
Answer:
The empirical formula of compound is CH2O, if vapour density is 45. Then the molecular formula is
First Step: We are going to find the Molecular Weight (MW) of the Empirical Formula (EF), let's see:
if: CH2O
C = 1*(12 a.m.u) = 12 a.m.u
H = 2*(1 a.m.u) = 2 a.m.u
O = 1*(16 a.m.u) = 16 a.m.u
-------------------------------------
Second Step: Knowing that it is given in vapor density of 45, we will find the Molecular Weight of the Molecular Formula "MW (M.F)", see:
Molecular Weight (M.F) = 2 * Vapour density
MW (M.F) = 2 * 45
Third Step: Knowing that the Molecular Weight of the Molecular Formula is 70 units of atomic mass and that the Molecular Weight of the Empirical Formula is 30 units of atomic mass, then we will find the number of terms (n) for the molecular formula of the compound, let us see:
Therefore, the Molecular Formula is the Empirical Formula times the number of terms (n), then, we have:
Answer:
C3H6O3
Read more on Brainly.in - https://brainly.in/question/1167188#readmore