A compound has the following percentage composition by mass : Carbon = 54.55%, Hydrogen = 9.09% and Oxygen = 36.26%.
Its vapour density is 44. Find the Empirical and molecular formula of the compound.
(H = 1, C = 12,0 = 16)

A compound has the following percentage composition by mass: Carbon - 54.55%, Hydrogen - 9.09% Its vapour density is 50. Find the Empirical formula. (H = 1; C = 12; O = 16).

A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1, C = 12, CI = 35.5)

In a compound Carbond =52.2% , Hydrogen =13% , Oxygen =34.8% are present vapour density of the compound is 46. Calculate molecular formula of the compound ?

A compound on analysis gave the following percentage composition by weight: hydrogen = 9.09, oxygen = 36.36 carbon = 54.55 Its VD is 44. Find the molecular formula of the compound.

A compound has the following percentage composition : C = 40%, H = 6. 7%, O = 53.3%, the vapour density of the compound is 30. Calculate its molecular formula (C = 12, H = 1, O = 16).

The compound A has the following percentage composition by mass : carbon 26. 7%, oxygen 71.1 %, hydrogen 2.2%. Determine the empirical formula of A. (Work to one decimal place) (H = 1, C = 12, O = 16). If the relative molecular mass of A is 90, what is the molecular formula of A ?

A compound contains 40% carbon, 6.6% hydroaen and 53.33% oxy1en._Its vapour density is 30. Calculate its empirical and molecular formulae.

An organic compound has an empirical formula CH_2 O . Its vapour density is 45. The molecular formula of the compound is

If the empirical formula of a compound is CH and its vapour density is 13, find the molecular formula of the compound.

A compound on analysis gave the follwing result C=54.54%, H=9.09% and vapour density of compound = 88.Determine the molecular formula of the compound.