Calculate the empirical and molecular formula of the compoud having the following percentage coposition:
Na = 36.5%, H = 0.8%, P = 24.6% ,O = 38.1%. The molecular mass of the compound is 126 a.m.u. Also name the compound.

Write the empirical formulae of the compounds with the following percentage composition : Zn = 47.8% Cl = 52.2%

Determine the empirical formula of a compound with following percentage composition Cu= 47.3, Cl= 52.7.

Write the empirical formula of the compounds having the molecular formulas : C6H12O6

An organic compound has the following percentage composition : C= 48%, H= 8% N= 28% and rest of oxygen. Calculate the empirical formula of the compound.

What is the simplest formula of the compound which has the following percentage composition carbon 80% , hydrogen 20% , if the molecular mass is 30. calculate its molecular formula ?

A substance on analysis gave the following percentage composition Na = 43.4% C = 11.3% O = 45.3% Calculate the empirical formula.

An organic compound on analysis gave the following percentage composition of C= 57.8, H= 3.6 and the rest is oxygen. Vapour density of the compound is 83. Find out the molecular formula of the compound.

An inorganic salt gave the following percentage composition : Na= 29.11%, S= 40.51% and O= 30.38%. Calculate the empirical formula of the salt. (At. Mass Na= 23, S=32)

An organic substance containing carbon, hydrogen and oxygen gave following percentages, C= 40.68%, H_(2) = 5.085% and O_(2) = 54.22%. The vapour density of the compound is 59. Calculate the molecular formula of the compound.

An organic compound on analysis gave the following data : C = 57.82% , H = 3.6% and O = 38.58% . Find its empirical formula.