If a compound on analysis was found to contain $C = 18.5\%$,$H = 1.55\%$,$Cl = 55.04\%$,and $O = 24.81\%$,then its empirical formula is:

On combustion,$0.210 \ g$ of an organic compound containing $C$,$H$,and $O$ gave $0.127 \ g$ $H_2O$ and $0.307 \ g$ $CO_2$. The percentages of hydrogen and oxygen in the given organic compound respectively are $:$

Find the empirical formula of an organic compound if it contains $18.6 \% C$,$1.55 \% H$,and $55.04 \% Cl$. (Atomic masses: $C=12$,$H=1$,$Cl=35.5$,$O=16$)

The formula which represents the simplest ratio of atoms of different elements present in a molecule of a substance is called:

The percentage of nitrogen in urea is about

$A$ welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives $3.38 \, g$ of carbon dioxide,$0.690 \, g$ of water,and no other products. $A$ volume of $10.0 \, L$ (measured at $STP$) of this welding gas is found to weigh $11.6 \, g$. Calculate:
$(i)$ empirical formula,
$(ii)$ molar mass of the gas,and
$(iii)$ molecular formula.