$60 \ g$ of a compound on analysis gave $C = 24 \ g$,$H = 4 \ g$ and $O = 32 \ g$. Its empirical formula is

An ornamental of gold having $75\%$ of gold,it is of .............. carat.

An organic compound contains $C = 74.0\%$,$H = 8.65\%$ and $N = 17.3\%$. Its empirical formula is:

An organic compound containing carbon and hydrogen has an empirical formula of $CH_2$. The mass of $1 \, L$ of this organic gas is equal to the mass of $1 \, L$ of $N_2$ gas at the same temperature and pressure. What is the molecular formula of the organic gas?

The simplest formula of a compound containing $50\%$ of element $X$ (at. wt. $= 10$) and $50\%$ of element $Y$ (at. wt. $= 20$) is

By usual analysis, $1.00 \ g$ of compound $(X)$ gave $1.79 \ g$ of magnesium pyrophosphate. The percentage of phosphorus in compound $(X)$ is: (nearest integer)
(Given, molar mass in $g \ mol^{-1}$; $O=16, Mg=24, P=31$)