In an organic compound,the percentage of $C$ is $92.3 \%$. What is its empirical formula?

$A$ compound contains $C = 90\%$ and $H = 10\%$. The empirical formula of the compound is:

An organic compound containing $C$,$H$,and $N$ gave the following analysis: $C = 40\%$,$H = 13.33\%$,and $N = 46.67\%$. Its empirical formula would be:

The molecular formula of sulphur is

$0.30 \ g$ of an organic compound containing $C$,$H$ and $O$ on combustion yields $0.44 \ g \ CO_2$ and $0.18 \ g \ H_2O$. If $1 \ mol$ of the compound weighs $60 \ g$,then the molecular formula of the compound is:

An organic compound gave $C = 92.31\%$ and $H = 7.69\%$. If the molecular weight of the compound is $78$,its molecular formula is