Find the empirical formula of the compound if $M = 68 \%$ (Atomic mass $= 34$) and the remaining $32 \%$ is oxygen:

$A$ compound contains $69.5\%$ oxygen and $30.5\%$ nitrogen and its molecular weight is $92$. The molecular formula of the compound is

$A$ compound containing $Xe$ and $F$ has $63.8\%$ of $Xe$ by mass. Calculate the oxidation number of $Xe$ in this compound.

Complete combustion of $X$ g of an organic compound gave $0.25$ g of $CO_2$ and $0.12$ g of $H_2O$. If the percentage of carbon is $25\%$ and of hydrogen is $4.89\%$, then $X = \_\_\_\_ \times 10^{-3}$ g. (Nearest integer) (Molar mass of $C$, $H$ and $O$ are $12, 1$ and $16$ g mol$^{-1}$ respectively.)

$A$ gaseous hydrocarbon gives upon combustion $0.72 \ g$ of water and $3.08 \ g$ of $CO_{2}$. The empirical formula of the hydrocarbon is:

$A$ sample of calcium carbonate $(CaCO_3)$ has the following percentage composition: $Ca = 40\%$; $C = 12\%$; $O = 48\%$. If the law of constant proportions is true,then the weight of calcium in $4 \, g$ of a sample of calcium carbonate obtained from another source will be $... \, g$.