An organic compound has the percentage of $C$ and $H$ in the ratio $6 : 1$ and the percentage of $C$ and $O$ in the ratio $3 : 4$. The compound is:

Calculate the percentage composition of the following elements:
$(a)$ Sulphur in $CN_4H_4S$
$(b)$ Carbon in $C_2H_4O_2$
$(c)$ Nitrogen in $CN_2H_2O$
$(d)$ Hydrogen in $C_6H_6$
Given atomic masses: $C = 12, H = 1, O = 16, N = 14, S = 32$.

$A$ 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 $10.0 \ L$ of this welding gas (measured at $STP$) is found to weigh $11.6 \ g$. Calculate: $(i)$ Empirical formula,$(ii)$ Molar mass of the gas,$(iii)$ Molecular formula.

An organic compound on analysis gave $C = 48 \,g$, $H = 8 \,g$ and $N = 56 \,g$. The volume of $1.0 \,g$ of the compound was found to be $200 \,mL$ at $NTP$. What is the molecular formula of the compound?

An organic compound is estimated through the Dumas method and is found to evolve $6$ moles of $CO_2$,$4$ moles of $H_2O$,and $1$ mole of nitrogen gas. The formula of the compound is

An organic compound has $42.1 \%$ carbon,$6.4 \%$ hydrogen and the remainder is oxygen. If its molecular weight is $342$,then its molecular formula is: