Question

A compound containing $Ca,C,N$ and $S$ was subjected to quantitative analysis and formula mass determination. A 0.25 g of this compound was mixed with $N{a}_{2}CaC{O}_3$ to convert all $Ca$ into 0.16 g $CaC{O}_3$. A 0.115 gm sample of compound was carried through a series of reactions until all its $S$ was changed into $S{O}_4^{2-}$ and precipitated as 0.344 g of $BaS{O}_4$. A 0.712 g sample was processed to liberated all of its $N$ as $N{H}_3$ and 0.155 g $N{H}_3$ was obtained . The formula mass was found to be 156. Determine the empirical and molecular formula of the compound.

Answer 1

Let the molecular formula of the compound ${Ca}_x{C}_y{N}_z{S}_p$.

$1^{st}$ reaction :

${Ca}_x{C}_y{N}_z{S}_p+x{Na}_2{CO}_3=xCa{CO}_3+{Na}_{2x}{C}_y{N}_z{S}_p$

So, $(40x+12y+14z+32p)\times 0.16=0.25\times 100⟶\left(1\right)$

2nd reaction : ${Ca}_x{C}_y{N}_z{S}_p⟶pBa{SO}_4\downarrow$

So, $(40x+12y+14z+32p)\times 0.334=0.115\times 233p⟶\left(2\right)$

3rd reaction : ${Ca}_x{C}_y{N}_z{S}_p⟶{xNH}_3\uparrow$

$\therefore$ $(40x+12y+14z+32p)\times 0.155=0.712\times 17z⟶\left(3\right)$

After solving the system, $x:y:z=1:2:2:2$

$\therefore$ Molecular formula will be ${CaC}_2{N}_2{S}_2$ or ${Ca\left(SCN\right)}_2$ having molecular weight $156$.