Question

A compound of Ca, C, N, and S was subjected to quantitative analysis and formula mass determination, and the following data were recorded. A 0.250-g sample was mixed with Na2CO3 to convert all of the Ca to 0.160-g of CaCO3. A 0.115-gm sample was carried through a series of reactions until all of its S was changed to 0.334-g of BaSO4. A 0.712-gm sample was processed to liberate all of its N as NH3, and 0.155-g NH3 was obtained. The formula mass was found to be 156-g/mol. Determine the empirical formula of this compound.

Answer 1

0.250 0.160
Ca_xC_yN_zS_p + xNa₂CO₃ = xCaCO₃ + Na_2xC_yN_zS_p
(40x+12y+14z+32p) x*100

0.115 0.334
Ca_xC_yN_zS_p ----------------------> pBaSO4
(40x+12y+14z+32p) 233p

0.712 0.155
Ca_xC_yN_zS_p ---------------------------------> zNH₃
(40x+12y+14z+32p) 17z


[(40x+12y+14z+32p)*0.160 = 0.250*x*100

{(40x+12y+14z+32p)*0.334 = 0.115 * 233p

[(40x+12y+14z+32p) *0.155 = 0.712*17z

After solving of this system: x ; y:z:p = 1:2:2:2

it means that empiric formula is CaC₂N₂S₂, or Ca(SCN)₂ with molecular weight 156