A monoacid organic base gave the following data on analysis (a) 0.2790g of the base gave 0.7920g of CO_(2) and 0.1890g of H_(2)O (b) 0.1163g of the base gave 14mL of dry nitrogen at NTP (c ) 0.2980g of the platinichloride left 0.0975g of Pt Calculate the molecular formula of the base.

A monoacid organic base gave the following data on analysis (a) 0.279

1.	A monoacid organic base gave the following data on analysis (a) 0.2790g of the base gave 0.7920g of CO_(2) and 0.1890g of H_(2)O (b) 0.1163g of the base gave 14mL of dry nitrogen at NTP (c ) 0.2980g of the platinichloride left 0.0975g of Pt Calculate the molecular formula of the base.
Answer» Solution :MOLES of `C=1 xx` moles of `CO_(2)= (0.792)/(44)= 0.018` Moles of `H= 2XX` moles of `H_(2)O =2 xx (0.1890)/(18)= 0.021` Moles of `N=2 xx` moles of `N_(2)=2 xx (14)/(22400)= 0.00125` Moles of N in 0.2790g of the base `=(0.00125)/(0.1163)xx 0.2790` = 0.003 `therefore` moles of `C: H: N= 0.018: 0.021: 0.003` `=18:21:3` `=6: 7:1` Hence, the empirical formula is `C_(6)H_(7)N or C_(6)H_(5)NH_(2)` Suppose the monoacid base is B `{:(B,+,H_(2)PtCl_(6),rarr,B_(2)H_(2)PtCl_(6),rarr,Pt),("Base",,"Acid",,"Chloroplatinate",,),(("Monoacid"),,,,("Platinichloride"),,),(,,,,0.2980g,,0.0975g):}` Applying POAC for Pt atoms: `1xx` moles of `B_(2)H_(2)PtCl_(6)`= moles of Pt in product `(0.2980)/("mol. wt of" B_(2)H_(2)PtCl_(6))=(0.0975)/(195)` `therefore` mol. wt of `B_(2)H_(2)PtCl_(6)=596` Molecular weight of B `=("mol. wt. of" B_(2)H_(2)PtCl_(6)-"mol. wt of" H_(2)PtCl_(6))/(2)= (596-410)/(2)= 93` Since empirical formula weight is also 93, therefore, the molecular formula of the base is `C_(6)H_(5)NH_(2)`