#2Fe(HCO_3)_3rarrFe_2(CO_3)_3+2H_2O+3CO_2#
I will assume that the #H_2O# and #CO_2# are driven off as gases.
So the mass of #H_2O# + #CO_2=1.394-0.978=0.416"g"#
We can work back and find how much iron(III) hydrogen carbonate would be needed to produce this amount of #H_2O# and #CO_2#:
2 mol #Fe(HCO_3)_3rarr# 2mol #H_2O# + 3mol #CO_2#
Converting to grams using the #M_r# values #rArr#
#(2xx238.9)"g"rarr(2xx18.01)"g"+(3xx44.00)"g"#
#:.477.8"g"rarr168.02"g"#
So this tells us that:
#168.02"g"# of #CO_2# and #H_2O# are produced from #477.8"g"# iron(III) hydrogen carbonate.
#:.1"g"# from #(477.8)/(168.02)"g"#
#:.0.416"g"# from #(477.8)/(168.02)xx0.416=1.183"g"#
This is the mass of iron(III) hydrogen carbonate.
So the % purity is:
#(1.183)/(1.394)xx100=84.86#
