A #"0.4596 g"# sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as #"AgCl"# by the addition of an excess of silver nitrate. The mass of the resulting #"AgCl"# is found to be #"0.6326 g"#...?

Question

Henry Antrim · Accepted Answer

#%Cl=34.1%# by mass..........

Excellent query. To start, we measure the amount of chloride that is there. We found that..... #Ag^+ + Cl^(-) rarr AgCl(s)darr# #"Moles of AgCl"=(0.6326*g)/(143.32*g*mol^-1)=4.414xx10^-3*mol#. Keep in mind that using silver chloride gravimetrically would not be very practical because it tends to photo-oxidize and is a very curdy material that is difficult to isolate. Now here, chloride anion was the material in deficiency; i.e. silver ion was present in excess.....there was thus a mass of #4.414xx10^-3*molxx35.45*g*mol^-1=0.1565*g# in the original sample. And so #%Cl=(0.1565*g)/(0.4595*g)xx100%=34.1%#

Cooper Adams · Answer

undefined The mass of AgCl is 143.32 g/mol. The mass of AgCl can be used to calculate the moles of AgCl produced, which is approximately 0.00441 mol. Since AgCl and chloride ions (Cl^-) are a 1:1 ratio compound, the moles of chloride ions in the original sample are also approximately 0.00441 mol. The molar mass of chloride ions (Cl^-) is approximately 35.45 g/mol. The mass of the original chloride compound can be calculated using the moles of chloride ions, which can be found by multiplying the mass of AgCl by the molar mass of AgCl. This yields a mass of 0.1562 g.