Convert 273.7 mg CO2 to mg C, then to %C.
273\.7 mg x 12/44 = ?, then
(?/255) x 100 = about 29.3% C.
Convert 130.6 mg H2O x (2/18) = ?
Then (?/255)*100 = about 5.7% H.
The halogen must be 100%-29.3%-5.7% = about 65% halogen. Which halogen?
The cream colored ppt PROBABLY MEANS it is Br but it might also be Cl or I. X = halogen. Convert 133 mg AgX to the factor that makes the conversion; i.e. (atomic mass X/molar mass AgX). That is done this way and I will call the factor just F.
%X = (133 x F x 100/87)*100 = about 65%. Solve for F. I get about 0.425
Then 0.425 = X/(108+X) and solve for X. I get about 80 which means Br.
The analysis of halogenated hydrocarbon gave the following results:
A sample of mass m=255mg produces 273.7 mg carbon dioxide and 130.6 mg of water .A second sample of mass 87 is treated by hot molten sodium .The obtained mixture is dissolved in distilled water and filtered .To the aqueous filtrate an excess of silver nitrate solution is added. A cream color precipitate is formed which weighs 133 mg.
Identify the halogen
Thank You!!
1 answer
1 answer