A 12.5843g sample of ZrBr4 was dissolved and, after several steps, all of the combined bromine was precipitated as AgBr. The silver content of AgBr was found to be 13.2160g. Assume the atomic masses of silver and bromine to be 107.868 and 79.904. What value was obtained for the atomic mass of Zr from this experiment?
(I have the correct answer but I need to know the procedure for solving the problem)
2007-07-29 10:41:56 · 3 answers · asked by Tastytaint 1 in Science & Mathematics ➔ Chemistry
Divide the Ag weight by its atomic mass to get g-atoms of Ag.
Multiply this g-atoms by atomic mass wt of Br to get Br weight in ZnBr4.
Subtract Br weight from wt of ZnBr4, this gives you weight of Zn for ZnBr4.
The Zn g-atoms is ONE-FOURTH that of the g-atoms of Ag.
Divide weight of Zn by the Zn g-atoms
2007-07-29 11:24:27 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
13.1260 g of Ag in AgBr
13.1260 g / 107.868 = 0.121686 moles of Ag (must have 0.121686 moles of Br also).
12.5843 g ZrBr4 less 0.122168x79.904 = 2.8611 gm Zr.
Since you have 0.121686 moles of Br you have 0.121686 / 4 moles of Zr or 0.03042 moles
2.8611 gm / 0.03042 moles = 94.05 gm/mole of Zr
2007-07-29 11:24:35 · answer #2 · answered by skipper 7 · 0⤊ 0⤋
(14 / 14.01) x 100 and (0.01 / 14.01) x 100 99.93%, N14 and 0.07% N15 gets a bit tricky if not given the isotopes.
2016-05-17 06:59:24 · answer #3 · answered by ? 3 · 0⤊ 0⤋