[jkulier]

Hi there, I've been stuck on the first part of this problem for a while.

A solid sample of a complex salt, with ideal composition Ni(NH3)6(NO3)2, was analyzed for nickel by dissolution in slightly acidic solution and precipitation as the dimethylglyoximate, Ni(DMG)2. A sintered glass crucible of mass 12.7309 g was used to filter off the precipitate. The dried crucible and precipitate had a combined mass of 12.8239 g.

Calculate the theoretical mass of the original Ni(NH3)6(NO3)2 sample.

The only thing I could come up with was 12.8239 - 12.7309 = 0.0930 g
However, that was wrong.

Any help would be much appreciated,
thanks.

[Borek]

You need to convert mass of Ni(DMG)₂ precipitate to moles. For that you need molar mass of DMG. Try either your textbokk, google or wikipedia.

[jkulier]

Alright, so I have done this:

0.0930 g Ni(DMG)2 * (1 mol / 288.91 g Ni(DMG)2) = 3.1966 x 10^-4 moles of Ni(DMG)2

Since Ni(DMG)2 and Nickel are in a 1:1 ratio, there are 3.1966 x 10^-4 moles of Ni

and thus, 3.1966 x 10^-4 moles Ni * (58.693 g Ni / 1 mole) = 0.0188 grams of Ni in the sample.

I am wondering if that is correct up to this point? If so, what would I do next ?

Thanks again.

[Borek]

Looks OK, although you don't need mass of Ni - convert moles of Ni to moles of Ni(NH₃)₆(NO₃)₂, then calculate its mass.

[xoto]

the difference in the two masses is the mass of the ppt (Ni(DMG)2).
there exist a 1:1 mole ratio btn the sample and the ppt. using this mass and the molar masses, you can get what you want. i had 0.2255g. check if am correct.

[bramptonboy]

Hello, I a apologize for replying to this, but I am dealing with the same analysis of a nickel 2 complex. However, I am having trouble with the following question:

If the actual mass of the original sample in this experiment was 'X'g (instead of the theoretically calculated mass) -- indicating that the original Ni(NH3)6(NO3)2 complex had lost some ammonia -- what would be the corresponding value of n in the revised formula Ni(NH3)n(NO3)2. Enter your answer to 2 decimal places.

How woud you go on solving this?

What i had tried already is subtracting the actual mass from the theoretical, and dividing it by the molar mass of Ni(NH3)6(NO3)2 to get the mols and multiply that value by 6 due to the amount of 6 ammonia molecules in the formula.

However, I am confused on what to do next. Any insight will be extremely helpful.