A solid sample of a complex salt, with ideal composition Ni(NH3)6(NO3)2, was ana

Question

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 14.1564 g was used to filter off the precipitate. The dried crucible and precipitate had a combined mass of 14.2534 g. Calculate the theoretical mass of the original Ni(NH3)6(NO3)2 sample.If the actual mass of the original sample in this experiment was 0.0859 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 placesA 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 14.1564 g was used to filter off the precipitate. The dried crucible and precipitate had a combined mass of 14.2534 g. Calculate the theoretical mass of the original Ni(NH3)6(NO3)2 sample.If the actual mass of the original sample in this experiment was 0.0859 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

Explanation / Answer

weight of crucible = 14.1564g

weight of crucb + sample = 14.2534g

Therefore weight of sample = 0.097g

mol weight of given Ni(NH3)6(NO3)2 = 284.8863 g

mol weight of Ni(DMG)2 = 288.9146 g

thus 288.9164 g will be obtained from 284.8862 g Ni(NH3)6(NO3)2

therefore 0.097 g will be obtained from = 0.09564 g of Ni(NH3)6(NO3)2

Now, 284.8863 g of Ni(NH3)6(NO3)2 contains 102 g of ammonia

so, 0.09564 g Ni(NH3)6(NO3)2 will contain .034242 g of ammonia

but the actual weight of Ni(NH3)6(NO3)2 is given 0.0859 g

the difference in weight of actual and theoretical value = 0.00974 g

So formula should be Ni(NH3)4(NO3)2

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