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 13.2379 g was used to filter off the precipitate. The dried crucible and precipitate had a combined mass of 13.3319 g.

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

0.0927 g

You are correct. Your receipt no. is 152-9832

Previous Tries

Calculate the mass of nickel in this sample.

Incorrect.

Tries 4/5

Previous Tries

If the actual mass of the original sample in this experiment was 0.0733 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.

Tries 0/5

Calculate the percent loss of ammonia which accounts for the difference in the theoretical and actual mass of the initial sample.
Enter your answer to three significant figures.

%

Answer 1

Ni (NHino (Nos)? - Ni COMG) DI ZH ON Secu vo Ni Sound VIL CHACON on = 58+ 14+3) X6 vis (As shown in the question) Molecular weight of sample 7 + [(14+ # 16x33 20 egunot 158 + 102 + 124) gimot 2.8.4 g i motsatta Molecular weight of NiCO.MG)2 / ECG HIU NA 04 N = {(1,2x8) + 14 (14х4) + (16*4) +58) gimos s. 288 g/mol A gain, weight of crucible = 13.23799 weight of crucible & preecipitate : .13.33199 is weight of precipitati = (18.3319-13.2379) g- 010949

. Amount of NiCDMG) 2 produced = .094 g = 094 mole = 3:26AX10A Again from the equation, it is clear that I mole sample with precipitate je 1 mole Ni(DMG) . sample amount is made 3:264.8104 mole =(3,264*10*4* 284) g -0.0927g. Theoretical field of Ni (N.14%) 6 (No, lis 0.09277 Mass of Ni' in 1 mole sample is - - 2. mass of 'Ni' in sample is 3.264 x 10-4, mole 13.264 x 10 9 458) g 1) = 0.0189 g.

a Actual mass of original sample is 0.0733g and theoretical yield is 0:0927g " the amount of lost ad ammonia is 10.0.927.-0107 33) g. 010194 g AY! O cenile The no. of in this ammonia molecule 0.019,4 g NH3 of 31264 inole sample if. roso r 010194,8 u 1 4 more 2 17 g/mole x 3.264 x10-4 more 3.50 000 > The remaining NH3 molecule en 0= (6-3,5) = 2:50504; ...Ni (NH3)3.50 (NO3)2 25

- A t Loss of ammonia I 0.0194 g : 6 mDe x 17 g / mole x 3:264x10 motel As initially in si264 x 10-4 molle sample suolecules of NH3 will to be present to trapian :: Loss of ammoniated Xloo 0.0194 6*17 *.3.264*104 = 58.3%,