Solve Chemistry Problem III: Find n in [Ni(NH3)n](NO3)2

[higherme]

Question:

Given
mass of hexaammine complex [Ni(NH3)6](NO3)2 is 0.0937 g
mass of Ni2+ is 0.0193 g

if the mass of [Ni(NH3)6](NO3)2 was 0.0886 g instead of 0.937g, this would indicate that the hexaammine complex was not pure and some ammonia had been lost. Calculate the value of n in the formula [Ni(NH3)n](NO3)2


my attempt:

- Take the mass of Ni2+ and find the moles of it using Ni2+ molar mass. This mole would then be the same as the moles of the complex (1:1 ratio) --> i found to be 0.0003288 moles

- then take the grams of the complex 0.0886 g and divide by that mole: 0.0886g/0.0003288moles = 264.4404 g/mol which would be the molar mass of the complex

- then i subtracted the molar masses of Ni2+ and (NO3)2 from that molar mass and what is left is the amount of (NH3)n. Then i took that amount and divided by the molar mass of one NH3 to see how many NH3 I have.

am I thinking right?

 

[chemisttree]

You are well on your way.

 

[higherme]

i keep getting the wrong answer...what am i doing wrong?

 

[higherme]

so here is what i did

0.0193g Ni x (1 mol Ni/58.693g) x (1mol [Ni(Nh3)6](NO3)2/ 1 mol Ni) = 3.288 E-4 moles of [Ni(Nh3)6](NO3)2

then i took 0.0886g and divide by this moles to get the molar mass
0.0886g/3.288 E-4 moles [Ni(Nh3)n](NO3)2
= 269.44 g/mol

then i substracted Ni (58.693g/mol) and (NO3)2 (124.008g/mol) from that molar mass
269.44 - 58.693 - 124.008
86.739 g/mol

then take this number and divide by (NH3) molar to see how many NH3 there are:
(86.739g/mol) / (17.031g/mol)
= 5.09 NH3

so n is equal to 5.09... this answer is still incorrect when i entered into the computer for grading ><

what am I doing wrong!?

 

[eli64]

did you enter 5 or 5.09? n should be a whole number

 

[chemisttree]

How many significant figures are you asked to provide? Generally you would use an integer value if you were trying to determine the empirical formula for a pure substance. In this case the sample may be a mixture of the hexamine, pentamine, etc... and so a non-integer value for n may be correct.