14. Calculate the mass percent of Ni2+(aq)in the unknown solution.
    
    
    • .
    
    
    
    

.

gNi NH Cl

gNi

0 350

0 0892 ()aq

(^32) n
2
^h
× 100% = 25.5%
Empirical Formula and Percent Yield of Ni(NH 3 )nCl 2

15. Determine the mass percent of Cl–in Ni(NH 3 )nCl 2

= 100% – (mass % Ni2+(aq)) (mass % NH 3 )

= 100% – (25.5% + 43.1%) = 31.4%

16. Determine the empirical formula of Ni(NH 3 )nCl 2.

mass % Ni2+= 25.5%

mass % NH 3 = 43.1%

mass % Cl–= 31.4%

Assuming 100. g Ni(NH 3 )nCl 2

+

+

. + +
.

.

.

..
.
..

gNi

gNi

mol Ni mol Ni

gNH

gNH

mol NH mol NH

gCl

gCl

mol Cl mol Cl

1

25 5

58 69

(^1) 0 434
1
43 1
17 04
(^1253)
1
31 4
35 45
(^1) 0 886
2
2
(^22)
3
3
(^33)

=

=

• 
  
  
  
  • 
    
  
  
  
  


• 
  
  
  
  • 
    
  
  
  
  

The Lowest Common Multiplier (LCM) is 0.434, which gives an actual ratio of

+

.

. :
.
. :
.
Ni NH Cl. .:.:.
0 434

0 434

0 434

253

0 434

0 886 1 00 5 83 2 04

3 =

2 -

Lowest whole number ration = 1 Ni2+: 6NH 3 : 2 Cl–

Therefore, the empirical formula is Ni(NH 3 ) 6 Cl 2

MM of Ni(NH 3 ) 6 Cl 2 = 231.83 g ⋅mol–1

17. Determine the theoretical yield of product in grams.

.

..

g NiCl H O

g NiCl H O

mol NiCl H O mol

1

700

237 71

(^61) 0 0294
6
(^226)
22
: # 22
:
: =

mols of NiCl 2 ⋅6 H 2 O = moles Ni(NH 3 ) 6 Cl 2 = 0.0294 mol

..
.

mol Ni NH Cl

mol Ni NH Cl

gNi NH Cl

1 gNi NH Cl

0 0294

1

231 83

682

3 6 2

3 6 2

3 6 2

# = 3 6 2

^

^

^

^

h

h

h

h

18. Determine the actual % yield of the impure product.

%

.

.

%.%

theoretical yield of product

actual yield of product

gNi NH Cl

gNi NH Cl

100

683

550

100 80 5

pure

impure

3 6 2

3 6 2

#

#

=

==

^

^

h

h

Laboratory Experiments