Question

16. You synthesized 3.204 g of Ni(NH3)nCl2. The mass percent of NH3 and Ni2+ in the synthesized Ni(NH3)nCl2 were 44.1% and 25.3%, respectively. Molar mass of Ni2+ , NH3 and Cl2 are 58.69 g/mol, 17.04 g/mol, and 70.90 g/mol, respectively.

a) (1 pts) Calculate the mass percent of Cl2.

b) (1 pts) Calculate the masses of Ni2+ , NH3 and Cl2.

c) (1 pts) Calculate the number of moles of Ni2+ .

d) (1 pts) Calculate the number of moles of NH3.

e) (1 pts) Calculate the number of moles of Cl2.

f) (2 pts) What is the empirical formula for this compound?

Answer 1

1)

The percentage of NH₃ as 44.1 % and percentage of Ni²⁺ as 25.3% in Ni(NH₃)_nCl₂

the coordination number of Ni²⁺ is 4 so the value of n = 4 in Ni(NH₃)_nCl₂

Mass percent must add up to 100.

So we have,

44.1 + 25.3 + x = 100

69.3 + x = 100

x =30.6

So percentage of Cl2 is 30.6

2)

Masses of Ni²⁺ , NH₃ and Cl₂

Given

Molar mass of Ni2+ as 58.69 g/mol,

NH3 as 17.04 g/mol,

Cl2 as 70.90 g/mol.

Mass percent = [ Molar mass of element / Total molecular mass of compound ] x 100

the molecular mass of Ni(NH₃)₄Cl₂ is 58.69 + 17.04 x 4 + 70.90 = 146.63 g/mol

Outof 3.204 g of Ni(NH₃)₄Cl₂  44.1% is NH3 = 3.204 x (44.1/100) = 1.413 g

25.3% is  Ni2+ = 3.204 x (25.3 / 100) = 0.811 g

30.6% is Cl2 = 3.304 x (30.6 / 100) = 0.980 g

c)

Number of moles of Ni2+

= 0.811 g / 58.69  g/mol

= 0.0138 mol

d)

Number of moles of NH3

= 1.413 g / 17.04 g/mol

= 0.0829 mol

e)

Number of moles of Cl2

0.980 g / 70.90 g/mol

= 0.0138 mol