The refining of nickel begins with the conversion of solid metallic nickel to ga

Question

The refining of nickel begins with the conversion of solid metallic nickel to gaseous tetracarbonyl nickel by reaction with carbon monoxide.

(a) Tetracarbonyl nickel is 34.38% Ni, 28.14% C, and 37.48% O by mass. What is the empirical formula of

tetracarbonyl nickel?

(b) A sample of tetracarbonyl nickel at 745 torr and 100. °C has a density of 5.47 g/L. What is the molecular formula

of nickel tetracarbonyl?

(c) Which would be faster, a molecule of tetracarbonyl nickel gas at STP or a molecule of fluorine gas at STP. How much faster would that molecule be?

(d) Write a balanced chemical reaction showing the formation of tetracarbonyl nickel from metallic nickel and carbon monoxide.

(e) A 5.000 g sample of 83.5% pure nickel ore is reacted with carbon monoxide according to the equation in (d). Assume the impurities in the ore are inert to carbon monoxide. If the percent yield of the reaction is 77.1%, how many mL of tetracarbonyl nickel will be produced at STP?

Explanation / Answer

a) Tetracarbonyl nickel has 34.38% Ni, 28.14% C and 37.48% O by mass. Therefore, 100.0 g tetracarbonly nickel has 34.38 g Ni, 28.14 g C and 37.48 g O.

The atomic masses are: Ni = 58.6934 g/mol; C = 12.01 07 g/mol and O = 15.9994 g/mol.

Mole(s) of Ni corresponding to 34.38 g Ni = (38.34 g)/(58.6934 g/mol) = 0.6423 mole.

Mole(s) of C corresponding to 28.14 g C = (28.14 g)/(12.0107 g/mol) = 2.3429 mole.

Mole(s) of O corresponding to 37.48 g O = (37.48 g)/(15.9994 g/mol) = 2.3426 mole.

The ratio of the mole(s) of Ni:C:O = 0.6423:2.3429:2.3426 = (0.6423/06423):(2.3429/06423):(2.3426/06423) = 1:3.6476:3.6472.

Approximate the above ratio to a simplest ratio (small whole integers) as 1:4:4; therefore, tetracarbonyl nickel will contain 1 atom of Ni, 4 atoms of C and 4 atoms of O. The empirical formula is Ni(CO)4 (ans).

b) Assume tetracarbonyl nickel to behave as an ideal gas and apply the ideal gas law. We have, pressure P = 745 torr = (745 torr)*(1 atm/760 torr) = 0.9803 atm; temperature T = 100°C = (100 + 273) K = 373 K and density d = 5.47 g/L.

Let M be the molecular weight of tetracarbonyl nickel.

We have

P = d/M*R*T

Plug in values and obtain

0.9803 atm = (5.47 g/L)/M*(0.082 L-atm/mol.K)*(373 K)

====> M = (5.47 g/L)*(0.082 L-atm/mol.K)*(373 K)/(0.9803 atm)

====> M = 170.6675 g/mol.

Find out the empirical formula mass as EM = (1*58.6934 + 4*12.01017 + 4*15.9994) g/mol = 170.7338 g/mol.

Use the relation

(EM)n = M

====> (170.7338 g/mol)*n = 170.6675 g/mol

====> n = 0.9996 1.000

The molecular formula of tetracarbonyl nickel is Ni(CO)4 (ans).

c) Use Graham’s law of diffusion as below

R1/R2 = M2/M1

where R1 and R2 are the rates of diffusion of fluorine and tetracarbonyl nickel gases and M1 and M2 are the molecular weights. We know that M2 = 170.6675 g/mol (as obtained from part b above) and M1 = 2*18.9994 g/mol = 37.9988 g/mol.

Plug in values and obtain

R1/R2 = (170.6675 g/mol)/(37.9988 g/mol) = 4.49139 = 2.1192 2.12

====> R1 = 2.12*R2

The rate of diffusion of fluorine gas is 2.12 times faster than that of tetracarbonyl nickel gas (ans).

d) The balanced chemical equation for the formation of tetracarbonyl nickel is

Ni (s) + 4 CO (g) -------> Ni(CO)4 (g)

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