Question

a 0.2076 g sample of an oxide of cobalt on analysis was found to contain 0.1476g of cobalt.Calculate the empirical formula of the oxide

Answer 1

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◆ Answer -

Empirical formula of given cobalt oxide = Co2O3

◆ Explanation -

Percentage of cobalt in the oxide -

w/w % = 0.1476 / 0.2076

w/w % = 0.7110 = 71.10 %

That is 100 g of oxide contains 71 g Co & 29 g O.

No of moles of Co -

n(Co) = 71/59 = 1.2

No of moles of O -

n(O) = 29/16 = 1.8

Simplest ratio of moles of Co & O can be 1.2:1.8 = 2:3 .

Hence, empirical formula of given cobalt oxide is Co2O3.

Answer 2

Answer:

The Emphirical Formula will be C0203.

Explanation:

total gram of cobalt oxide = 0.2076g.

and total gram of cobalt is =0.1476g.

so the total gram of oxigen will be = 0.2076g -- 0.1476g = 0.06g.

so oxigen = 0.06g

now relative number of atoms in cobalt = 0.1476g/59g = 0.0025g.

similarly for oxygen = 0.06g/16g = 0.00375g.

now for simplest ratio of cobalt = 0.0025g/0.0025g = 1.

similarly, in oxygen = 0.00375g/0.0025g = 1.49.

finally, for simplest whole number ratio = 1*2 = 2.

and for oxygen = 1.49*2 = 3.

so by above result the empirical formula is = CO2O3.

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