Question

Cobalt forms two different oxides. The mass percentages of cobalt in the pure oxides are measured as being 78.65% and 71.06%.

A. Calculate the formula of each oxide using the percent by mass, then name them.

B. Calculate the mole fraction of cobalt and mass percentage of cobalt in a mixture containing 4.00kg of the first oxide and 6.50kg of the second oxide.

C. A sample of mixed cobalt oxides with a mass of 5.283kg is completely reduced until the total mass is 4.025kg of pure cobalt metal. What percentage of cobalt ions in the original sample had an oxidation state of +2? (by mole)

I need a full explanation of ALL the steps including how to calculate molar ratio in letter A. Please do not skip any steps. I am very confused

Answer 1

A)

Sample (1);

% of Cobalt in first oxide = 78.65%   

% of oxygen in first oxide = 100 - 78.65 = 21.35%

Moles of Cobalt = percentage/molar mass = 78.65/58.933 = 1.335

Moles of O = 21.35/16 = 1.334

Mole ratio;

Co : O = 1.335 : 1.334

We can get the simplest mole ratio by diving smallest mole number in calculation. Here smallest mole number is 1.334. So, dividing each mole by same number i.e.

Co : O = (1.335/1.334) : (1.334/1.334)

Co : O = 1 : 1

So, formula is Co1O1 OR simply = CoO .....Answer

Sample (2);

% of Cobalt in second oxide = 71.06%  

% of O = 100 - 71.06 = 28.94 %

Moles of Cobalt = 71.06/58.93 = 1.206

Moles of O = 28.94/16 = 1.80875

Mole ratio;

Co : O = 1.206 : 1.80875

Simplest mole ratio;

Co : O = (1.206/1.206) : (1.80875/1.206)

Co : O = 1 : 1.5

To get the integer number, multiplying moles by smallest possible number. So, number will be 2.

Co : O = 2*1 : 2*1.5

Co : O = 2 : 3

Formula is Co2O3 ......Answer

B)

Mass of Co in first oxide = 4*78.65/100 = 3.146 Kg = 3146 grams

Moles of Co in first oxide (n1) = 3146/58.93 = 53.3854

Mass of O in first oxide = 4000 - 3146 = 854 grams

Moles of O in first oxide (n2) = 854/16 = 53.375

Mass of Co in second oxide = 6.50*71.06/100 = 4.6189 Kg = 4618.9 grams

Moles of Co in second oxide (n3) = 4618.9/58.93 = 78.3794

Mass of O in second oxide = 6500 - 4618.9 = 1881.1 grams

Moles of O in second sample (n4) = 1881.1/16 = 117.5687

Total mass of cobalt = 3146 + 4618.9 = 7764.9 g

Total mass of both oxides = 4000 + 6500 = 10500 g

% by mass of Co = 7764.9*100/10500 = 73.95 % ....Answer

Total moles of Co = 53.3854 + 78.3794 = 131.7648

Total moles in sample = n1 + n2 + n3 + n4 = 53.3854 + 53.375 + 78.3794 + 117.5687 = 302.7085

Mole fraction of Co = 131.7648 / 302.7085 = 0.4353 ....Answer

Let me know if any doubts/answer is not matching.