## Correct Answer for Audio-Visual Focus-Topics Empirical and Molecular Formulas Pre-Module Assessment

Jasmine Reddy DIS 1E
Posts: 34
Joined: Fri Sep 28, 2018 12:17 am

### Correct Answer for Audio-Visual Focus-Topics Empirical and Molecular Formulas Pre-Module Assessment

What is the correct answer for #14?
#14.
339.20 g of a cobalt metal is reacted with fluorine gas to produce 996.08 g of a compound. What is the empirical formula of the new compound?

How can we find the correct answers for the pre/post assessments?

Swetha Ampabathina1I
Posts: 64
Joined: Fri Sep 28, 2018 12:18 am

### Re: Correct Answer for Audio-Visual Focus-Topics Empirical and Molecular Formulas Pre-Module Assessment

The pre/post assessment only grades your work. It only tells you if you got it correct or not

Swetha Ampabathina1I
Posts: 64
Joined: Fri Sep 28, 2018 12:18 am

### Re: Correct Answer for Audio-Visual Focus-Topics Empirical and Molecular Formulas Pre-Module Assessment

The correct answer for that question was CoF6

Laura Gong 3H
Posts: 89
Joined: Fri Sep 28, 2018 12:26 am
Been upvoted: 1 time

### Re: Correct Answer for Audio-Visual Focus-Topics Empirical and Molecular Formulas Pre-Module Assessment

To answer your second question first, I noticed that for the pre-assessment it doesn't tell you how many you got correct/wrong.
However, after submitting the post-assessment, they do tell you how many you got correct/wrong.

For #14, you need to use the idea of conservation of mass to find the mass of fluorine gas.
So 996.08g-339.20g=656.88g of fluorine gas (F2).

From there, you can determine the mass percent composition of the compound by dividing the mass of cobalt and the mass of F2 by the mass of the compound formed.
Cobalt:339.20g/996.08g=.3405 or 34.05%
F2:656.88g/996.08g=.6595 or 65.95%

Now assume that the total mass of the sample formed is 100g, that way you can easily translate the mass % into grams of Cobalt and F2.

Using the molar mass of Cobalt (58.93g/mol) and of F2 (38.00g/mol) and stoichiometric ratios, you can find the number of mols of Co and F in the compound.

34.05g Co/58.93g*mol^-1 Co=.5778 mol Co
65.95g F2/38.00g*mol^-1 F2 x 2 mol F/1 mol F2=3.471 mol F

Then divide the values above by the smallest value (.5778) to get a ratio of 1 Co: 6 F, so the empirical formula is CoF6.