## E29 Part D

Aj1156
Posts: 34
Joined: Fri Sep 28, 2018 12:22 am

### E29 Part D

A chemist measured out 8.61 g of copper (II) chloride tetrahydrate CuCl2 4H20 (d) What fraction of the total mass of the sample is due to oxygen?

What would be the proper conversion to find the fraction of the mass due to oxygen in the total sample?

I tried 8.61g CuCl2 4H20 / 206.45 x 1m/4m x 16g but didn't get the correct answer and I need some help.

Samantha Ito 2E
Posts: 64
Joined: Fri Sep 28, 2018 12:29 am

### Re: E29 Part D

Since there are 4 moles of oxygen, you just multiply the molar mass of the oxygen by 4. Then, you divide this value by the molar mass of the entire sample.

Gillian Ward 1F
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

### Re: E29 Part D

I believe it is like doing percent composition without multiplying by 100 to get a percent. (percent composition is part mass/whole mass x 100)
In this case, part mass/whole mass = fraction
The amount of oxygen is (16.00 x 4), the 16.00 being the molar mass and the four comes from the number of moles of oxygen in the sample. That is the "part mass." To find the "whole mass" you would find the mass of the entire sample which ends up being 206.532 g/mol.
64.00 divided by 206.532 is going to end up being .3099 which is the fraction of the total mass that is oxygen.