E 29 part c

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venning-1J
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Joined: Fri Sep 29, 2017 7:04 am

E 29 part c

Postby venning-1J » Tue Oct 03, 2017 11:26 pm

Im having trouble figuring out part c.

E.29 A chemist measured out 8.61 g of copper(II) chloride tetrahydrate, CuCl2 4H2O.
(a) How many moles of CuCl2 4H2O were measured out? 0.0417 mol
(b) How many moles of Cl ions are present in the sample? .0834 mol
[i](c) How many H2O molecules are present in the sample? [/i]
(d) What fraction of the total mass of the sample was due to oxygen? 0.31

I attempted to multiply 0.0417 x 4 x 6.0022x10^23 but I am not getting the answer, which the book states is 1.00 x 10^23.

RahilVaknalliDis3A
Posts: 17
Joined: Fri Sep 29, 2017 7:06 am

Re: E 29 part c

Postby RahilVaknalliDis3A » Tue Oct 03, 2017 11:32 pm

You are on the right track. I believe there is a simple calculation error since you are multiplying the number of moles of water molecules with the wrong number. Avogradro's constant is 6.022 x 1023 and not 6.0022 x 1023

Christy Zhao 1H
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Joined: Fri Sep 29, 2017 7:07 am

Re: E 29 part c

Postby Christy Zhao 1H » Wed Oct 04, 2017 11:12 pm

I'm confused with part d. How would I go about to find what fraction of the total mass of the sample was due to oxygen?

jorineraymundo2h
Posts: 3
Joined: Sat Jul 22, 2017 3:01 am

Re: E 29 part c

Postby jorineraymundo2h » Thu Oct 05, 2017 12:36 pm

It's easier to answer the question if you think of the problem's pathway.

Mass (CuCl2 4H2O) -> Moles (CuCl2 4H20) -> Moles (4H20) -> Molecules

(8.16g/206.53gmol^(-1)) x 4molesH20 x (6.0221x10^(23)) = 1.00 x 10^(23)

I hope this helps!

Aman Sankineni 2L
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Joined: Fri Aug 30, 2019 12:17 am

Re: E 29 part c

Postby Aman Sankineni 2L » Tue Oct 01, 2019 11:32 pm

For part d, you must figure out the mass of oxygen as part of the formula. Since there is 4 oxygen atoms, the mass of oxygen is 16x4=64. By dividing that by the total mass of the formula, 206.446, you can figure out what fraction of the total mass was due to oxygen. 64/206.446=.310008 or 31%.

Amy Kumar 1I
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Re: E 29 part c

Postby Amy Kumar 1I » Thu Oct 03, 2019 10:16 am

Aman Sankineni 3E wrote:For part d, you must figure out the mass of oxygen as part of the formula. Since there is 4 oxygen atoms, the mass of oxygen is 16x4=64. By dividing that by the total mass of the formula, 206.446, you can figure out what fraction of the total mass was due to oxygen. 64/206.446=.310008 or 31%.


Do you not have to take into account that there were 8.61 g of copper(II) chloride tetrahydrate initially?

Kassidy Ford 1I
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Re: E 29 part c

Postby Kassidy Ford 1I » Thu Oct 03, 2019 10:47 am

Because it is just the percentage of mass of oxygen, then the original mass of 8.61g copper(II) chloride tetrahydrate doesn't matter. No matter how many grams of the sample you have the percentage mass of oxygen should always be the same because of the molar ratios. Correct me if I'm wrong!

Kassidy Ford 1I
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Re: E 29 part c

Postby Kassidy Ford 1I » Thu Oct 03, 2019 10:48 am

Because it is just the percentage of mass of oxygen, then the original mass of 8.61g copper(II) chloride tetrahydrate doesn't matter. No matter how many grams of the sample you have the percentage mass of oxygen should always be the same because of the molar ratios. Correct me if I'm wrong!

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

Re: E 29 part c

Postby Aman Sankineni 2L » Thu Oct 03, 2019 11:06 pm

Kassidy Ford 1J wrote:Because it is just the percentage of mass of oxygen, then the original mass of 8.61g copper(II) chloride tetrahydrate doesn't matter. No matter how many grams of the sample you have the percentage mass of oxygen should always be the same because of the molar ratios. Correct me if I'm wrong!

Yup, this is right. Molar ratios always remain the same. No matter the mass of the copper(II) chloride tetrahydrate, the percent composition of oxygen remains the same.


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