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Offline knowledge

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Need help with formulas! percent abundance, proportions?
« on: May 24, 2008, 02:36:25 PM »
My chem class is going completely over my head and I am not grasping the formulas at all! I really want to be able to understand what the questions are asking so I can set up the formulas with ease.

What are the simplest equations to figure out these questions

The element copper has two naturally occurring isotopes with masses of 62.94 amu and 64.93 amu. What is the percent abundance of the first isotope?

A sample of Cl2O was composed of 7.8493 g of oxygen. How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine?

Is that a law of multiple proportion questions? I am getting lost with the Cl2O to CLO2 difference.

I appreciate anyone's responses!

Offline DevaDevil

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Re: Need help with formulas! percent abundance, proportions?
« Reply #1 on: May 24, 2008, 04:19:12 PM »
The element copper has two naturally occurring isotopes with masses of 62.94 amu and 64.93 amu. What is the percent abundance of the first isotope?

what is the resulting mass for the element copper?
The resulting mass (the one listed in tables as THE molar mass of copper) is made up by the sum of the fractions of the isotopes times the mass of the isotopes. Or in formula: MCu = Sum { xi * mi}, with xi mole fraction of the isotope and mi mass of the isotope. Remember that the sum of the mole fractions has to be 1
this will give you the mole fractions (mole fraction * 100% = percentage)

A sample of Cl2O was composed of 7.8493 g of oxygen. How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine?

Convert gram to mole and then to gram of chlorine, I have no idea why they mention Cl2O

Offline knowledge

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Re: Need help with formulas! percent abundance, proportions?
« Reply #2 on: May 24, 2008, 09:11:15 PM »
OK, I think this is where I struggle with formulas. I need a key that tells me what xi, mi, etc stands for.

I'm not even quite sure how to find the molar mass. Is it the atomic weight times the number of atoms present in the element? In this case there is only one. Therefor, the molar mass of Cu would just be its atomic weight of 63.546? Correct?

The element copper has two naturally occurring isotopes with masses of 62.94 amu and 64.93 amu. What is the percent abundance of the first isotope?

So I do 63.546 = 1 (??? * 100)

I'm just really confused :/


A sample of Cl2O was composed of 7.8493 g of oxygen. How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine?


5.3846 g of Cl / 35.453 g/mol ...is that how I set it up?

Offline Astrokel

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Re: Need help with formulas! percent abundance, proportions?
« Reply #3 on: May 24, 2008, 11:06:02 PM »
The element copper has two naturally occurring isotopes with masses of 62.94 amu and 64.93 amu. What is the percent abundance of the first isotope?

So I do 63.546 = 1 (??? * 100)

You are almost there!

Copper molar mass is made up of two of its isotopes, Cu-62.94 and Cu-64.93

As DevaDevil has said, the x is the value you want to solve.

If you let x% be the percentage abundance of Cu-62.94 in Cu
then, the percentage abundance of Cu-64.93 in Cu is (x-100)%

together the made up the molar mass of copper in the periodic table.

Adjust your equation on the right hand side


A sample of Cl2O was composed of 7.8493 g of oxygen. How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine?


5.3846 g of Cl / 35.453 g/mol ...is that how I set it up?

You are right, now find the moles of Cl2 and O2?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline knowledge

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Re: Need help with formulas! percent abundance, proportions?
« Reply #4 on: May 24, 2008, 11:35:54 PM »
The element copper has two naturally occurring isotopes with masses of 62.94 amu and 64.93 amu. What is the percent abundance of the first isotope?

63.546 = 127.87 (x% * 62.94)

Ugh, I don't know why I'm so dense :( I'm really sorry I'm not comprehending this. I'm a bio major, so I know I have a couple more years of chem to come and it's very frustrating that I can't get down such a basic concept.

So far all I've figured out is how to check the right answer. I would do something like (x1 * 62.94) + (x2 * 64.93) and see if it equates to the molar mass, correct? X1 being the percentage for isotope 1 and x2 being the percentage for isotope 2.



A sample of Cl2O was composed of 7.8493 g of oxygen. How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine?

Cl: 35.453
O: 15.9994

How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine

Cl    5.3846 g of Cl/35.453 g/mol = .1518799537

Cl2  10.7692 g of Cl2/35.453 g/mol = .3037599075

O   7.8493 g of O/15.994 g/mol = .490765287

O2   15.6989 g of O2/15.9994 g/mol = .9812180457

Eek, now what?

Offline Astrokel

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Re: Need help with formulas! percent abundance, proportions?
« Reply #5 on: May 25, 2008, 09:18:02 AM »
So far all I've figured out is how to check the right answer. I would do something like (x1 * 62.94) + (x2 * 64.93) and see if it equates to the molar mass, correct? X1 being the percentage for isotope 1 and x2 being the percentage for isotope 2.

Yes!! but you can change x2 to (100-x1)% so you have one unknown and can solve for x1

A sample of Cl2O was composed of 7.8493 g of oxygen. How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine?

Cl: 35.453
O: 15.9994

How many grams of oxygen would there be in a sample of ClO2 that contains  5.3846 g of chlorine

Cl    5.3846 g of Cl/35.453 g/mol = .1518799537

Cl2  10.7692 g of Cl2/35.453 g/mol = .3037599075

O   7.8493 g of O/15.994 g/mol = .490765287

O2   15.6989 g of O2/15.9994 g/mol = .9812180457

Eek, now what?

no, you can't just simply multiply the mass by 2 because of O2

Oxygen exists as diatomic particles, therefore, the molar mass of oxygen is always 15.9994 x 2

so moles of O2 = 7.8493/(15.9994x2)

same applies to chlorine which exists as Cl2.

Once you get this, it is just proportionality. My guess is this question is referring to ClO2 and not Cl2O. Check it with the answer sheet.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

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