Calulating Atomic Percentages given moles and weights of compounds

In summary, the conversation discusses a problem with calculating atomic percentages in a paper's recipe for a slurry of ZnS with Cu and NH4Cl solutions. The problem arises from the use of strange units and notations, making it difficult to determine the correct ratios of atoms in the mixture. The conversation concludes that the issue lies with the original recipe's use of unconventional units and not with the person attempting to solve it.
  • #1

Homework Statement

This is the statement out of a paper:

Slurry 10 grams of ZnS with 3mL of .2N Cu(CH3COO)2 solution (corresponding to 0.6 atom % Cu) and 1.7 mL of .2N NH4Cl solution (corresponding to 0.34 atom % Cl)

So my question is how did they calculate the .6 at. % and .34 at. % from the information given

Homework Equations

The only equations I could find were for mole fractions and converting from weight percent to atomic percent, but by implementing these techniques I was still not able to reproduce the stated percentages.

The Attempt at a Solution

I attempted to solve this using excel by calculating wt. percent then converting it to atomic percentages and my numbers are smaller than the ones given.

I have also done mole fractions and that doesn't seem to be the right way either

My at. % of Cu was only .145 and for Cl it was .165
I accounted for all species present

I will appreciate any guidance for I have been stuck on this for days
Thank You
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  • #2
There is a problem with a very strange concept of atoms % that has been used in the paper. It is easy to see where they get their figures from: 3 mL of 0.2 N makes 0.6 meq of copper ion, while 1.7 mL of 0.2 N makes 0.34 mmol of chloride ion that is being slurried with the 10 g of zinc powder.

But atoms % should mean a ratio of atoms per 100 atoms. 100 atoms of what? there is no clear relationship between mmol and 100 mmol of anything in particular. Ant there is also the problem of use of an outdated and ambiguous concentration unit. Because copper(II) is formally a doubly charged ion, is an equivalent equal to 1 mol or 0.5 mol. If we are looking to reduce copper(II) to metallic copper, it should be 0.5 mol, but reading between the lines I suspect that the recipe you are looking at is for formation of some copper(I) chloro- or ammino- complex, and that we can read meq as mmol.

The problem does not seem to rest with you, it is with some very strange units and notations in the original recipe.

In the recipe as you have presented it, we have 0.6 mmol of copper and 0.34 mmol of chloride in a slurry with 10 g of Zn powder -- obviously a vast excess. atom % is a weird fiction, the way the authors have used it.

1. How do you calculate atomic percentages given moles and weights of compounds?

To calculate atomic percentages, first determine the molar mass of each element in the compound. Then, divide the molar mass of each element by the total molar mass of the compound. This will give you the decimal equivalent of the atomic percentage for each element. Finally, multiply each decimal by 100 to get the atomic percentage as a percentage value.

2. What is the purpose of calculating atomic percentages?

Calculating atomic percentages allows us to determine the relative abundance of each element in a compound. This is important in understanding the chemical composition and properties of a substance.

3. Can atomic percentages be greater than 100%?

No, atomic percentages cannot be greater than 100%. This is because the total atomic percentage of all elements in a compound must add up to 100%.

4. How can I use atomic percentages to find the empirical formula of a compound?

To find the empirical formula of a compound, you can use the atomic percentages to determine the ratio of elements present. This ratio can then be used to write the empirical formula by dividing each element's atomic percentage by its atomic weight and simplifying the resulting ratio.

5. Is it necessary to know the molecular formula of a compound to calculate atomic percentages?

No, it is not necessary to know the molecular formula of a compound to calculate atomic percentages. As long as you have the moles and weights of each element in the compound, you can calculate the atomic percentages using the method mentioned in the first question.

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