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Chemistry help

Guest Anonymous


I know this isn't what this is for, but I have this one Chemistry question I can't understand. Google is being unhelpful and it's for a class I'm taking online so the teaching is all through prerecorded videos, so I can't ask a teacher or something. Hopefully someone here has a decent understanding of 11th grade chemistry? This is the question, I would really appreciate an answer:

A sample of pure acetic acid, CH3COOH, contains 1.40 × 1023 carbon atoms. How many moles of acetic acid are in the sample?

The part I'm specifically confused on is how you get from there being a certain amount of CARBON atoms to there being an amount of TOTAL atoms. From there I can find the mole pretty easily

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Bro. This is Arocalypse, not brainly lol.

but I can give you an answer with Bing Chat.

Bing: To find the number of atoms in a molecule, you need to determine the number of each type of element and the subscript after the element symbol describes the number of atoms for that element. If you have a sample that contains only atoms of a particular element, weigh the sample in grams and divide by the atomic weight of the element. The quotient tells you the number of moles. Multiply that by Avogadro’s number, and you’ll find out how many atoms the sample contains 1.

In your case, you have 1.40 × 10^23 carbon atoms in a sample of pure acetic acid. Acetic acid has two carbon atoms, four hydrogen atoms, and two oxygen atoms. Therefore, you can calculate the total number of atoms in acetic acid by multiplying the number of carbon atoms by two and adding it to the sum of hydrogen and oxygen atoms. That is:

Total number of atoms = (2 × 1.40 × 10^23) + (4 × 1.40 × 10^23) + (2 × 1.40 × 10^23) = 14 × 10^23

Now that you know the total number of atoms in acetic acid, you can calculate the number of moles by dividing it by Avogadro’s number (6.022 × 10^23). That is:

Number of moles = (14 × 10^23) / (6.022 × 10^23) = 2.33 moles

I hope this helps! Let me know if you have any other questions.


Edited by SwiftySpeedy
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