I have a chemistry assignment that I just can't seem to figure out, my brain is too tired at this time of day (it's nighttime here, I need some sleep).
So, let's say that I have a sample of an element. I know how much its mass (0.815 grams) and I know how many atoms are in the sample (4.16x10^21). How would I figure out or calculate what element this is? I've tried to look through my notes and some previous lecture presentations, but haven't found a formula that I can use :/
So if anyone knows how to solve problems like that, I would appreciate some help :)
I have a chemistry assignment that I just can't seem to figure out, my brain is too tired at this time of day (it's nighttime here, I need some sleep).
So, let's say that I have a sample of an element. I know how much its mass (0.815 grams) and I know how many atoms are in the sample (4.16x10^21). How would I figure out or calculate what element this is? I've tried to look through my notes and some previous lecture presentations, but haven't found a formula that I can use :/
So if anyone knows how to solve problems like that, I would appreciate some help :)
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i don't really know anything about chemistry because all of my knowledge is reserved for pointless gatling gun fun facts.
my terrible advice? i think you could start some multi-billion world-class science project to discover the exact weight of a single atom of each element. then by using said data, you can then divide 0.815 by 4.16x10^21 and then compare it to the results of the exact atom weights and tada
wait do we already know what atoms weigh?
im too tired for this. imma go back to watching cannon vids. peace?
i don't really know anything about chemistry because all of my knowledge is reserved for pointless gatling gun fun facts.
my terrible advice? i think you could start some multi-billion world-class science project to discover the exact weight of a single atom of each element. then by using said data, you can then divide 0.815 by 4.16x10^21 and then compare it to the results of the exact atom weights and tada
wait do we already know what atoms weigh?
im too tired for this. imma go back to watching cannon vids. peace?
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Moell I've never had an assignment like that & I don't know what level yr taking, so this might be useless advice
but if you know the weight & numbers of atoms, can't you first calculate the weight of one atom & using that calculate the weight of one mole? Then look for elements w a similar molar mass.
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Moell I've never had an assignment like that & I don't know what level yr taking, so this might be useless advice
but if you know the weight & numbers of atoms, can't you first calculate the weight of one atom & using that calculate the weight of one mole? Then look for elements w a similar molar mass.
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Moell
So I did what any self-respecting adult who doesn't understand science does: I googled it.
Someone on yahoo answers actually asked the same thing and they were given a break down on how to figure it out
here
The second answer provided in the link gives you the actual answer. If you'd like to try out the formula first and then check your answer to theirs I'd recommend not scrolling down too much.
And if you don't feel like clicking the link I'll just quote both answers
Quote:
First answer:
Now 0.815 / X = moles from above. X is the molecular mass of the unknown. Once you have that, look it up in the periodic table
Second answers sans bottom line:
The other answer was correct. Avagadro's number is 6.02x10^23. If you divide 4.6x10^21 by 6.02 x10^23 you have ~0.00764 moles of the element. 1 mole of an element weighs its atomic wieght in grams
@
Moell
So I did what any self-respecting adult who doesn't understand science does: I googled it.
Someone on yahoo answers actually asked the same thing and they were given a break down on how to figure it out
here
The second answer provided in the link gives you the actual answer. If you'd like to try out the formula first and then check your answer to theirs I'd recommend not scrolling down too much.
And if you don't feel like clicking the link I'll just quote both answers
Quote:
First answer:
Now 0.815 / X = moles from above. X is the molecular mass of the unknown. Once you have that, look it up in the periodic table
Second answers sans bottom line:
The other answer was correct. Avagadro's number is 6.02x10^23. If you divide 4.6x10^21 by 6.02 x10^23 you have ~0.00764 moles of the element. 1 mole of an element weighs its atomic wieght in grams
I do physics, not chemistry, so please correct me if I'm wrong...
A mole of a substance can be defined as the number of atoms divided by Avogadro's number, so
mol = 4.16x10^21/6.022x10^23 = 0.0069 moles
So we have 0.0069 moles of some mystery substance.
We can use that to find the atomic mass. There is a formula for converting between number of moles and atomic mass (
this website explains it pretty well)
So atomic mass = mass/moles = 0.815/0.0069 = 118.
Looking at
this table, it looks like that's Tin (Sn). Again, take it with a grain of salt because I haven't taken a single chemistry class :D
Edit: looking at the above answer, looks like it's actually Palladium! ignore this lol
Edit2: it's different because their number of atoms is 4.6, not 4.16. So Tin should be right.
I do physics, not chemistry, so please correct me if I'm wrong...
A mole of a substance can be defined as the number of atoms divided by Avogadro's number, so
mol = 4.16x10^21/6.022x10^23 = 0.0069 moles
So we have 0.0069 moles of some mystery substance.
We can use that to find the atomic mass. There is a formula for converting between number of moles and atomic mass (
this website explains it pretty well)
So atomic mass = mass/moles = 0.815/0.0069 = 118.
Looking at
this table, it looks like that's Tin (Sn). Again, take it with a grain of salt because I haven't taken a single chemistry class :D
Edit: looking at the above answer, looks like it's actually Palladium! ignore this lol
Edit2: it's different because their number of atoms is 4.6, not 4.16. So Tin should be right.
Thanks a lot, everyone! I managed to find the right answer in the end, my brain was just tired to remember how to do this correctly. Even when I actually was doing it right I thought I was doing it wrong xD
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Coordinator: That might have worked. Haven't tried to do it that way, and I'm not sure if my calculator can handle such small numbers... But if I'm stuck in the future and can't remember how to do it an easier way, that might just work too :)
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ellobomalo: Would you look at that, exactly the same as my question. Maybe this yahoo-person and I use the same text book xD Found the right answer, so thank you so much!
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Jebii: But if you do physics, you're good at math and breaking down formulas so I'd say you're still qualified to help out :D
I actually switched the digits in my post, it was supposed to be 4.61 and not 4.16... So you're absolutely right, the answer was palladium!
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Longneckllama: Well, that would be quite the project xD Too bad I would've had to handle in the assignment way before that, but at least no-one could say I didn't put any effort into it.
Thanks a lot, everyone! I managed to find the right answer in the end, my brain was just tired to remember how to do this correctly. Even when I actually was doing it right I thought I was doing it wrong xD
@
Coordinator: That might have worked. Haven't tried to do it that way, and I'm not sure if my calculator can handle such small numbers... But if I'm stuck in the future and can't remember how to do it an easier way, that might just work too :)
@
ellobomalo: Would you look at that, exactly the same as my question. Maybe this yahoo-person and I use the same text book xD Found the right answer, so thank you so much!
@
Jebii: But if you do physics, you're good at math and breaking down formulas so I'd say you're still qualified to help out :D
I actually switched the digits in my post, it was supposed to be 4.61 and not 4.16... So you're absolutely right, the answer was palladium!
@
Longneckllama: Well, that would be quite the project xD Too bad I would've had to handle in the assignment way before that, but at least no-one could say I didn't put any effort into it.
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