General Chemistry Moles Question

[in the rye]

This is using dimensional analysis, and I'm stuck on this question, but I'll show what I have so far.

The questions is:
If there is 3.34x10²² atoms in a CH₃OH sample. What is the mass of this sample, how many moles are there of this sample, and how many molecules are in this sample?

Initially, I decided to try to solve for moles, but I'm REALLY lost. What i did was is figured out the percent of oxygen, carbon, and hydrogen present, then multiplied the percentages by the number of atoms to figure how many of each element's atoms were present in the sample. My idea was to use this data to solve for how many moles of the sample I had, so that from there I could calculate the mass and molecules.

Here's the calculations I did, but it doesn't seem right to me, I am going to use E for x*10^p. So for example, using the above number 3.34x10²², I would type it as 3.34E22.

O atoms present = 1.67E22
H atoms present = 4.175E21
C atoms present = 1.2525E22

(1.67E22 O atoms)*(1 mole O/6.02E23 O atoms)*(1 mole CH₃OH/1 mole O)*(3 mole H/1 mole CH₃OH)*(6.02E23 atoms H/1 mole H)*(1 mole CH₃OH/4.175E21 H atoms)*(1 mole C/1 mole CH₃OH)*(6.02E23/1 mole C)*(1 mole CH₃OH/1.2525E22 atoms) = 5.77E-3.

I feel like this is completely wrong, but I am at a total loss. Any suggestions on how to approach this question? I don't necessarily want someone to solve it for me, just nudge me in the right direction. Thanks.

[edit]

Would what I'm trying to do be just as simple as dividing the amount of atoms by the number of atom species? So 3.34x10²²/6 to get the molecules?

 

[SteamKing]

This is using dimensional analysis, and I'm stuck on this question, but I'll show what I have so far.

The questions is:
If there is 3.34x10²² atoms in a CH₃OH sample. What is the mass of this sample, how many moles are there of this sample, and how many molecules are in this sample?

Initially, I decided to try to solve for moles, but I'm REALLY lost. What i did was is figured out the percent of oxygen, carbon, and hydrogen present, then multiplied the percentages by the number of atoms to figure how many of each element's atoms were present in the sample. My idea was to use this data to solve for how many moles of the sample I had, so that from there I could calculate the mass and molecules.

Here's the calculations I did, but it doesn't seem right to me, I am going to use E for x*10^p. So for example, using the above number 3.34x10²², I would type it as 3.34E22.

O atoms present = 1.67E22
H atoms present = 4.175E21
C atoms present = 1.2525E27

(1.67E22 O atoms)*(1 mole O/6.02E23 O atoms)*(1 mole CH₃OH/1 mole O)*(3 mole H/1 mole CH₃OH)*(6.02E23 atoms H/1 mole H)*(1 mole CH₃OH/4.175E21 H atoms)*(1 mole C/1 mole CH₃OH)*(6.02E23/1 mole C)*(1 mole CH₃OH/1.2525E27 atoms) = 5.77E-3.

I feel like this is completely wrong, but I am at a total loss. Any suggestions on how to approach this question? I don't necessarily want someone to solve it for me, just nudge me in the right direction. Thanks.

Remember, however which way you calculate the number of atoms of C, O, and H, the total number of atoms in the sample is 3.34×10²².

Is there any way you can add three numbers together, one of which is 1.2525×10²⁷, and arrive at a sum of 3.34×10²² ?
(Hint: Look at the exponent in the factor of 10.)

It's not at all clear how you calculated the numbers of atoms of the individual elements, C, O, and H. Just looking at the chemical formula for the substance, you should have the numbers of C : O : H atoms in the ratio of 1 : 1 : 4.

 

[Chestermiller]

Would what I'm trying to do be just as simple as dividing the amount of atoms by the number of atom species? So 3.34x10²²/6 to get the molecules?

Yes. This is definitely the place to start. From the result you get, how many moles are there (based on Avagodro's number)? What is the molecular weight of CH3OH? What is the mass of the sample?

Chet

 

[in the rye]

Okay, thank you. I wasn't taught that trick, so-to-speak, in lecture.. We, actually, never encountered an example like that. I was able to resolve the rest based on that. Thank you.

As for my percentages, I was mistaken on them. The 1.2525x20²⁷ was supposed to be 1.2525x20²². I couldn't read my writing haha. But, with my adjustment, they do add up. It didn't help me with my solution, though.

Will dividing the the total atoms by the number of chemical species always yield the total amount of molecules? Also, the method I used above with proper adjustments to the exponents gives me the right digits, but wrong decimal place -- where is my calculation error?

 

[Chestermiller]

Welcome to Physics Forums.

 

[in the rye]

Welcome to Physics Forums.

Thanks :)

I'm sure I'll be here the next few years, at least, ha. :D

 

[SteamKing]

Will dividing the the total atoms by the number of chemical species always yield the total amount of molecules?

Well, if one molecule of your compound contains 6 atoms, and there are 36 atoms total, how many molecules could you have?

Also, the method I used above with proper adjustments to the exponents gives me the right digits, but wrong decimal place -- where is my calculation error?

It doesn't look like your calculations give the proper ratio of C : O : H atoms. You should have equal numbers of C and O atoms, and 4 times the number of H atoms as either C or O atoms.

FYI, there's no need to calculate the number of H atoms in CH₃ separately from the number of H atoms in the OH. It's all the same H.

 

[in the rye]

The calculations I had were based on mass, which is why they'd vary in the manner they do. I guess I wasn't really finding how many atoms I had basing it off mass which is why I wasn't getting the correct answer. I found the mass of 1 mole of the molecule, then divided it by the mass of each atom present to get those numbers. So, I really don't know what I figured out, ha. I guess I only figured out the mass ratio, but nothing else.

 

[Borek]

I wasn't taught that trick

Hardly a trick, counting atoms is not different from counting fingers or marbles. When you have to convert between number of atoms and mass is where the intuition sometimes fail - at least for those seeing it done for the first time.

 

[in the rye]

Hardly a trick, counting atoms is not different from counting fingers or marbles. When you have to convert between number of atoms and mass is where the intuition sometimes fail - at least for those seeing it done for the first time.

Right, well, the way we were showed to calculate the amount of atoms in a molecule was by use of dimensional analysis. We calculate the number of each chemical species, then add them together. Thus something that takes one step, I was shown to use about 6 different ratios to calculate one atom, and then add it together. So prior to this method, it took literally about 21 different operations depending on the molecule. Therefore, making the multiplication method, to me, a "trick" -- even though the method behind it is very straightforward. In fact, prior to posting this, I had 3 other examples having me calculate the amount of atoms per molecule, so when I saw that I was given the amount of atoms first, it threw me completely off because following the steps backwards doesn't result in an answer you can use (which is why I was wondering if there is a more complicated way of solving this through D.A.). It makes me wonder why we weren't shown this to begin with. However, since we were using dimensional analysis, the latter part was more intuitive to me. I also have a way of over thinking things, which leads me to trouble as my aforementioned method will show.