anonymous
  • anonymous
How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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zepp
  • zepp
Ethanol: \(\large \text{C}_2\text{H}_6\text{O}\)
zepp
  • zepp
So first question, how many moles does 2.04 x 10^20 molecules represent?
anonymous
  • anonymous
its not CH(Sub 6)O?

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anonymous
  • anonymous
why did u give Carbon2?
zepp
  • zepp
Because that's the formula for Ethanol
zepp
  • zepp
It's a 2-carbon alcohol.
anonymous
  • anonymous
\[(6.02 \times 10^{23})(2.04 \times 10^{20})?\]
anonymous
  • anonymous
oh ok
zepp
  • zepp
That's incorrect; \[\large 1 \text{ mole}= 6.02×10^{23}\]
zepp
  • zepp
\[\large ? \text{ mole}=2.04×10^{20}\]
zepp
  • zepp
That's a proportion and you should have this: \(\LARGE \frac{2.04×10^{20}}{6.02×10^{23}}\)
anonymous
  • anonymous
oh so its divide instead of multiply?
zepp
  • zepp
Yes, that's algebraic manipulation.
zepp
  • zepp
Once you get that, you'll get the number of moles this quantity of molecules represent, are afterward, you may convert that easily to its mass through the molar mass.
anonymous
  • anonymous
|dw:1349138080842:dw|
zepp
  • zepp
Why 1000?
anonymous
  • anonymous
for milligrams?
zepp
  • zepp
Our units are in moles...
zepp
  • zepp
\[\large \frac{1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=\frac{x \text{ mole}}{2.04 * 10^{20}\text{ atoms}}\]Multiply both sides by 2.04*10^20\[\large \frac{2.04*10^{20}\text{ atoms}*1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=x \text{ mole}\]
zepp
  • zepp
atoms and atoms simplify out, leaving the mole on both sides.\[\large \frac{2.04*10^{20}}{6.02*10^{23}}\text{mole}=x \text{ mole}\]
anonymous
  • anonymous
ok i see
anonymous
  • anonymous
i got .0003 correct?
zepp
  • zepp
You'll get as answer: \(\large x\text{ moles}\) Now you may proceed to transform this using \(\LARGE n=\frac{m}{M}\), where \(\large \text{n}\) is the number of moles, \(\large \text{m}\) is the mass in \(\large \textbf{grams}\) (so you'll need to retransform it into milligrams as you're asked to do) and \(\large \text{M}\) is the molar mass.
anonymous
  • anonymous
Molar mass is 46.07?
zepp
  • zepp
Yes
anonymous
  • anonymous
well for ethanol it self right?
anonymous
  • anonymous
would i multiply that by .0003
zepp
  • zepp
Exact.
anonymous
  • anonymous
46.68->46.7 sorry
anonymous
  • anonymous
so i got .0140 because i used 46.68*.0003
anonymous
  • anonymous
thats just grams right?
zepp
  • zepp
Yep
anonymous
  • anonymous
.0140 x 10^-3 ?
zepp
  • zepp
Good, now you'll need to set up another proportion:\[ \frac{\text{Mass of Carbon in 1 mole of Ethanol}}{\text{Molar mass of Ethanol}}=\frac{\text{Mass of Carbon (Unknown) }}{\text{Mass of the Ethanol given to you}}\]
anonymous
  • anonymous
(?/46.68)=(?/.0140x10^-3?)
zepp
  • zepp
Mass of carbon in 1 mole of ethanol is a known number, 2*Molar mass of carbon, 12, 2*12 = 24.
anonymous
  • anonymous
oh ok i see where you got that from
anonymous
  • anonymous
(24/46.68) = (?/.0140x10^-3?)
anonymous
  • anonymous
was i correct on the second half too?
zepp
  • zepp
I don't know, I didn't push the values in my calculator, I've shown you how to do these kind of problem, all the other mathematical manipulations/calculations should be done by you; the right method is shown so you can take down any question that looks like this.
anonymous
  • anonymous
I was asking about for the mass of the ethanol that the answer i found before right?
zepp
  • zepp
Like I said, I don't know :)
anonymous
  • anonymous
Ok so now i simple mult across top and bottom and cancel out to get my answer
anonymous
  • anonymous
simply*
anonymous
  • anonymous
i think i have it
anonymous
  • anonymous
thank you
zepp
  • zepp
np

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