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wondering_
Group Title
How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?
 one year ago
 one year ago
wondering_ Group Title
How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?
 one year ago
 one year ago

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zepp Group TitleBest ResponseYou've already chosen the best response.2
Ethanol: \(\large \text{C}_2\text{H}_6\text{O}\)
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
So first question, how many moles does 2.04 x 10^20 molecules represent?
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
its not CH(Sub 6)O?
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
why did u give Carbon2?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
Because that's the formula for Ethanol
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
It's a 2carbon alcohol.
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
\[(6.02 \times 10^{23})(2.04 \times 10^{20})?\]
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
That's incorrect; \[\large 1 \text{ mole}= 6.02×10^{23}\]
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
\[\large ? \text{ mole}=2.04×10^{20}\]
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
That's a proportion and you should have this: \(\LARGE \frac{2.04×10^{20}}{6.02×10^{23}}\)
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
oh so its divide instead of multiply?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
Yes, that's algebraic manipulation.
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
dw:1349138080842:dw
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
for milligrams?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
Our units are in moles...
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
\[\large \frac{1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=\frac{x \text{ mole}}{2.04 * 10^{20}\text{ atoms}}\]Multiply both sides by 2.04*10^20\[\large \frac{2.04*10^{20}\text{ atoms}*1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=x \text{ mole}\]
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
atoms and atoms simplify out, leaving the mole on both sides.\[\large \frac{2.04*10^{20}}{6.02*10^{23}}\text{mole}=x \text{ mole}\]
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
ok i see
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
i got .0003 correct?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
Molar mass is 46.07?
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
well for ethanol it self right?
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
would i multiply that by .0003
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
46.68>46.7 sorry
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
so i got .0140 because i used 46.68*.0003
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
thats just grams right?
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
.0140 x 10^3 ?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
Good, now you'll need to set up another proportion:\[ \frac{\text{Mass of Carbon in 1 mole of Ethanol}}{\text{Molar mass of Ethanol}}=\frac{\text{Mass of Carbon (Unknown) }}{\text{Mass of the Ethanol given to you}}\]
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
(?/46.68)=(?/.0140x10^3?)
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
Mass of carbon in 1 mole of ethanol is a known number, 2*Molar mass of carbon, 12, 2*12 = 24.
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
oh ok i see where you got that from
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
(24/46.68) = (?/.0140x10^3?)
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
was i correct on the second half too?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
I was asking about for the mass of the ethanol that the answer i found before right?
 one year ago

zepp Group TitleBest ResponseYou've already chosen the best response.2
Like I said, I don't know :)
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
Ok so now i simple mult across top and bottom and cancel out to get my answer
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
simply*
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
i think i have it
 one year ago

wondering_ Group TitleBest ResponseYou've already chosen the best response.0
thank you
 one year ago
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