wondering_
How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?
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zepp
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Ethanol: \(\large \text{C}_2\text{H}_6\text{O}\)
zepp
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So first question, how many moles does 2.04 x 10^20 molecules represent?
wondering_
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its not CH(Sub 6)O?
wondering_
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why did u give Carbon2?
zepp
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Because that's the formula for Ethanol
zepp
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It's a 2-carbon alcohol.
wondering_
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\[(6.02 \times 10^{23})(2.04 \times 10^{20})?\]
wondering_
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oh ok
zepp
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That's incorrect;
\[\large 1 \text{ mole}= 6.02×10^{23}\]
zepp
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\[\large ? \text{ mole}=2.04×10^{20}\]
zepp
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That's a proportion and you should have this: \(\LARGE \frac{2.04×10^{20}}{6.02×10^{23}}\)
wondering_
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oh so its divide instead of multiply?
zepp
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Yes, that's algebraic manipulation.
zepp
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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.
wondering_
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|dw:1349138080842:dw|
zepp
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Why 1000?
wondering_
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for milligrams?
zepp
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Our units are in moles...
zepp
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\[\large \frac{1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=\frac{x \text{ mole}}{2.04 * 10^{20}\text{ atoms}}\]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
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atoms and atoms simplify out, leaving the mole on both sides.\[\large \frac{2.04*10^{20}}{6.02*10^{23}}\text{mole}=x \text{ mole}\]
wondering_
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ok i see
wondering_
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i got .0003 correct?
zepp
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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.
wondering_
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Molar mass is 46.07?
zepp
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Yes
wondering_
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well for ethanol it self right?
wondering_
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would i multiply that by .0003
zepp
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Exact.
wondering_
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46.68->46.7 sorry
wondering_
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so i got .0140 because i used 46.68*.0003
wondering_
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thats just grams right?
zepp
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Yep
wondering_
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.0140 x 10^-3 ?
zepp
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Good, now you'll need to set up another proportion:\[ \frac{\text{Mass of Carbon in 1 mole of Ethanol}}{\text{Molar mass of Ethanol}}=\frac{\text{Mass of Carbon (Unknown) }}{\text{Mass of the Ethanol given to you}}\]
wondering_
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(?/46.68)=(?/.0140x10^-3?)
zepp
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Mass of carbon in 1 mole of ethanol is a known number, 2*Molar mass of carbon, 12, 2*12 = 24.
wondering_
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oh ok i see where you got that from
wondering_
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(24/46.68) = (?/.0140x10^-3?)
wondering_
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was i correct on the second half too?
zepp
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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.
wondering_
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I was asking about for the mass of the ethanol that the answer i found before right?
zepp
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Like I said, I don't know :)
wondering_
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Ok so now i simple mult across top and bottom and cancel out to get my answer
wondering_
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simply*
wondering_
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i think i have it
wondering_
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thank you
zepp
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np