How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?

- anonymous

How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?

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- schrodinger

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- zepp

Ethanol: \(\large \text{C}_2\text{H}_6\text{O}\)

- zepp

So first question, how many moles does 2.04 x 10^20 molecules represent?

- anonymous

its not CH(Sub 6)O?

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## More answers

- anonymous

why did u give Carbon2?

- zepp

Because that's the formula for Ethanol

- zepp

It's a 2-carbon alcohol.

- anonymous

\[(6.02 \times 10^{23})(2.04 \times 10^{20})?\]

- anonymous

oh ok

- zepp

That's incorrect;
\[\large 1 \text{ mole}= 6.02×10^{23}\]

- zepp

\[\large ? \text{ mole}=2.04×10^{20}\]

- zepp

That's a proportion and you should have this: \(\LARGE \frac{2.04×10^{20}}{6.02×10^{23}}\)

- anonymous

oh so its divide instead of multiply?

- zepp

Yes, that's algebraic manipulation.

- 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

|dw:1349138080842:dw|

- zepp

Why 1000?

- anonymous

for milligrams?

- zepp

Our units are in moles...

- 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

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

ok i see

- anonymous

i got .0003 correct?

- 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

Molar mass is 46.07?

- zepp

Yes

- anonymous

well for ethanol it self right?

- anonymous

would i multiply that by .0003

- zepp

Exact.

- anonymous

46.68->46.7 sorry

- anonymous

so i got .0140 because i used 46.68*.0003

- anonymous

thats just grams right?

- zepp

Yep

- anonymous

.0140 x 10^-3 ?

- 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

(?/46.68)=(?/.0140x10^-3?)

- zepp

Mass of carbon in 1 mole of ethanol is a known number, 2*Molar mass of carbon, 12, 2*12 = 24.

- anonymous

oh ok i see where you got that from

- anonymous

(24/46.68) = (?/.0140x10^-3?)

- anonymous

was i correct on the second half too?

- 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

I was asking about for the mass of the ethanol that the answer i found before right?

- zepp

Like I said, I don't know :)

- anonymous

Ok so now i simple mult across top and bottom and cancel out to get my answer

- anonymous

simply*

- anonymous

i think i have it

- anonymous

thank you

- zepp

np

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