## wondering_ 3 years ago How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?

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1. zepp

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

2. zepp

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

3. wondering_

its not CH(Sub 6)O?

4. wondering_

why did u give Carbon2?

5. zepp

Because that's the formula for Ethanol

6. zepp

It's a 2-carbon alcohol.

7. wondering_

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

8. wondering_

oh ok

9. zepp

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

10. zepp

$\large ? \text{ mole}=2.04×10^{20}$

11. zepp

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

12. wondering_

oh so its divide instead of multiply?

13. zepp

Yes, that's algebraic manipulation.

14. 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.

15. wondering_

|dw:1349138080842:dw|

16. zepp

Why 1000?

17. wondering_

for milligrams?

18. zepp

Our units are in moles...

19. 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$\large \frac{2.04*10^{20}\text{ atoms}*1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=x \text{ mole}$

20. zepp

atoms and atoms simplify out, leaving the mole on both sides.$\large \frac{2.04*10^{20}}{6.02*10^{23}}\text{mole}=x \text{ mole}$

21. wondering_

ok i see

22. wondering_

i got .0003 correct?

23. 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.

24. wondering_

Molar mass is 46.07?

25. zepp

Yes

26. wondering_

well for ethanol it self right?

27. wondering_

would i multiply that by .0003

28. zepp

Exact.

29. wondering_

46.68->46.7 sorry

30. wondering_

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

31. wondering_

thats just grams right?

32. zepp

Yep

33. wondering_

.0140 x 10^-3 ?

34. zepp

Good, now you'll need to set up another proportion:$\frac{\text{Mass of Carbon in 1 mole of Ethanol}}{\text{Molar mass of Ethanol}}=\frac{\text{Mass of Carbon (Unknown) }}{\text{Mass of the Ethanol given to you}}$

35. wondering_

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

36. zepp

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

37. wondering_

oh ok i see where you got that from

38. wondering_

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

39. wondering_

was i correct on the second half too?

40. 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.

41. wondering_

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

42. zepp

Like I said, I don't know :)

43. wondering_

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

44. wondering_

simply*

45. wondering_

i think i have it

46. wondering_

thank you

47. zepp

np