Topic: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element (Read 3572 times)

esmith351 · « **on:** March 07, 2015, 09:52:56 AM »

Working on a take-home exercise for my 113 class, just want to check if a couple of the problems I'm not too sure about are right:

Concerning Xylocaine (C₁₄H₂₂N₂)

1. How many carbon atoms are there in 0.80 mol of Xylocaine?

What I have written down:

0.80 mol Xylocaine * 6.02*10²³ molecules / 1 mol Xylocaine = 4.8 * 10²³ molecules. (14)4.8*10²³ (14 being the number of carbon atoms in 1 molecule) = 6.7*10²⁴ carbon atoms.

2. How many grams of carbon are there in 0.80 mol of Xylocaine?

What I have written down:

0.80 mol Xylocaine * 168.14 g C / 1 mol Xylocaine = 1.3*10² grams of carbon.

Hunter2 · « **Reply #1 on:** March 07, 2015, 11:27:16 AM »

The first one should be calculated with more digits. But the calculation way is correct.

The second one is correct.

billnotgatez · « **Reply #2 on:** March 07, 2015, 11:51:43 AM »

@Hunter2
I am not sure of your second response
WIKI has the xylocaine Molecular mass= 234.34 g/mol
(Formula C₁₄H₂₂N₂O)
If you have 0.80 mol would you have ~~117.17~~ 187.472 g of xylocaine?

I ball parked the carbon grams to be 134 which is similar to @esmith351

As far as significant digits for both answers, I always get that wrong.

Maybe I am missing something

EDIT: fixed typo at strikeout

Hunter2 · « **Reply #3 on:** March 07, 2015, 11:54:38 AM »

The original post has different formula C14H22N2. The 134 g is ok because 0,8 mol * 12 g/mol *14 = 134.4 g. What is also different to 130 g from the posters result. 4.4 g difference is for me significant.

billnotgatez · « **Reply #4 on:** March 07, 2015, 12:01:09 PM »

Something is wrong with my calculations
I can not have more carbon than xylocaine
for .8 mol

Hunter2 · « **Reply #5 on:** March 07, 2015, 12:05:39 PM »

0.8 mol xylocaine C₁₄H₂₂N₂O is with the molare weight of 234 g/ mol = 187.2 g. So 134 g fit to it.

billnotgatez · « **Reply #6 on:** March 07, 2015, 12:09:13 PM »

If you have 0.80 mol would you have 187.472 g of xylocaine?
should have been what I typed
Which is what you caught as my mistake as well

billnotgatez · « **Reply #7 on:** March 07, 2015, 12:19:51 PM »

I still probably have the significant digits wrong but the values are ballpark.

Borek · « **Reply #8 on:** March 07, 2015, 02:35:15 PM »

No need for approximations., even no need for the exact formula of the xylocaine, as long as we get the number of carbons per molecule right. 0.80 moles with 14 carbon atoms per molecule means we have 0.80*14 moles of C. To convert to mass just multiply by 12 g/mol. 0.80*14*12 = 134.4 = 1.3×10² g.

Topic: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element (Read 3572 times)

esmith351

Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

Hunter2

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

billnotgatez

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

Hunter2

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

billnotgatez

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

Hunter2

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

billnotgatez

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

billnotgatez

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

Borek

Re: Moles --> Atoms of an Element in a compound and Moles --> Grams of an Element

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