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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: chimias223 on October 28, 2020, 02:04:41 PM

Title: Simple question
Post by: chimias223 on October 28, 2020, 02:04:41 PM

Hi,
I need a little help with a little question.

The empirical formula of the mineral spodumene is: LiAlSi2O6
Given that the relative prevalence of the 6-li isotope in nature is 7.4%
How much moles of Li-6 atoms are found in 14 kg of spodumene?

Note: To get accurate results the molar masses should be taken with 3 digits after the decimal point.

Title: Re: Simple question
Post by: billnotgatez on October 28, 2020, 02:10:58 PM

You have to show your attempts or thoughts at solving the question to receive help.
This is a forum policy.
Click on the link near the top center of the forum page.
Forum Rules: Read This Before Posting.
http://www.chemicalforums.com/index.php?topic=65859.0

Title: Re: Simple question
Post by: chimias223 on October 28, 2020, 02:26:46 PM

Quote from: billnotgatez on October 28, 2020, 02:10:58 PM

You have to show your attempts or thoughts at solving the question to receive help.
This is a forum policy.
Click on the link near the top center of the forum page.
Forum Rules: Read This Before Posting.
http://www.chemicalforums.com/index.php?topic=65859.0

Sorry. This is waht I tried.
https://pasteboard.co/JxLUA6y.jpg

Title: Re: Simple question
Post by: AWK on October 28, 2020, 04:29:37 PM

Only 75.23 moles are OK

Title: Re: Simple question
Post by: chimias223 on October 28, 2020, 04:58:44 PM

Quote from: AWK on October 28, 2020, 04:29:37 PM

Only 75.23 moles are OK

what's about the information ''Given that the relative prevalence of the 6-li isotope in nature is 7.4%''.?  should not use it?

I think I need to multiply 75.23 with 7.4/100. but I'm not sure.

Title: Re: Simple question
Post by: Borek on October 28, 2020, 05:04:59 PM

Quote from: chimias223 on October 28, 2020, 04:58:44 PM

I think I need to multiply 75.23 with 7.4/100. but I'm not sure.

Idea is sound, but that's not what you did.

Title: Re: Simple question
Post by: chimias223 on October 28, 2020, 05:14:09 PM

Quote from: Borek on October 28, 2020, 05:04:59 PM

Quote from: chimias223 on October 28, 2020, 04:58:44 PM

I think I need to multiply 75.23 with 7.4/100. but I'm not sure.

Idea is sound, but that's not what you did.

Yea I know. I think It's supposed to be the final soloution but I dont sure (5.567). can you help me with this,please?

Title: Re: Simple question
Post by: AWK on October 29, 2020, 03:54:44 AM

Have you ever heard of significant figures?

Title: Re: Simple question
Post by: Borek on October 29, 2020, 04:32:21 AM

Question is poorly formulated in this aspect, it asks for using atomic masses to 3 decimal points, yet gives ⁶Li abundance to only two sigfigs.

That being said, the number is OK, it is just a matter of a correct representation of its uncertainty.

Title: Re: Simple question
Post by: chimias223 on October 29, 2020, 06:03:41 AM

Quote from: AWK on October 29, 2020, 03:54:44 AM

Have you ever heard of significant figures?

No

Quote from: Borek on October 29, 2020, 04:32:21 AM

Question is poorly formulated in this aspect, it asks for using atomic masses to 3 decimal points, yet gives ⁶Li abundance to only two sigfigs.

That being said, the number is OK, it is just a matter of a correct representation of its uncertainty.

The answer will be 5.567?

Title: Re: Simple question
Post by: Borek on October 29, 2020, 07:59:10 AM

Quote from: chimias223 on October 29, 2020, 06:03:41 AM

The answer will be 5.567?

Yes and no.

Yes - that the answer displayed by the calculator when you correctly key in all numbers given and do the correct calculations.

No - you are told abundance of ⁶Li is 7.4%. That basically means it is somewhere between 7.35% and 7.449999% as all these numbers will be rounded down to 7.4. So, your final answer should take it into account. https://en.wikipedia.org/wiki/Significant_figures are an approximate (but better than nothing) way of doing so. 5.567 contains too many digits and should be rounded down to a correct number of sigfigs.