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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: mr cool on February 16, 2009, 06:00:59 PM

Title: Graham's law practice
Post by: mr cool on February 16, 2009, 06:00:59 PM: Ok so i gave this problem a try... an unknown gas is found to travel twice as fast as CO₂, which traveled 14.00 cm in 12.00 sec. What is the molecular mass of the unknown gas?

My work....

(28.00 cm / 16.00 sec). _____
------------------------ = \| UK. (square root)
(14.00 cm / 16.00 sec). ---------
\| 44.0095 g/mol
_______ _____________
( \| UK ) (14.00 cm) (\|44.0095 g/ mol) (28.00 cm)
--------------------------- = --------------------------------
16.00 sec. 14.00 cm

____ ____________
\| UK = (\|44.0095 g/mol) (28.00 cm)
------------------------------
14.00 cm

I dont know what to do with my final set-up to get the answer :( help?

UK = unknown btw
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 06:10:10 PM: Dear reezyfbaby;

Do it analog the example in: http://en.wikipedia.org/wiki/Graham's_law
(Wiki is "always" your friend.)

You did wrong rearrange your equation!

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 07:24:23 PM: Quote from: ARGOS++ on February 16, 2009, 06:10:10 PM
Dear reezyfbaby;

Do it analog the example in: http://en.wikipedia.org/wiki/Graham's_law
(Wiki is "always" your friend.)

You did wrong rearrange your equation!

Good Luck!
ARGOS⁺⁺

So which part did i do wrong? I followed the example elwe did in class.
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 07:32:17 PM: Dear reezyfbaby;

Please check your first equation for which gas has the higher speed?
(And check also if 16 sec or 12 sec will be the correct one.)

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 07:36:07 PM: Quote from: ARGOS++ on February 16, 2009, 07:32:17 PM

Dear reezyfbaby;

Please check your first equation for which gas has the higher speed?
(And check also if 16 sec or 12 sec will be the correct one.)

Good Luck!
ARGOS⁺⁺

Oh poo, it would help if i read the problem more closely. The UK is lighter. But why is it 12 because dont you take the total ammount of time it took?
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 07:41:32 PM: Dear reezyfbaby;

The total time of 12 sec. would be ok, but it seems that you changed it from your question till your first equation into 16 sec, but why?

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 07:44:17 PM: Quote from: ARGOS++ on February 16, 2009, 07:41:32 PM

Dear reezyfbaby;

The total time of 12 sec. would be ok, but it seems that you changed it from your question till your first equation into 16 sec, but why?

Good Luck!
ARGOS⁺⁺

I think i meant to make it 18 not 16 but because 12.00 sec + 6.00 sec (time it took for UK because it was 2x faster, or can you not do that?
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 07:49:10 PM: Dear reezyfbaby;

No!, - that’s definitely not allowed! Why not using the normal speed?

Additionally Hint: Why not first rearrange the law for MW_UK?

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 08:01:21 PM: So i did it again and wound up with this as my final equation....
_____________
(\|44.0095 g/ mol)(14.00 cm)
---------------------------------
12.00 sec

I got 7.7 g when i did it all out.
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 08:06:43 PM: Dear reezyfbaby;

Sorry!, - I get another value for MW_UK.
Please check at least your result with the formula of the example in the link I gave you.

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 08:09:20 PM: Quote from: ARGOS++ on February 16, 2009, 08:06:43 PM

Dear reezyfbaby;

Sorry!, - I get another value for MW_UK.
Please check at least your result with the formula of the example in the link I gave you.

Good Luck!
ARGOS⁺⁺

I qoild but i didnt understand what the ²/₁ and 2/2 were
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 08:22:46 PM: Dear reezyfbaby;

Let’s start to rearrange the original law for MW_UK:
______ ______
v_UK / v_CO2 = \| MW_CO2 / \| MW_UK

Are you able to square the law to remove all square roots?

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 08:26:13 PM: Quote from: ARGOS++ on February 16, 2009, 08:22:46 PM

Dear reezyfbaby;

Let’s start to rearrange the original law for MW_UK:
______ ______
v_UK / v_CO2 = \| MW_CO2 / \| MW_UK

Are you able to square the law to remove all square roots?

Good Luck!
ARGOS⁺⁺

What is v on the left?
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 08:29:10 PM: Dear reezyfbaby;

v on the left side are the speeds.

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 08:35:46 PM: So....
____ _____
V uk / V co2 = \| uk / \|44.0095 g co2
V uk / v co2 = uk g / 44.0085 g co2
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 08:42:31 PM: Dear reezyfbaby;

No!, - your equation is not balanced anymore!

It’s simple algebra, so I simply get:
v_UK² / v_CO2² = MW_CO2 / MW_UK

Can you now solve the equation for MW_UK?

Good Luck!
ARGOS⁺⁺
Title: Re: Graham's law practice
Post by: mr cool on February 16, 2009, 08:56:17 PM: 6.00 UK² / 12 CO2² = 44.0095 g CO2 / mw uk

I have no clue what to do next. We had square roots in the whole problem we did in class.
Title: Re: Graham's law practice
Post by: ARGOS++ on February 16, 2009, 09:06:44 PM: Dear reezyfbaby;

You inserted values instead of rearranging the law for MW_UK!
And you used the wrong values for the speeds.

Please try to rearrange the formula before you insert any value!!

Think: How fast is CO₂, and what means twice so fast?

As you are asked for the MW so you have to remove/solve the square roots!!
Good Luck!
ARGOS⁺⁺