High School Chemistry: Stoichiometry

[Blonde_At_Heart]

I have to do an assignment that deals with stoichiometry. I know it's basically just converting from one thing to another, but a problem baffles me. I have to change from grams to molecules, and I can't remember what kind of conversion that is. The problem says this:

 

Using the equation provided:

2 NO + O2 -> 2 NO2

 

How many molecules of NO will react with 27.7 g O2?

 

I know how to set up the problem, but I don't know what conversion factor to use to change from grams to molecules, or even to moles if that's possible.

 

Any tips would be greatly appreciated

Edited January 9, 2009 by Blonde_At_Heart

[ecoli]

Some important information:

 

Molar mass = g/mole (look this up on your periodic table)

 

Avogadro number is 6.022*10^23 = the number of molecules in one mole (of anything)

[Blonde_At_Heart]

So, would it be something like this?

 

27.7 g O2 x 2 (34.014) g NO x 1 mol x 6.022 x 10^23

__________1 (31.998) g O2__34.014g___1 mol

 

(Ignore the underscores in denominator)

 

So the g O2 would cancel, then the g for NO, and finally the mol?

Or is that not the correct way to set it up?

 

31.998 is the molar mass of O2, and 34.014 is the molar mass of NO

[Cap'n Refsmmat]

Yes, that'll get you the right answer. But you could have made it a bit simpler:

 

[math]\frac{27.2 \mbox{g O2}}{1} \times \frac{1 \mbox{mol O2}}{31.998 \mbox{g O2}} \times \frac{2 \mbox{mol NO}}{1 \mbox{mol O2}} \times \frac{6.02 \times 10^{23} \mbox{molecules}}{1 \mbox{mol NO}}[/math]

 

Convert to moles, use the mole ratio (coefficients in the original chemical equation) to go to NO, then turn that into molecules. I think this method makes a bit more sense.

Edited January 9, 2009 by Cap'n Refsmmat
whoopsies

[Blonde_At_Heart]

Oh, I see. Thank you so much! That helps a ton!

[ecoli]

I'm not sure exactly what your math is there...

 

start: mass (m) = 27.7 g O2

 

Molar mass (M) of O2 is 32 g/mole.

 

Mole of O2 = m/M = g/(g/mole)

= 27.7 g / 32 g/mole

= .87 moles of O2

 

Looking back at the original chemical equation, every mole of O2, reacts with 2 moles of NO. Therefore:

 

.87 moles * 2 = 1.74 moles of NO.

 

Now you have to convert 1.74 moles of NO into the number of molecules.

 

There are 6.022*10^23 molecules in one mole of a substance.

 

1.74 * 6.022*10^23 = 1.05 * 10^24 molecules.

 

I realize your teacher probably taught you the 'mole central station' method, which involves doing all the conversions in one step, but personally that always confuses me. I prefer to to a 'roadmap' method where each step is separate. It helps me keep track of where I need to get to and where I've been. Hope this helps.

[Cap'n Refsmmat]

Just noticed I screwed up the mole ratio (swapped a 2 and a 1) in my previous post. Sorry about that. I just fixed it.

[ecoli]

Just noticed I screwed up the mole ratio (swapped a 2 and a 1) in my previous post. Sorry about that. I just fixed it.

exactly why I don't like that method... it's too easy to screw up (at least for me).

[Cap'n Refsmmat]

I usually don't have trouble, but writing it out in LaTeX requires a bit of extra brainpower.

[Blonde_At_Heart]

Okay, just to make sure I've got it right, the next problem is this:

 

(using the same equation 2 NO + O2 -> 2 NO2)

How many molecules of O2 are needed to make 3.76 x 10^22 molecules NO2?

 

Would it be like this:

 

[math]\frac{3.76 \times 10^{22} \mbox{molecules NO2}}{1} \times \frac{1 \mbox{mol}}{6.022 \times 10^{23} \mbox{molecules NO2}} \times \frac{34.014 \mbox{g NO2}}{1 \mbox{mol}} \times \frac{1\times 31.998 \mbox{g O2}}{2 \times 34.014 \mbox{g NO2}} \times [/math] (same problem, just too long to fit in one [math ])

[math]\frac{1 \mbox{mol}}{31.998 \mbox{g}} \times \frac{6.022 \times 10 ^ {23} \mbox{molecules O2}}{1 \mbox{mol}}[/math]

 

And I know that is overly complicated and almost everything cancels out, but I was only taught one way to do this. I will simplify how I do it later - all I want to know is if that is one way of getting the correct answer.

 

So the molecules NO2 would cancel, the mol would cancel, the 34.014g NO2 would cancel, the 31.998 g would cancel, the mol would cancel, and the 6.022 x 10^23 would cancel. Leaving:

 

[math]\frac{3.76 \times 10^{22} \mbox{molecules}}{2}[/math]

 

Which is [math]1.88 \times 10^{22} \mbox{molecules O2}[/math], right?

 

EDIT: PS, I hope you don't mind that I used your post above to learn how to use the LaTeX, Cap'n Refsmmat.

EDIT, again: Oh, I see how I could have simplified the entire equation. I only had ten minutes to learn how to do this chapter and another, so I only learned how to apply the coefficients by grams. Duh, looking back, I could have saved myself a lot of work if I had realized it worked with molecules as well. Oh well, it took so long to write that, I think I'll leave it.

Edited January 9, 2009 by Blonde_At_Heart
oops, that was stupid