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### Topic: Mass to mole ratio  (Read 9878 times)

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#### lou_skywalker

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##### Mass to mole ratio
« on: September 15, 2009, 03:01:14 PM »
Elemental sulfur occurs as octatomic molecules, S8. What mass of fluorine gas is needed to react completely with 17.9 g of sulfur to form sulfur hexafluoride?
I am confused with the diatomic and octatomic stuff. what's the actual chemical reaction we are supposed to use?

#### Tin Man

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##### Re: Mass to mole ratio
« Reply #1 on: September 15, 2009, 03:17:19 PM »
Diatomic means that in pure form, the atoms will be paired. For instance under normal circumstances, you will not have pure oxygen consisting of individual O atoms; rather, it will consist of many O2 atoms paired together. The same logic applies to octatomic substances.

So, knowing that these substances are not going to be single atoms, can you figure out how you're going to have to adjust your mole-to-gram calculations when solving for the mass of each substance needed?
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#### lou_skywalker

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##### Re: Mass to mole ratio
« Reply #2 on: September 15, 2009, 03:19:43 PM »
Diatomic means that in pure form, the atoms will be paired. For instance under normal circumstances, you will not have pure oxygen consisting of individual O atoms; rather, it will consist of many O2 atoms paired together. The same logic applies to octatomic substances.

So, knowing that these substances are not going to be single atoms, can you figure out how you're going to have to adjust your mole-to-gram calculations when solving for the mass of each substance needed?

so the equation I use is S8 + F2 = SF6  If I balance it it would be S8 +24F2 = 8SF6?
and if I calculate the % mass of 17.9 grams do I do the % of one sulphur or 8?

#### Tin Man

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##### Re: Mass to mole ratio
« Reply #3 on: September 15, 2009, 03:45:14 PM »
Concentrate on your reactants. You aren't being asked for product information, you're being asked what you need to MAKE the product. That's all reactant stuff. The formula on the other side isn't your concern in this problem, except as far as balancing the equation (which is very important here).

If you know the mass of your sulfur, you can figure out the moles. To find the moles, you would do the same procedure as you would do with NaCl, or CH4, or any other atom. Add up the mass of all atoms involved. For fluorine, that would be both atoms of fluorine, and for sulfur, that's all eight atoms of sulfur. It makes sense. If you were finding the mass of table salt, you would add up one sodium and one chlorine, right?

Once you have your moles of sulfur, you use your mole ratio for the reactants. How many F2's per every unit of S8? There's your mole ratio (and why you figured out the balanced equation). This will allow you to find the number of F2's you need.

From there, do the reverse equation of what you did earlier to find the grams of fluorine. Remember that you're dealing with two fluorine atoms, not one.

Make sense?
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#### lou_skywalker

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##### Re: Mass to mole ratio
« Reply #4 on: September 15, 2009, 03:54:10 PM »
yeah now I understand it. I got 63.62g of Flourine. Now I have to deal with this stupid webassign (who is not accepting my answers..) many thanks!

#### Borek

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##### Re: Mass to mole ratio
« Reply #5 on: September 15, 2009, 04:11:13 PM »
Sig fig most likely.
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#### Tin Man

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##### Re: Mass to mole ratio
« Reply #6 on: September 15, 2009, 04:15:16 PM »
Sig fig most likely.

Yeah, a lot of those web assignment programs are brutal about those significant figures. The OP's answer wasn't too far off of mine.
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#### lou_skywalker

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##### Re: Mass to mole ratio
« Reply #7 on: September 15, 2009, 08:24:10 PM »
Sig fig most likely.

yes definitely. And this makes me feel stupid since I think I am following all the possible rules....

#### Tin Man

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##### Re: Mass to mole ratio
« Reply #8 on: September 15, 2009, 08:39:13 PM »
Did it not accept the answer?

Check to make absolutely sure of what the question was asking. I don't know if you copied and pasted or whatnot, but it would be strange for a general chemistry homework assignment (which I'm assuming this is) to be so picky, especially if you're following the basic rules of sig figs.
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#### lou_skywalker

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##### Re: Mass to mole ratio
« Reply #9 on: September 15, 2009, 10:02:28 PM »
I copied the question to this thread, the problem is our instructor only gives us 6 trials for each question. Now I have like two left. I'm sure I am missing something VERY obvious. Maybe I just need to relax nd come back in a couple of days.

#### Tin Man

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##### Re: Mass to mole ratio
« Reply #10 on: September 16, 2009, 12:04:17 AM »
Maybe you need to add units?

It is amazing what taking a break can do for you. I would have failed many physics assignments had it not been for my ability to walk away from an assignment before reaching critical temperatures.
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#### lou_skywalker

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##### Re: Mass to mole ratio
« Reply #11 on: September 16, 2009, 01:22:30 PM »
LOL got it now its 63.6 I did indeed make a mistake when determining sigfigs.Many thanks everyone!