A Question about Mole?
- Thread starter Kumar
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Moles are generally used because a mole of atoms equals the atomic mass in grams. In other words 1 mole of H = 1 gram and one mole of O = 16 grams.
A mole is just a number, 6.02 x 10^23. So lets just look at single atoms for the reaction you gave:
2H2 + 1O2 --> 2H2O
2H2 = 4H
1O2 = 2O
2H2O = 4H and 2O
Now lets look at the moles of the atoms and molecules
12.04 x 10^23 H2 = 24.08 x 10^23 H
6.02 x 10^23 O2 = 12.04 x 10^23 O
12.04 x 10^23 H2O = 24.08 x 10^23 H and 12.04 x 10^23 O
So the ratios are exactly the same for the reaction. When working in real world cemistry you will be working with grams or miligrams and of course not single atoms, so that is why moles are used. It is also much easier to write mol than it is to write out 6.02 x 10^23.
Hope that helps.
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Fundamentally, the mole is just a way to avoid using big numbers all the time.
1 mole of anything (atoms, cars, bananas, blood cells) is just $$6.02\times 10^{23}$$ of those somethings.
Atomic masses are defined in such a way that a carbon 12 atom has a mass of exactly 12 atomic mass units. 1 mole is defined to be the number of carbon 12 atoms needed to have a total mass of exactly 12 grams, as origin explained above. It turns out that to get 12 grams of carbon 12, we need $$6.02\times 10^{23}$$ carbon-12 atoms, so this is where the particular value of 1 mole comes from.
1 mole of anything (atoms, cars, bananas, blood cells) is just $$6.02\times 10^{23}$$ of those somethings.
Atomic masses are defined in such a way that a carbon 12 atom has a mass of exactly 12 atomic mass units. 1 mole is defined to be the number of carbon 12 atoms needed to have a total mass of exactly 12 grams, as origin explained above. It turns out that to get 12 grams of carbon 12, we need $$6.02\times 10^{23}$$ carbon-12 atoms, so this is where the particular value of 1 mole comes from.
Origon & James R, Thanks. It is clear to me. However, I want to know, how molecules actually remain present in solid, liquid or gas form? Are these remain present in their individual form or in clubbed form?
In following quote from above link, mole is also applicable to Electron & Photons. How these can be expressed in mole(simply by their numbers?)?
In following quote from above link, mole is also applicable to Electron & Photons. How these can be expressed in mole(simply by their numbers?)?
Take water as an example. If you have a given mass of water, it will always have the same number of water molecules, regardless of whether it is solid (ice), liquid (water) or gas (water vapour).
Ice, liquid water and water vapour are just different ways of arranging water molecules. So, if you start with 1 mole of ice and you heat it up until it melts, you end up with 1 mole of liquid water.
1 mole of anything is just $$6.02\times 10^{23}$$ of those things.
1 mole of water molecules means $$6.02\times 10^{23}$$ water molecules.
1 mole of electrons means $$6.02\times 10^{23}$$ electrons.
1 mole of photons means $$6.02\times 10^{23}$$ photons.
James R,
Thanks. It is quite clear to me. Mole belongs to same numbers of all things but not to their weght. Like for example, suppose, 100 apples weigh 10 Kgs but 100 strawberry just 2 kg.
I was also willing to understand, in what form a substance will be present in water. Whether in macromolecular or in individual molecular or in clubbed molecular form? I think, it depend on solubilty of a particular substance.
Thanks. It is quite clear to me. Mole belongs to same numbers of all things but not to their weght. Like for example, suppose, 100 apples weigh 10 Kgs but 100 strawberry just 2 kg.
I was also willing to understand, in what form a substance will be present in water. Whether in macromolecular or in individual molecular or in clubbed molecular form? I think, it depend on solubilty of a particular substance.
exchemist
Valued Senior Member
When a substance dissolves in a solvent, its constituent molecules (or ions) dissociate from one another. In the solid there are bonds of various sorts between them and during dissolving these bonds are broken and become replaced by new bonds, formed between the molecules (or ions) of the solid and the molecules of the solvent. From this, you may be able to understand that a solute substance will dissolve readily if the solute-solvent bonding is as great as, or stronger than, the bonding between the solute molecules (or ions). If it is less strong, then the solubility in that solvent will be low.
What you mean by "clubbed" molecular form sounds to me as if you are asking whether you get clumps of solid molecules still bound together, in the dissolved state. I think the answer to this is generally no. The dissolving process generally involves attack of the solid surface by the solvent molecules, bonding to individual solute molecules (or ions) and thus detaching them progressively, one by one. There is not - in general - a process for sub-assemblies of solute molecules to break off and float away in the solvent. However, there may be some exceptions, though I cannot offhand think of any.
exchemist,
Thanks.
Let us take an example of platic piece or a SiO2 crystal. If we put these in water I think, these never get dissolved. Thanks for correcting; Clubbed to clump.
My furthur question will be, if mole equal to Avogadro number and is just a unit but not actual quantity of a substance, how this number can be related to diluting out of a substance completely from a solution?
Thanks.
Let us take an example of platic piece or a SiO2 crystal. If we put these in water I think, these never get dissolved. Thanks for correcting; Clubbed to clump.
My furthur question will be, if mole equal to Avogadro number and is just a unit but not actual quantity of a substance, how this number can be related to diluting out of a substance completely from a solution?
exchemist
Valued Senior Member
I do not know what you mean by "diluting out" of a substance from solution. Can you rephrase?
But, as has already been explained, a mole of a substance is Avogadro's number of atoms or molecules of that substance. So it most certainly is a measure of the quantity of substance, but expressed in terms of the number of atoms or molecules present, rather than, say, by the mass or volume.
It is arguably the most fundamental measure of quantity it is possible to have. It is the only thing that tells you the actual number of atoms or molecules you are dealing with.
Say you have a mole of Na+ ions in DI water, each time you do a 10:1 dilution using DI water you would cut the amont by a factor of 10. So the quantity of Na+ ions would go from 6.02 x 10^23 to 6.02 x 10^22. So 24 10:1 dilutions would leave you with a theoretical number of .6 Na+ ions. Obviously there is no such thing as .6 of a Na+ ion. One more dilution would leave you with .06 Na+ ions so you can be relatively sure there are no ions in the water. So that is about 25 dilutions.
It sounds like you are investigating homeopathy. Is that why you are asking?
Just to clarify - the dilution would entail removing say 10 mils from the original solution and adding it to 90 mil of DI water for the 10:1 dilution. Then remove 10 mils of that solution and again add to 90 mils of DI water, etc.
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exchemist,
Let us take Salt Nacl and Gulucose. If we put these seprately in a glass of water, in what form, Nacl, Glucose and water will there after that. We are taking these as these are easily soluble in water.
Origin,
Thanks for telling and understanding me.
I think you are not considering the following things while diluting out a substance from a solution completely.
1. Total quantity of substance: it can be less or more than a mole quantity.
2. Nature of substance: how it react with water?
3. Degree of Solubility: how a substance is soluble in water.
4. Adsorption: How much a substance is adsorbed on walls of a container.
5. Predictibilty: it may not be necessary that a substance is diluted out at every step homogenously unless a substance is completely dissolved and present everywhere solution with same concentration.
6. Others??
I think, above factors can change the presence of a substance put in water and mixed. I am not mentioning dissolved in water since it is not necessary all substances are equally and homogenously dissolved in water.
4.
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Let us take Salt Nacl and Gulucose. If we put these seprately in a glass of water, in what form, Nacl, Glucose and water will there after that. We are taking these as these are easily soluble in water.
Origin,
Thanks for telling and understanding me.
I think you are not considering the following things while diluting out a substance from a solution completely.
1. Total quantity of substance: it can be less or more than a mole quantity.
2. Nature of substance: how it react with water?
3. Degree of Solubility: how a substance is soluble in water.
4. Adsorption: How much a substance is adsorbed on walls of a container.
5. Predictibilty: it may not be necessary that a substance is diluted out at every step homogenously unless a substance is completely dissolved and present everywhere solution with same concentration.
6. Others??
I think, above factors can change the presence of a substance put in water and mixed. I am not mentioning dissolved in water since it is not necessary all substances are equally and homogenously dissolved in water.
4.
4
In the casa you have to know the solubility of the compound , say salt NaCl I believe is soluble about 14.7 % so the ions in solution will not be 1 molar 54 gm/liter but 54 x 14.7/100 but about 0.14 molar
or you might say you have 0.14 mole of NaCl but as far glucose ,which is soluble 100% , you 186. gm/186mw / 1000 ml. you have one molar solution
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NaCl has a solubility of 356 g/L, so I am not sure where you got your number of 14% or what that means. One mole of NaCl is approximately 58.5 grams not 54 grams.
A saturated liter of water will have 356 grams of NaCl which is about 6.1 moles or a molarity of 6.1.
So of course 58.5 grams of NaCl in a liter of water will give you a molarity of 1 for both Na and Cl.
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Of course it can, if you dissolve 0.5 moles of salt in a liter of water you will have a molartiy of 0.5. So what is the problem?
I did take that into account that is why I used the example I did because the salt will disassociate in water. If you have an ionic substance like NaCl it will not disassociate in an nonpolar solvant. That was not your questions as far as I could determine.
I did consider that but that was not part of your question was it?
It would be negligable
Of course. The assumption is complete mixing.
Perhaps you could be more specific in your questions. If you have something specfic you want answered then ask away. Don't ask general questions if you do not want general answers.
My dear friend at this point I bow to you . The 14.7 % NaCl is in methanol, It is my error . the 54 Mw come out of my head, but it is in the ball park
exchemist
Valued Senior Member
NaCl in the crystal is a lattice (what we call a "giant structure", as opposed to one composed of molecules) of Na+ ions and Cl- ions, arranged in a regular pattern that allows the electrostatic attraction between +ve and -ve ions to be maximised. This is called "ionic" bonding and it is quite strong, hence the high melting point of salt. When it dissolves, each Na+ ion and each Cl- ion become surrounded by water (H-O-H) molecules. Although the water molecule is electrically neutral the oxygen atom carries a partial -ve charge, while the hydrogen atoms carry a partial +ve charge. We say that the water molecule is "polar" and that water is a "polar" solvent. This enables water to form electrostatic bonds with both types of ion and this enables the salt to dissolve. It is worth noting that NaCl will not dissolve very well in a hydrocarbon such as hexane, because hexane molecules are not polar.
With glucose, the solid consists of molecules, each of which is a 6 membered ring with hydroxyl (-OH) groups attached at most vertices. The molecule is quite big and has 6 carbon atoms, 6 oxygen atoms and 12 hydrogen atoms. The bonding between the molecules is weaker, hence the low melting point of glucose, and comprises attractions due to the fact that -OH groups, like water, are polar and attract one another (though I' m simplifying a bit). You may not be surprised to learn that water molecules can form similar bonds with these -OH groups and this enables whole glucose molecules to detach and wander through the solution.
So in the first case you have in solution individual ions i.e individual atoms with a missing or extra electron, while in the second case you have whole molecules (each with 24 atoms) in solution. In both cases the dissolved entity is surrounded by a cage or shell of water molecules, oriented so as to maximise the attraction to the dissolved ion or molecule.
Ok, that makes sense now. Getting 54 grams without looking it up is pretty good!
I meant, while diluting out a substance completely from a substance we can not base mole or Avogdro Constant because actual quantity of a substance mixed in water can be less and more than a mole quantity and it may take less or more sequences of dilutions.
Sorry, I meant nature of substance, how it will behave(not react) in water. Like Nacl, glucose, Sio2 can behave differently on mixing these in water.
Degree of solubilty of a substance in water can be important consideration in making "completely diluting out Calculation". A plastic ball can be removed from water from a container either on first step or may never be.
But still it need to be considered in "completely diluting out Calculation" esp where lower quantity is more important than higher ef Homeopathy.
This can make "completely diluting out Calculation in referance to Avogdro Constant/mole invalid. Moreover it is not necessary that at every step of shifiting solution from one container to other it will be of same concentration as of full solution. If we shift 10% water from a glass full of water and one ball, it is not necessary that ball will also be shifted to other container. Moreover when we mix glucose in water, we do note diffeent tastes of sugar concentration on diffrent sips.
For convinience sake, you can consider above for homeopathic dilution procedure, if can be misunderstood unless all above(may be more) thought are considered.
exchemist
Valued Senior Member
Look Kumar, Origin has explained this already. One mole of salt crystals has a mass of 58.5g. So if you dissolve, say, 5.85g in a litre of water you have a solution of 0.1 moles/litre. That is a fact. So you do know what you have got: 6.02 x 10²² Na+ ions and the same number of Cl- ions.
Furthermore it is the work of a few moments with a stirrer to make a solution homogeneous. So your notion that you don't know what you have got after a dilution is wrong. You do know exactly, provided you measure everything properly and provide you do a bit of mixing to ensure the solutions are homogeneous at each step. (In fact, as you may or may not know, even if you do not stir the solution, diffusion will eventually cause any inhomogeneities to disappear.)
I agree of course that if you get down to a solution with less than a hundred or so solute molecules in it, then statistical averaging may no longer be enough to ensure you know exactly the number of molecules present if you dilute further. But so what? At such dilutions you effectively have pure water anyway, because the level of solute is far far below any detectable limit, or equally, any level at which any biological effect would remain.
So, this is about homeopathy. In that case none of this really matters since homeopathy is complete and utter nonsense.