Stankovic123's chem q's

[zvezda]

In order to determine the molecular formula of a compound known to contain only carbon and hydrogen, two experiments were carried out. In the first experiment, a 1.122g sample of the compound was burnt in excess oxygen. When the gases evolved were passed through anhydrous CaCl2, its mass increased by 1.442g. The remaining gases, when bubbled through NaOH solution, increased its mass by 3.521g. In the second experiment a 2.000g sample of the compound was vaporized. The vapour occupied a volume of 797mL at STP. Calculate the empirical formula and hence the molecular formula of the compound.

Ok, first of all, I'm really stumped by the bolded part of the question, what would I do with such info?
I managed to find the mol of the compound (0.01996058 mol) and the molar mass of the compound (56.21079069 g mol^-1). What next?

oh and btw, I only convert to sig figs for the final answer so that's why I've got so many decimal places ^^

[teletubbies_95]

I saw similar questions in nelson.
I read that cacl2 absorbs h20 , while naoh , absorbs c02 . But other than that I'm stumped too. 

[Reckoner]

Good question 

Firstly we need to look at the first experiment. When the compound is burnt, it will produce CO₂ and H₂O (combustion equations are very common, get used to them!). Now it's fairly safe to assume that one of these products reacts with the CaCl₂ and the other reacts with the NaOH.

Once you've figured out which reacts with which, you can then use the mass increases to determine the amount in moles of each product that was produced (as the increase in mass would come from the evolved gases).

Then, determine the mole ratio of Hydrogen and Carbon  (as the Oxygen comes from the air, not the compound we are testing!) This is your empirical formula. Then use Experiment 2 to finish the question.

See how you go from here, answer is below

Spoiler

The H₂O reacts with the CaCl₂, meaning that 1.442g of H₂O was evolved (0.08011 mol)
The CO₂ reacts with the NaOH; meaning the 3.521g of CO₂ was produced (0.08002 mol)

From looking at this, it is ~1:1 mole ratio of CO₂:H₂O, so the C:H ratio is 1:2, giving an empirical formula of CH₂

Using the molar mass (calculated from experiment 2), we see that ~4 of the CH₂'s "fit" into the Mr of 56, so we multiply by a factor of 4, making the compound C₄H₈ 

   

   

[zvezda]

Thanks for that, understood everything. Except, how would one know that H2O would react would react with CaCl2 and CO2 would react with NaOH

[Reckoner]

I'm not entirely sure on the actual reasons behind why they react, (I just knew, I don't actually know much chem, just how to do VCE chem haha). I'll give it a go though! I knew that NaOH reacts with CO₂ from titration procedure (NaOH reacts with CO₂ from the atmosphere, so can't be used as a standard solution or whatever its called, I forget). And The anhydrous CaCl₂ is soluble in water, so water would get absorbed by it (also the word anhydrous suggests that there is a hydrated form, thus will absorb water). As you do more chem you will learn more about what reacts with what.

I don't think you need to know that actual theory behind it though, just know the reactions. I don't really know how to explain it though 

Hope you can make sense of that haha

[teletubbies_95]

Thank you. If this sorta question was given in a vcaa exam , how much time should it take to do it?

[zvezda]

lol, but how did you then arrive at the conclusion that 1.442 grams of H2O was evolved if it is absorbed by the CaCl2??? How would the mass of something increase after being absorbed??

[zvezda]

By the way, how does something like CaCl2 absorb H2O when it is soluble? Why isn't it dissolved instead?

[Reckoner]

lol, but how did you then arrive at the conclusion that 1.442 grams of H2O was evolved if it is absorbed by the CaCl2??? How would the mass of something increase after being absorbed??

Think of it this way; if 1.442 grams of chocolate was put next to you, you would eat it because chocolate is nice (you "react" with it) and your mass would increase by 1.442 grams. Law of conservation of mass. However if 3.521 grams of licorice (which you don't like or "react with"!) was given to you, you wouldn't "eat" it, so it wouldn't increase your mass. So if you weigh yourself after this, your increase in mass was because of the chocolate (H₂O) instead of the licorice (CO₂) I also made the assumption that all of the H₂O had been absorbed, and only H₂O had been absorbed.

By the way, how does something like CaCl2 absorb H2O when it is soluble? Why isn't it dissolved instead?

This is where my knowledge falls down. But remember that the H₂O is a gas at this stage, so the water wouldn't wash away the solid, but the solid would capture the gas. I think it still sort of "dissolves" the CaCl₂, and the remaining solution stays on the CaCl₂, thus increasing its mass. 

Thank you. If this sorta question was given in a vcaa exam , how much time should it take to do it?

This I have no idea haha, but I'd guess 5-8 mins depending on how long it takes you to figure out how to do it. As the ratios are fairly simple, once you get going it won't take all that long. I didn't show my working while doing it, so I can't really judge sorry 

[zvezda]

Think of it this way; if 1.442 grams of chocolate was put next to you, you would eat it because chocolate is nice (you "react" with it) and your mass would increase by 1.442 grams. Law of conservation of mass. However if 3.521 grams of licorice (which you don't like or "react with"!) was given to you, you wouldn't "eat" it, so it wouldn't increase your mass. So if you weigh yourself after this, your increase in mass was because of the chocolate (H₂O) instead of the licorice (CO₂) I also made the assumption that all of the H₂O had been absorbed, and only H₂O had been absorbed.

This is where my knowledge falls down. But remember that the H₂O is a gas at this stage, so the water wouldn't wash away the solid, but the solid would capture the gas. I think it still sort of "dissolves" the CaCl₂, and the remaining solution stays on the CaCl₂, thus increasing its mass. 

This I have no idea haha, but I'd guess 5-8 mins depending on how long it takes you to figure out how to do it. As the ratios are fairly simple, once you get going it won't take all that long. I didn't show my working while doing it, so I can't really judge sorry 

Even if it is a gas, the molecules are still highly polarised arent they? Appreciating the help here

And by the way, another question lol: When finding the molecular formula for example and you divide Mr of compound by Mr of empirical formula, what if you got 1.91127? safe to assume that the molecular formula is 2 times the empirical formula?

[teletubbies_95]

Thanks again! The chocolate thing really helped!

I guess it's got to do with the bonding . Maybe the lone pairs of O2 bond with ca ( postive) . And therefore solid state ! I'm not entirely sure!

Stankovic , I'm pretty sure it's safe to say its 2 times the e.f.

[Reckoner]

Even if it is a gas, the molecules are still highly polarised arent they? Appreciating the help here

Yep, still the same molecules (although the properties are different, I'm not too sure on the details)

And by the way, another question lol: When finding the molecular formula for example and you divide Mr of compound by Mr of empirical formula, what if you got 1.91127? safe to assume that the molecular formula is 2 times the empirical formula?

Hmm I'd check my working if I got something like that. I'd say most likely yes, that would be fine. I doubt VCAA would give you one that worked out to be that borderline though.

Thanks again! The chocolate thing really helped!

Haha I'm glad 

[Mao]

By the way, how does something like CaCl2 absorb H2O when it is soluble? Why isn't it dissolved instead?

We are here assuming that there is a lot more CaCl2 than H2O. Dissolution occurs when there is a LOT of water compared to the solute. A highly concentrated solution of 1M has 55 water molecules to one solute. That is not the case here, we are giving small up a small number of water molecules to a large number of salt ions.

Many anhydrous salts form what is called a "hydrated" structure, for example, the blue crystals of CuSO4 (http://en.wikipedia.org/wiki/Chalcanthite). This means contrary to the simple crystal lattice for NaCl that you see in textbooks, the crystal lattice includes spots for extra water molecules held in place by the ions. The precise reason for this is too complicated to discuss here, but many salts prefer hydration, and thus their anhydrous form readily absorbs moisture.

It is impossible (at this point in time) to predict the degree of hydration (i.e. the number of water molecules). We don't really know what will and will not form structural hydrates, but rest assured you will be told if you are dealing with a hydrate, and a formula would be given if it is a quantitative analysis.

[zvezda]

We are here assuming that there is a lot more CaCl2 than H2O. Dissolution occurs when there is a LOT of water compared to the solute. A highly concentrated solution of 1M has 55 water molecules to one solute. That is not the case here, we are giving small up a small number of water molecules to a large number of salt ions.

Many anhydrous salts form what is called a "hydrated" structure, for example, the blue crystals of CuSO4 (http://en.wikipedia.org/wiki/Chalcanthite). This means contrary to the simple crystal lattice for NaCl that you see in textbooks, the crystal lattice includes spots for extra water molecules held in place by the ions. The precise reason for this is too complicated to discuss here, but many salts prefer hydration, and thus their anhydrous form readily absorbs moisture.

It is impossible (at this point in time) to predict the degree of hydration (i.e. the number of water molecules). We don't really know what will and will not form structural hydrates, but rest assured you will be told if you are dealing with a hydrate, and a formula would be given if it is a quantitative analysis.

Ok, that's cleared things up. Thanks

[zvezda]

Came across a question which asked me to draw the structural formula of 1-chlorohex-3-ene???
How is that possible, shouldn't the double bond take precedence over functional groups when naming hydrocarbons??