ATAR's y11 Chemistry Qs!

[pi]

1.85g of Na2CO3 and 3.15g of H2O, you know the molar masses of each, so work like you just did

[iNerd]

Q:

The combustion reaction of a hydrocarbon can be written as:

C_xH_y + excess O₂ ------> x CO₂ + y/2 H₂0

10g of the unknown hydrocarbon was burnt in excess oxygen and 29.3g of carbon dioxide was formed.

Assuming all the carbon and hydrogen was converted into carbon dioxide, determine the mass of carbon and hydrogen present in the hydrocarbon. Also determine the empirical formula of C_xH_y




All I can do is n(CO₂) = 0.6659 mol

[pi]

Hints:
'determine the mass of carbon'

all the mass of carbon will be in the carbon dioxide, so work from there

[Dr.Lecter]

then you find mass (C atoms)

[iNerd]

I don't get it :/

[Water]

CxHy + excess O2 ------> x CO2 + y/2 H20


29.3g of carbon dioxide was formed.


Hi ATAR,


From this information, it is possible to piece together the molecule structure of the hydrocarbon.

Similarly to what pi and Dr.Lecter have said, you'll need to find the 29.3g.


So basically, from CxHy ----> x CO2 ,   

We need X amount(mass) of Carbon to make X amount(mass) of CO2. The X's are the same amount. They also have 1:1 ratio.

Law of Chemistry: Matter can neither be created or destroyed. So the mass of Carbon is merely being converted to create a different molecule




So basically, we need to find the mole of CO2 First.

n(CO2) = 29.3g/ 44gram mol   = Y (I'm not going to use calculator, so we'll call the answer Y)


Now we want to find the mass in CO2. Remember, To make CO2, you'll require the same equivalent mass from the carbon in the hydrocarbon. They'll work hand in hand.


n(C) = n(CO2)

m(C) = Y x 12.

This m(C) works for both the CO2 and Hydrocarbon. Since you needed the mass of C in hydrocarbon to make the mass of C in CO2. So they are the same.




Now we know the mass of C.


Ignore the Oxygen, since it is not part of the hydrocarbon molecule. And we only want to find the mass of Hydrogen. So , we have the mass of carbon, and the mass of sample, it is a matter of elimination

It is basically 10g (sample) - m(carbon) = m(Hydrogen)



After this, you use empirical formula to find the structure of your hydrocarbon.

Moderator action: removed real name, sorry for the inconvenience

[iNerd]

^

If you read above I gave your 'Y' a value of 0.6659 mol which was the only step I could get

Thanks for the rest! :smitten:

[Water]

Yeah, I did read it, but was too lazy to do all the calculations xD  :2funny:

[iNerd]

n(CO2) = 0.6659 mol.

m(C) = 0.6659 x 12 = 7.99g

m(O) = 10 - 7.99 = 2.01

C 7.99      O 2.01

   0.6658      0.1256

   5.3           1

    16            3

Therefore C₁₆H₃

Is that right?

[pi]

No (whenever you get an answer, just check to see if it would exist, I'm pretty sure C16H3 does not exist), I think what Water meant was:

It is basically 10g (sample) - m(carbon) = m(hydrogen)



After this, you use empirical formula to find the structure of your hydrocarbon.

[iNerd]

I'm lost -.-

Can someone provide me with a worked solution?

[Water]

Sorry, my error on explanation, after you found your carbon mass, you minus it from your sample, to get mass of hydrogen. Then repeat your empirical formula again.

I'll do the question as well, and see if my method gives the right answer

[Water]

I got CH3.


Btw, bought 6 Buenos . They are on sale in Safeway for 1 dollar each.



                             (C)              (H)

mass(in grams)          8                2
In mol :                 0.67            2

Simplest Mol R:        1                3


EF: CH3


PS: Error on my part, didn't read question properly. There is no O inside the CH, haha.

[pi]

I'm lost -.-

Can someone provide me with a worked solution?

What I remember from Unit 1:
































EDIT: beaten, but my way looks better

[iNerd]

Haha yeah pi's looks sexier

Thanks to both!!

Moderator action: removed real name, sorry for the inconvenience