Post by: iNerd on March 21, 2011, 06:04:29 pm
Q1: Find the relative atomic mass of nickel if 3.370 g nickel was obtained by reduction of 4.286 g of the oxide (NiO).
Q2: 4.150 g tungsten was burned in chlorine and 8.950 g tungsten chloride (WCl6) was formed. Find the relative atomic mass of tungsten.
Q3: If 3.72 g of element X reacts with exactly 4.80 g of oxygen to form a compound whose molecular formula is shown, from other experiments, to be X4O10, what is the relative atomic mass of X?
Answers:
A1: 58.9
A2: 184.2
A3: 31
Post by: iNerd on March 21, 2011, 06:09:07 pm
N = M x Na
And I think you mean N = n x Na
Post by: iNerd on March 21, 2011, 06:45:49 pm
Anyone with fully worked examples?
Post by: nacho on March 21, 2011, 06:54:34 pm
It's a very odd question, from my understanding you figure out the relative atomic mass by averaging the weights of different isotopes or something along those lines
..or refer to the periodic table O_O
Post by: iNerd on March 21, 2011, 07:00:46 pm
questions : what textbook is this from?Heinemann.
It's a very odd question, from my understanding you figure out the relative atomic mass by averaging the weights of different isotopes or something along those lines
..or refer to the periodic table O_O
Ah I'm so lost :(
Post by: |ll|lll| on March 21, 2011, 07:10:22 pm
Solve for x, x = 58.864628 ~ 58.87 (4 s.f.)
2. Let x = Ar(W)
Solve for x
x= 184.15625
~ 184.2 (4 s.f.)
To solve these kind of questions, use proportion. Frankly speaking, you don't even need to know mole.
3. I used mole to solve this question though.
Write equation and balance it:
4X + 5O2 --> X4O10
n(O2) = m/M =
n(X) = 0.15/5 x 4 = 0.12
n(X) = m / Ar(X)
0.12 = 3.72/Ar(X)
Ar(X) = 31
Post by: iNerd on March 21, 2011, 07:25:43 pm
3. I used mole to solve this question though.I don't understand from n(X) onwards. Why did you use 0.15 (the mole of oxygen?) and where did you get 5 x 4? :S
Write equation and balance it:
4X + 5O2 --> X4O10
n(O2) = m/M =
n(X) = 0.15/5 x 4 = 0.12
n(X) = m / Ar(X)
0.12 = 3.72/Ar(X)
Ar(X) = 31
Post by: iNerd on March 21, 2011, 07:40:48 pm
n(X)= 0.15/5 x 4 = 0.123. I used mole to solve this question though.I don't understand from n(X) onwards. Why did you use 0.15 (the mole of oxygen?) and where did you get 5 x 4? :S
Write equation and balance it:
4X + 5O2 --> X4O10
n(O2) = m/M =
n(X) = 0.15/5 x 4 = 0.12
n(X) = m / Ar(X)
0.12 = 3.72/Ar(X)
Ar(X) = 31
n(X) is just based on mole proportion, which you could tell from the equation.
The no. of moles of X is four-fifths that of O2.
I was under the assumption that n = m/M
You're substituting the value of 0.15mol for m which stands for mass?
And by 4/5 you mean the ratio of X:O is 4:5 right?
I'm still lost -.-
Post by: Water on March 21, 2011, 07:45:29 pm
Empirical Formula Set Up:
X O
mass(grams) 3.72gram 4.80gram
in mol 0.3
Simplest Mol Ratio 4 10
EF: X4O10
Working Backwards Finding What Multiplied with 0.3 will give 10.
So 0.3 x Y = 10
Y = 10/0.3
= 33.33
Now We know what is multiplied to both, that will give the ratio 4:10, we can work backwards for the mol of X.
4/ y = mol for X
4/33.33 = 0.12 mol for X
Mass of X
0.12 Mol for X = 3.72gram
Molar mass of X = 3.72g/0.12 mol
= 31gram
X = Phosphorus
PS: 400th Post
Post by: iNerd on March 21, 2011, 08:00:20 pm
I see the light! < = > I get it!
However it's different to chery's. Why and which is better? Oh and what's the name of the two methods aswell?
Post by: Souljette_93 on March 21, 2011, 08:03:57 pm
n(X)= 0.15/5 x 4 = 0.123. I used mole to solve this question though.I don't understand from n(X) onwards. Why did you use 0.15 (the mole of oxygen?) and where did you get 5 x 4? :S
Write equation and balance it:
4X + 5O2 --> X4O10
n(O2) = m/M =
n(X) = 0.15/5 x 4 = 0.12
n(X) = m / Ar(X)
0.12 = 3.72/Ar(X)
Ar(X) = 31
n(X) is just based on mole proportion, which you could tell from the equation.
The no. of moles of X is four-fifths that of O2.
I was under the assumption that n = m/M
You're substituting the value of 0.15mol for m which stands for mass?
And by 4/5 you mean the ratio of X:O is 4:5 right?
I'm still lost -.-
You can either do it this way or Water's way, but I personally found this one better.
You are right about moles, since we found that the moles of Oxygen was 0.15, all you do it apply ratios to X, which it would be 4/5.
then multiply 4/5 with 0.15, you get 0.12 mols.
Now use the formula M=m/n, which you already know that m= 3.72 so that means 3.72/0.12=31
Post by: Water on March 21, 2011, 08:16:43 pm
Post by: iNerd on March 21, 2011, 08:25:30 pm
An interesting observation I had.
Both Water and I prefer the dumber method and we're both guys.
Both chery and souljette prefer the smarter method and are both girls.
-.-
Anyways I'm trying to be a girl (smarter method) and think I'm getting it.
If anyone could be bothered/find a Q that is like this so I can practice the new method? Much appreciated! :)
Post by: |ll|lll| on March 21, 2011, 09:19:53 pm
A compound of sulfur contains 2.4% hydrogen, 39.0% sulfur and 58.6% oxygen.
Find the empirical formula of the compound.
Ans: H2SO3
A harder one...
Washing soda crystals may be used to bleach linen. When crystallised from water, washing soda (sodium bicarbonate, Na2CO3) forms crystals of a hydrated ionic compound. When 5.00 g of washing soda crystals were dried in a desiccator, 1.85 g of sodium bicarbonate remained. Calculate the empirical formula of the hydrated compound.
Ans: Na2CO3 . 10H2O (this is a hydated compound)
I prefer using my method though because it's the method which you will use most commonly. You'll get used to it in stoichiometry.
Post by: iNerd on March 24, 2011, 04:18:31 pm
2.4 1.2 3.66
2 1 3
Therefore H2S03 which is Hydrogen sulfite!
---------------------------------------------------------------------------------------
How do you do the 2nd one? ???
Post by: pi on March 24, 2011, 04:23:24 pm
Post by: iNerd on March 26, 2011, 11:06:40 am
The combustion reaction of a hydrocarbon can be written as:
CxHy + 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 CxHy
???
All I can do is n(CO2) = 0.6659 mol
Post by: pi on March 26, 2011, 11:16:26 am
'determine the mass of carbon'
all the mass of carbon will be in the carbon dioxide, so work from there
Post by: Dr.Lecter on March 26, 2011, 11:19:35 am
Post by: iNerd on March 26, 2011, 11:22:19 am
Post by: Water on March 26, 2011, 12:23:34 pm
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
Post by: iNerd on March 26, 2011, 12:44:29 pm
If you read above I gave your 'Y' a value of 0.6659 mol which was the only step I could get :P
Thanks for the rest! :smitten:
Post by: Water on March 26, 2011, 12:46:44 pm
Post by: iNerd on March 26, 2011, 03:21:30 pm
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?
Post by: pi on March 26, 2011, 03:27:30 pm
It is basically 10g (sample) - m(carbon) = m(hydrogen)
After this, you use empirical formula to find the structure of your hydrocarbon.
Post by: iNerd on March 26, 2011, 03:41:59 pm
Can someone provide me with a worked solution?
Post by: Water on March 26, 2011, 03:50:19 pm
I'll do the question as well, and see if my method gives the right answer
Post by: Water on March 26, 2011, 03:52:52 pm
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.
Post by: pi on March 26, 2011, 03:54:40 pm
I'm lost -.-
Can someone provide me with a worked solution?
What I remember from Unit 1:
EDIT: beaten, but my way looks better
Post by: iNerd on March 26, 2011, 04:11:24 pm
Thanks to both!!
Moderator action: removed real name, sorry for the inconvenience
Post by: iNerd on March 28, 2011, 08:44:50 pm
Molar mass of Mg(OH)2 = 24.3 + (17x2) = 24.3 + 34 = 56.3 g/mol
n(HCl) = 0.242 divided by M(HCl) which is 36.5 so 0.242/36.5 = 6.63 x 10^-3
m[Mg(OH)2] = 6.63 x 10^-3 x 56.3 = 0.3732g
Answer is 0.193g.
Post by: luken93 on March 28, 2011, 08:57:49 pm
n(HCl) = .242/36.5 = 0.00663 mol
n(Mg(OH)2) = 1/2 x n(HCl) = 0.003315 mol
m(Mg(OH)2) = .003315 x 58.3 = 0.193g
Molar Mass of Mg(OH)2 is incorrect...
Post by: iNerd on March 28, 2011, 08:59:15 pm
Mg(OH)2 + 2HCl -> Mg(Cl)2 + 2H2OAh. How on earth am I meant to know that equation -.-
n(HCl) = .242/36.5 = 0.00663 mol
n(Mg(OH)2) = 1/2 x n(HCl) = 0.003315 mol
m(Mg(OH)2) = .003315 x 58.3 = 0.193g
Molar Mass of Mg(OH)2 is incorrect...
Thanks!
Post by: luken93 on March 28, 2011, 09:09:26 pm
All I'm doing is equating the coefficients, but I wouldn't worry too much about that, it will come with time..Mg(OH)2 + 2HCl -> Mg(Cl)2 + 2H2OAh. How on earth am I meant to know that equation -.-
n(HCl) = .242/36.5 = 0.00663 mol
n(Mg(OH)2) = 1/2 x n(HCl) = 0.003315 mol
m(Mg(OH)2) = .003315 x 58.3 = 0.193g
Molar Mass of Mg(OH)2 is incorrect...
Thanks!
Post by: iNerd on March 29, 2011, 07:15:41 am
I don't understand? By equating coefficients do you mean balancing 'cause I can balance...I meant how am I meant to know that acid + base ---> salt + water. That is Unit 2 knowledge I don't know :/All I'm doing is equating the coefficients, but I wouldn't worry too much about that, it will come with time..Mg(OH)2 + 2HCl -> Mg(Cl)2 + 2H2OAh. How on earth am I meant to know that equation -.-
n(HCl) = .242/36.5 = 0.00663 mol
n(Mg(OH)2) = 1/2 x n(HCl) = 0.003315 mol
m(Mg(OH)2) = .003315 x 58.3 = 0.193g
Molar Mass of Mg(OH)2 is incorrect...
Thanks!
Post by: Greatness on March 29, 2011, 07:22:14 am
n(unknown) = (coefficient of unknown)/(coefficient of known) * no. of mol reacting
Post by: iNerd on March 29, 2011, 07:27:40 am
You have to balance the equation first i.e. get all elements equal. Then you use mole ratios to figure out how much you need to multiply the mol of something by to find the mol of what you want.I know stoich -.-
n(unknown) = (coefficient of unknown)/(coefficient of known) * no. of mol reacting
I obviously couldn't do that because I had no idea what the equation would like :/
All we accelerate is Mass-mass stoich from U2 to U1.
I had to self-learn that acid equation thing.
Post by: Greatness on March 29, 2011, 07:34:14 am
Yeah you learn the acid base reactions in unit 2.
Post by: pi on March 29, 2011, 08:33:10 pm
Ohhh right. hahah
Yeah you learn the acid base reactions in unit 2.
The equation could have been found using yr10 Science knowledge though... I believe that acid + metal hydroxide equations are in the chem chapter.
Post by: pi on March 29, 2011, 08:47:23 pm
lol mole and stoich have got to be the easiest chem chapters -
2 formulas and your done :)
Easy for now... You obviously haven't met PV=nRT, or the 5 gas conversions between units (mmHg, Pa, ppm, etc.), or even back titrations...
Post by: Water on March 29, 2011, 08:48:36 pm
Year 12 Chemistry looks hard with the theory ):
Moderator action: removed real name, sorry for the inconvenience
Post by: Souljette_93 on March 29, 2011, 08:59:29 pm
Post by: iNerd on March 29, 2011, 09:02:05 pm
Post by: luken93 on March 29, 2011, 09:26:21 pm
you don't have ownership over your threads...well he kinda does, I don't see Bazza16's thread written anywhere here? Afterall, it is ATAR's chemistry questions. He is asking for help, and from the above posts, you aren't really giving any...
Post by: iNerd on March 29, 2011, 09:27:13 pm
I request a moderator to delete page 3 and 4...you're blowing out my threads :/you don't have ownership over your threads...well he kinda does, I don't see Bazza16's thread written anywhere here? Afterall, it is ATAR's chemistry questions. He is asking for help, and from the above posts, you aren't really giving any...
(not luken..luken has actually helped me in more ways than one).. :smitten:
Post by: Russ on March 30, 2011, 03:05:37 pm
Keep this thread on topic and answering questions, if you want to discuss the difficulty of the syllabus then do it elsewhere.
Post by: iNerd on April 11, 2011, 05:17:58 pm
How are you meant to know how to draw them? The book says to maximise repulsion and place them as far away as possible - there's so many places I could put them..
For eg - H2S.
I drew it:
H ------- S ------- H Linear
But it's like V-shaped with a lone pair at the top, one in the bottom right and two single covalent bonds with two hydrogen atoms.
???
Post by: nacho on April 11, 2011, 05:22:44 pm
'Shapes of molecules'here:
How are you meant to know how to draw them? The book says to maximise repulsion and place them as far away as possible - there's so many places I could put them..
For eg - H2S.
I drew it:
H ------- S ------- H Linear
But it's like V-shaped with a lone pair at the top, one in the bottom right and two single covalent bonds with two hydrogen atoms.
???
Quote
http://www.dynamicscience.com.au/tester/solutions/chemistry/molecules/molecularshapes.htmIf you don't get it, seriously do not worry, it's something I've been unable to pick up and it's completely unecessary for year 12 chem.
Edit:
Something more straight to the point:
Quote
http://www.dynamicscience.com.au/tester/solutions/chemistry/molecules/molecularshapes5.htm
Post by: iNerd on April 11, 2011, 05:25:51 pm
Post by: pi on April 11, 2011, 06:44:59 pm
If you don't get it, seriously do not worry, it's something I've been unable to pick up and it's completely unecessary for year 12 chem.
But you also don't want to be drawing things funnily in organic chem. Its good to know shapes.
Post by: iNerd on April 11, 2011, 06:45:59 pm
So can you explain it please? ???If you don't get it, seriously do not worry, it's something I've been unable to pick up and it's completely unecessary for year 12 chem.
But you also don't want to be drawing things funnily in organic chem. Its good to know shapes.
Moderator action: removed real name, sorry for the inconvenience
Post by: pi on April 11, 2011, 06:50:27 pm
So can you explain it please? ???If you don't get it, seriously do not worry, it's something I've been unable to pick up and it's completely unecessary for year 12 chem.
But you also don't want to be drawing things funnily in organic chem. Its good to know shapes.
Its not something you can easily explain. You've actually got to go out of your way to research (internet, uni chem books) and practice (uni chem books maybe?). Maybe even go beyond the course just to get it. Of course I can't draw everything, but I can draw most of the chemicals commonly used in yr12 if needs be.
Moderator action: removed real name, sorry for the inconvenience
Post by: nacho on April 11, 2011, 06:54:17 pm
I think he doesn't understand the spacing concepts. Otherwise, in year 12, everything drawn at right angles is acceptable (i.e w/o spacing concepts)If you don't get it, seriously do not worry, it's something I've been unable to pick up and it's completely unecessary for year 12 chem.
But you also don't want to be drawing things funnily in organic chem. Its good to know shapes.
Moderator action: removed real name, sorry for the inconvenience
Post by: Souljette_93 on April 11, 2011, 07:11:17 pm
Post by: Mao on April 11, 2011, 07:14:20 pm
in CO2, the valence electrons form 2 groups: two C=O bonds (total of 8 valence electrons about C). The two groups repel to a linear shape.
in CH2O (formaldehyde), the valence electrons form 3 groups: C=O bond (4 electrons), two C-H bonds (Total of 8 valence electrons about C), the three groups repel to the optimal geometry of a planar triagle.
etc.
Post by: iNerd on April 11, 2011, 07:38:12 pm
So in y12 if I did CH4 as right-angle to each other (wrong shape) would I lose marks?
Post by: Mao on April 11, 2011, 07:41:05 pm
Thanks guys! (esp. Mao)
So in y12 if I did CH4 as right-angle to each other (wrong shape) would I lose marks?
No. You are not taught about stereochemistry, so you are not expected to know the correct 3D structure. The convention for organic molecule is to draw zig-zags, but right angles are fine.