ATAR's y11 Chemistry Qs!

[iNerd]

Hey all - just thought I'd start a thread as Chem seems to be getting harder >.>

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

[iNerd]

Hah really? Why don't you help me then 'cause it aint working! -.-

N = M x Na

And I think you mean N = n x Na

[iNerd]

I'm not following that (any question) -.-

Anyone with fully worked examples?

[nacho]

questions : what textbook is this from?
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

[iNerd]

questions : what textbook is this from?
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

Heinemann.

Ah I'm so lost

[|ll|lll|]

1. Let x be Ar(Ni)

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

[iNerd]

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

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

[iNerd]

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

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

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.

n(X)= 0.15/5 x 4 = 0.12

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 -.-

[Water]

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?



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

[iNerd]

^

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?

[Souljette_93]

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

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

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.

n(X)= 0.15/5 x 4 = 0.12

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

[Water]

Either way is fine, since I'm dumb , and I like things simple, I like to use something thats systematic, and easy to follow. AKA Heineman's Way of working things out

[iNerd]

Thanks to all!

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!

[|ll|lll|]

2 Questions which you can solve using water's method.

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.

[iNerd]

H 2.4     S 39     O 58.6

   2.4        1.2         3.66

     2           1          3

Therefore H₂S0₃ which is Hydrogen sulfite!

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How do you do the 2nd one?