Empirical formula question

[nacho]

Q46. "Find the relative atomic mass of nickel if 3.370 nickel was obtained by reduction of 4.286 g of the oxide (NiO)."

Q47. "4.150 g tungsten was burned in chlorine and 8.950 g tungsten chloride (WCl6) was formed. Find the relative atomic mass of tungsten

Q48. "if 3.72 g of element X exactly reacts with 4.80g 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?"

I'm still really hazy with the following terms:
Relative atomic mass
Relative molecular mass
Relative formula mass
Molar mass
I know what they are denoted by, but i'm not quite sure when they are used, as some of them are quite similar..

What exactly does it mean when it says 'relative to Carbon-12' ?
What is the Carbon 12 scale? All i know is that it is a standard, now, to say the relative atomic mass of carbon is 12 exactly.
So then, if Boron has a relative atomic mass of 10.811, what on earth does this mean? (in concerning carbon-12) - i know it means that after averaging the relative isotopic masses, you get 10.811, but i don't see how this concerns carbon..
Thanks, these questions are from the hienemann chemistry 1 - textbook, and i'm not quite sure on how to solve them.

Edit: This is what my understanding of when to use these terms are so far:
"Find the relative atomic mass of Boron" = 10.811
The question would be incorrect if it said "Find the relative molecular mass of Boron, as Boron is not a molecule, but an atom.
Ar(B) = 10.811

"Find the relative molecular mass of Water" = H2O = 2 + 16 = 18
"Find the relative atomic mass of Water" would be incorrect as Water is a molecule and not an atom.
Mr(H2O) = 18
_ _ _ _ _ _ _ _
I know these are very basic questions, and i should have had the terms nailed down by term 1.. but as soon as I see the words 'relative to Carbon 12/ on the carbon 12 scale' i just get a mind black and want to drop out of VCE!!

[Greatness]

Relative atomic mass- the mass of an element
Relative molecular mass- the mass of all the atoms in a molecule
Relative formula mass- is used to find the mass of a ionic compund
Molar mass- is the mass of ONE mole of a substance
ill have a look through your other qns in a min
in regards to 'relative to carbon 12' it relates to the mole. A mole is defined as the standard amount of substance which ocntains the same no. of specified particles as there are atoms in exactly 12g of the carbon 12 isotope

[dptjandra]

The difference between "molar mass" and "relative atomic/molecular mass" is just the units.  Molar mass is "grams per mol" - that is, there are 10.811 g per mol of Boron (or 6.02 x 10^23 atoms of Boron - with each isotope occurring in natural proportions - weigh 10.811 g).  The relative atomic mass of boron is 10.811 "atomic mass units" - an "atomic mass unit" is just an arbitrary unit of mass, like a kilogram.

The idea of a relative formula mass pertains to an ionic compound - you don't see it much (not really at all) in 3/4.

So we generally use the molar mass values when we are doing a calculation of a mass in grams or kilograms, or anything involving moles.  We use the "relative mass" when we are just stating a weight.

As an analogy, if someone asked me my weight, I would say, "I weigh 70 kg" rather than "Each 18-year old me weighs 70 kg."
This is comparable to, "how much does boron weigh?", the answer would be "On average (due to different isotopes), Boron weighs 10.811 atomic mass units," [stating the weight as a relative atomic mass] rather than "each mole of boron weighs 10.811 g." [stating the weight as a molar mass].

So what is an atomic mass unit? This is where the "relative scale" comes in.  An atomic mass unit is a unit of mass (just like a kilogram), which derives its value relative to one atom of carbon-12.  It's a very confusing concept, I know...but consider this.
1. We take one atom of Carbon-12 and weigh it.  This has some value on the generic kg scale.  But we ignore that and we just say "we will call you 12 atomic mass units." (this is just a helpful convention - because Carbon-12 happens to have 12 nucleons.)
2. I now take one atom of element X.  My scale says it weighs twice as much as my one atom of carbon 12.  Therefore, I will say "element X has a RELATIVE mass of 24 atomic mass units" (24/12 = 2)
3. Now consider your boron.  The "relative atomic mass" is 10.811 atomic mass units after averaging.  The importance of carbon-12 is just to give "atomic mass units" meaning.  Relative to carbon-12, your atom is 10.811/12 = 0.901 times as heavy as one atom of carbon-12 --> this is what 10.811 atomic mass units means.  The only relevance to carbon-12 is that the unit you're giving to give your mass a meaning (atomic mass units), is COMPLETELY dependent on carbon-12 as a set reference point.

Hope this helps to clear things up - a bit of a tricky concept to explain over a computer!  Will look at your other questions later.

[dptjandra]

Q46. "Find the relative atomic mass of nickel if 3.370 nickel was obtained by reduction of 4.286 g of the oxide (NiO)."

Q46. m(Ni) = 3.370 g
m(NiO) = 4.286 g
So m(O) = 4.286 - 3.370 = 0.916 g
n(Ni) = n(O) since n(Ni) : n(O) = 1 : 1
n(O) = m/M = 0.916/16 = 0.05725 mol = n(Ni)
Now M(Ni) = m/n = 3.370/0.05725 = 58.86 g/mol [this is the molar mass]
The relative atomic mass of Nickel is 58.86 atomic mass units (provided I haven't made any arithmetic slip - my phone calculator isn't very good )
Do you understand the difference between molar mass and relative atomic mass?  They have the same numerical value, just different units.

[dptjandra]

Q48. "if 3.72 g of element X exactly reacts with 4.80g 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?"

Left out Q47 because it is basically the same process - thought you might like to give it a shot yourself.

Q48. This is basically the same again, in that you have:
m(X) = 3.72 g
m(O) = 4.80 g
n(O) : n(X) = 10 : 4 = 1 : 0.4
so n(X) = 0.4 * n(O)
n(O) = m/M = 4.60/16 mol
n(X) = 0.4 * 4.60/16 mol
And similarly to Q46, M(X) = m/n
Once you have M(X) in g/mol, you just change the units to amu to get the relative atomic mass of X.

The unifying theme is that you find the molar mass each time, and this is numerically identical to the relative mass, just with atomic mass units instead of g/mol.  The only relevance to carbon-12 in any of them is that the units - atomic mass units - are derived from assigning one atom of carbon-12 as 12 atomic mass units.

[nacho]

The difference between "molar mass" ...
Hope this helps to clear things up - a bit of a tricky concept to explain over a computer!  Will look at your other questions later.

Big thank you, such a relief to have such a thorough and simply-put explanation of a concept that I thought would haunt me for the rest of year 12.
Also, thanks swarley for creating the distinction between terms.

Edit: Just wanted to clarify my method for 47 as my answer is off by 1 or 2.

M(W) = m/n , 
m(W) = 4.15
n(W) = ?
n(W) = 1/6 n(Cl) as   W + 6Cl ------> WCl6
n(Cl) = m/M, m = 8.95 - 4.15 = 4.8
                  M = 35.45
m/M = 0.1354
Therefore n(Cl) = 0.1354
n(W) = 1/6 n(Cl) = 0.1354/6 = 0.0223
M(W) = m/n
        = 4.15/0.0223 = 186.09865
Therefore, answer is 186.09
My answer = 186.09
BoB's answer = 184.2

[dptjandra]

The book has probably used M(Cl) = 35.5, which from memory is the value used by VCAA.  I re-did your steps with 35.5 and it got the right answer.  Your technique was exactly right and if you were using the VCAA data booklet, you would've gotten the same answer as the book.

[ruchika5]

Can anyone please help me with this question relating to determining the molecular formula of a gas... thanks

"In one experiment an 18.99 g sample of the compound was burnt in excess oxygen. When the gases evolved were passed through anhydrous CaCl2, its mass increased by 11.38 g. The remaining gases, when bubbled through a NaOH solution, increased its mass by 27.83 g. In a separate experiment, a 6.21 g sample of the compound was vaporised. The vapour occupied 2.17 L at 200*C and 1.25 × 105 Pa. Calculate the molecular formula of the compound."

[dptjandra]

Does this help: http://vce.atarnotes.com/forum/index.php/topic,34800.0.html

If you still can't do it, let me know and I will go through the question step by step.  If you have the actual answer that would be helpful too (to make sure I don't make any slips along the way!)

[ruchika5]

Ah thanks, I understand it now