calculating the amount in moles


by Struggling
Tags: moles
Struggling
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#1
May25-05, 09:57 PM
P: 52
hi guys, need help with a question.

i know how to get the amount of a substance in moles from something like NaCl in 5.85g of salt you just use n = m/M

but would like to know how to:

calculate the amount(in mole) of:

cl ions in 13.4 g of nickel chloride(NiCl2)

i know the answer i just cant get it
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dextercioby
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#2
May26-05, 12:04 AM
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How many moles of chloride are there in a mole of nickel dichloride...?

Daniel.
Struggling
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#3
May26-05, 01:52 AM
P: 52
sorry i forgot to add that in 35.5 is the atomic mass of cholrine(Cl)

The Bob
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#4
May26-05, 02:35 AM
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calculating the amount in moles


Quote Quote by Struggling
calculate the amount(in mole) of:

cl ions in 13.4 g of nickel chloride(NiCl2)
There is very little difference to this from what you normally do. What do you do to find the number of moles from the mass and molar mass? Well you know that because you stated it. What is the molar mass of Cl2? That should answer your question.

The Bob (2004 )
Gokul43201
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#5
May27-05, 07:30 AM
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Quote Quote by The Bob
What is the molar mass of Cl2?
This is not required.

The question asks for the number of moles of Cl- ions. How many such ions are there in one molecule of NiCl2 ? Therefore, how many moles will there be in each mole of NiCl2 ? Finally, how many moles of NiCl2 are there ?
The Bob
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#6
May27-05, 09:30 AM
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Quote Quote by Gokul43201
This is not required.

The question asks for the number of moles of Cl- ions. How many such ions are there in one molecule of NiCl2 ? Therefore, how many moles will there be in each mole of NiCl2 ? Finally, how many moles of NiCl2 are there ?
Point made. Good call.

However I would find the percentage, by mass, between the two elements and then work out how many there are of chlorine. Then I would work out the number of moles this represented.

Your way, though, is simplier.

The Bob (2004 )
Struggling
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#7
May27-05, 09:56 PM
P: 52
i still cant get the right answer ive got:
the amount (in mole) of Cl ions in 13.4g of nickel chloride (NiCl2)

Cl molar mass = 35.5
Ni molar mass = 59
NiCl2 molar mass = 130

n= m/M --> n = 13.4/35.5 = 0.37746

and then i dont understand ive tried getting the percentage from 130 moles and all that but it comes nowhere near the answer.

the answer is 0.207 moles but i always seem to get .188 moles or .338 around those figures
dextercioby
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#8
May27-05, 10:48 PM
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Premise:
13.4 g of NiCl_{2}------------------->(13.4/130) moles of NiCl_{2}

*) 2 moles of Cl^{-}------------------->1 mole of NiCl_{2}
*) x moles of Cl^{-}-------------------->13.4/130 moles of NiCl_{2}

Daniel.
Struggling
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#9
May27-05, 11:06 PM
P: 52
arghh i still dont get it, this is killing me
dextercioby
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#10
May27-05, 11:16 PM
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I see you're really struggling Which line of the 3 i've written is fuzzy...?

Daniel.
Struggling
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#11
May28-05, 01:56 AM
P: 52
this line:

*) x moles of Cl^{-}-------------------->13.4/130 moles of NiCl_{2}

i cant seem to get the answer from it, ive been stuck on this thing for days
dextercioby
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#12
May28-05, 07:21 AM
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This is something i've learned in 6-th grade.Translating it mot--mot from Romanian,i'd call it "the rule of 3 simple".

Basically it is a ratio.U have to find the "x" by using logics to obtain the correct ratio.HINT:Check out the units.They should cancel.

For example a ratio:x moles/y moles=x/y which is just a number.

Daniel.
Struggling
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#13
May28-05, 09:43 AM
P: 52
eeeeek, i dont think were as well taught as europeans are here in australia.

im still not understanding, im sort of getting the cancelling out part but i shall try again in the morning its late here now. wat i understand is that the units cancel each other out that is moles, so the equation is x/y = x/y am i wrong here????
Gokul43201
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#14
May28-05, 12:06 PM
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Quote Quote by Struggling
im still not understanding, im sort of getting the cancelling out part but i shall try again in the morning its late here now. wat i understand is that the units cancel each other out that is moles, so the equation is x/y = x/y am i wrong here????
Understand the reasoning. Read post #5 first. Answer each of those 3 simple questions and you'll see that you are doing exactly what Dexter did in post #8.
isabelle2010
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#15
Jan23-11, 02:51 AM
P: 1
I also struggled with this in chemistry:

Find the number of cl- ions in 13.4 g of nickel chloride(NiCl2)

use the formula n=m/M
n(cl)=13.4g/M(Ni) +2M(Cl)
n(cl)=13.4/129.7
therefore n(cl)=0.103 mol

For the next step you have to look at the mole ratios
For every 1 mole of NiCl_2, there will be 2 moles of Cl- ions present
Therefore all you have to do is double the answer from above

2*0.103 = 0.206 or 2.6 * 10^-1 in scientific notation
Opus_723
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#16
Feb1-11, 05:17 PM
P: 179
You have 13.4g of nickel(II) chloride, or NiCl2 (13.4g NiCl2/1)

The molar mass of NiCl2 is, as you stated, about 130. (1mol NiCl2/130g NiCl2)

From this you can find the moles of NiCl2. (13.4g/1)(1mol/130g) = (13.4/130) mol NiCl2

Now the only question left is: How many moles of Cl- are in a single mole of NiCl2? As you can see, each molecule of NiCl2 has two chlorine ions, so each mole of NiCl2 will have 2 moles of chlorine ions.

You can set up a conversion factor with this: (2mol Cl-/1mol NiCl2)

Now you just convert from mol NiCl2 to mol Cl-.

The whole thing looks like this:
(13.4g NiCl2/1)(1mol NiCl2/130g NiCl2)(2mol Cl-/1mol NiCl2) = n mol Cl-

As you can see, the units cancel out. to give you mol Cl-.
Borek
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#17
Feb1-11, 05:32 PM
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Quote Quote by isabelle2010 View Post
Find the number of cl- ions in 13.4 g of nickel chloride(NiCl2)
You correctly calculated number of moles, but not number of ions.

You also successfully posted in almost 6 years old thread. This is called necroposting.
kemble
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#18
Feb19-11, 10:31 PM
P: 1
Quote Quote by Borek View Post
You correctly calculated number of moles, but not number of ions.

You also successfully posted in almost 6 years old thread. This is called necroposting.
Fortunately the curriculum hasn't changed much and the same question is on page 15 of the Heinemann Chemistry 2 4th Edition -Enhanced-
and I was having exactly the same problem as the VCE graduate before me, which is now solved. THANKS :D


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