Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: earthnation112 on April 29, 2016, 09:26:12 PM
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Just had some work relating to calculating moles for titration experiment, will be grateful if anyone can help pick out any mistakes where I have gone wrong. I spent quite a while trying to get the answers and the following is what I got.
Some data relating to the experiment:
Mean value of titration = 21.65 cm3
Weight of borax used: 4.9865
Equation of the reaction:
2HCl (aq) + Na2BO7.xH2O (aq) 2NaCl(aq) + 4H3BO3 yH2O
Molarity of HCl solution = 0.1773M
The questions which I answered were:
Number of moles of HCl used
= (21.65/1000) * 0.1173 = 0.002539545 moles
Number of moles of borax in 25 cm3 solution
= 0.0012697725 moles
Number of moles in 250cm3
= Same as 25cm3
Weight of Na2B4O7 (Molar Mass 201.22) in sample of borax
= 0.2555036225
Weight of water in the borax sample
=
Molar mass Na2B4O7·10H20 = 381.37 g/mol
Molar mass of 10H2O = 180.2 g/mol
Mass Percent H2O = (180.2 g/mol / 381.37g/mol) = 0.47251
Subtracting the total mass from the percentage of H2O in the mass itself.
4.9865g - (4.9865 x 0.47251) = 2.616865 g Borax remains
Weight of water in the borax sample:
4.9865 -2.616865 = 2.36965g
The number of molecules of water of crystallisation in Borax:
(2.36965/201.22) * 6.022 x 1023 = 7.0995 x 1021
First time I’m doing these calculation, is this right or have I gone horribly wrong?
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Wouldn't hurt to know what was the procedure and what you were trying to determine (I guess I know, but I can be wrong).
Number of moles in 250cm3
= Same as 25cm3
Already looks suspicious, as in typical case we would take 25 mL out of 250 mL of the borax solution, so while the concentration is the same in both 250 mL and 25 mL, number of moles is not. Unless you diluted 25 mL to 250 mL, in which case yes, number of moles is identical - but the dilution is irrelevant.
Molar mass Na2B4O7·10H20 = 381.37 g/mol
How do you know it is 10H2O? I thought it is up to you to determine this number?
You have correctly calculated mass of the anhydrous borax from the titration result, you know the initial mass of the sample. These two numbers allow easy calculation of mass of water present in the original sample.
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Thank you for the reply, not on a computer just yet so il add the procedure when I get the chance.
As for the the number being the same moles for 250cm3 and 25cm3 would I simply divide the answer i got for 250cm3 by 10?
As for knowing the value of x in the word equation I remember seeing the answer on yahoo answers when someone asked something similar. I totally forgot to expand that part and see how they got to that conclusion. How would I go about finding out that the x value is 10?
Thanks
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As for the the number being the same moles for 250cm3 and 25cm3 would I simply divide the answer i got for 250cm3 by 10?
Yes
Molarity of HCl solution = 0.1773M[/i]
The questions which I answered were:
Number of moles of HCl used
= (21.65/1000) * 0.1173 = 0.002539545 moles
You have made a careless mistake here, and it affected the whole calculation.
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Thank you I should really be proof reading this numerous times before posting. Thanks again 0.1173M is the actual Molarity of the HCl, I made a mistake when posting it under "some data relating to experiment" so the calculation haven't been affected. But thanks for pointing that out.
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You have correctly calculated mass of the anhydrous borax from the titration result, you know the initial mass of the sample. These two numbers allow easy calculation of mass of water present in the original sample.
Ok. As Borek pointed above, you can calculate the mass of water directly which the other masses. Also, try to think of an expression to represent the molar mass of Na2B4O7·xH2O (Hint: In terms of the x)
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You have correctly calculated mass of the anhydrous borax from the titration result, you know the initial mass of the sample. These two numbers allow easy calculation of mass of water present in the original sample.
Ok. As Borek pointed above, you can calculate the mass of water directly which the other masses. Also, try to think of an expression to represent the molar mass of Na2B4O7·xH2O (Hint: In terms of the x)
Weight of water in the borax sample:
4.9865 -2.616865 = 2.36965g
Doesn’t the above give me the answer to the amount of water present in the borax sample?
Also the expression I came up with to represent the molar mass of Na2B4O7·xH2O is:
\begin{equation} 381.3721 = Na_{2}B_{4}O_{7}.xH_{2}O \end{equation}
\begin{equation} 381.3721 = 1\times 201.22 + 18.01528x \end{equation}
\begin{equation} 180.1521 = 18.01528x \end{equation}
\begin{equation} x=\frac{180.1521}{18.01528} \end{equation}
\begin{equation} x = 10 (rounded) \end{equation}
So the moles of x.H2O is 10H2O
Is the following correct?
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The number 2.616865g and 381.3721 g mol-1 is based on the assumption that the formula of that compound is Na2B4O7·10H2O. How do you know that the number is 10?
By the way, do you really know what is mean by 'number of molecules of water of crystallisation'? It is the x in Na2B4O7·xH2O.
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The number 2.616865g and 381.3721 g mol-1 is based on the assumption that the formula of that compound is Na2B4O7·10H2O. How do you know that the number is 10?
By the way, do you really know what is mean by 'number of molecules of water of crystallisation'? It is the x in Na2B4O7·xH2O.
The reason I know the number is 10 is because the experiment was about Borax, and its formula is Na2B4O7·10H2O. Is it possible to calculate this number not knowing what is it?
Also I did not know the x in Na2B4O7·xH2O stood for "number of molecules of water of crystallisation", what does the y in yH2O stand for?
My attempt at calculating the number of molecules of water of crystallisation was:
(2.36965/201.22) * 6.022 x 1023 = 7.0995 * 1021
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https://en.wikipedia.org/wiki/Water_of_crystallization
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Where am I going wrong? Any input will be appreciated, feels like I'm at a dead end with this one.
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This is in a way funny, as you did many things correctly, just for some reason you do your best to ignore the experimental data.
Looks like the task here is to experimentally determine the x in the Na2B4O7·xH2O formula.
Use titration to determine the anhydrous borax mass (you almost did it in your opening post, you were just off by a factor of ten, as you ignored fact that your titrated sample is 1/10 of the total sample).
Use this mass to calculate mass of water in the original sample (calculate it not from the formula, but from the experimental data).
Then find x using exactly the same reasoning you used when you got masses of anhydrous part and mass of water from the formula - just use your experimental data for calculations.
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This is in a way funny, as you did many things correctly, just for some reason you do your best to ignore the experimental data.
Looks like the task here is to experimentally determine the x in the Na2B4O7·xH2O formula.
Use titration to determine the anhydrous borax mass (you almost did it in your opening post, you were just off by a factor of ten, as you ignored fact that your titrated sample is 1/10 of the total sample).
Use this mass to calculate mass of water in the original sample (calculate it not from the formula, but from the experimental data).
Then find x using exactly the same reasoning you used when you got masses of anhydrous part and mass of water from the formula - just use your experimental data for calculations.
I'm assuming that when you mentioned "off by a factor of 10" you were referring to:
Number of moles of borax in 25 cm3 solution
= 0.0012697725 moles
Number of moles in 250cm3
= 0.012697725 moles
This error has been fixed. From what I understand to calculate the:
Weight of Na22B4O7 (Molar Mass 201.22) in sample of borax
= 201.22g * 0.012697725= 2.555036225g
Then to calculate the mass of water in original sample I use the weight found above minus the weight of Borax which was 4.9865g.
Weight of water in the borax sample
4.9865 - 2.555036225 = 2.431463775g
Will have a go at calculating x now and post when I think I've got something, thanks for the help
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My attempt at solving for x is:
Na2BO7:xH2O
201.22 :18.01528
2.555036225:2.431463775
0.012699345:0.1349667
1:9.84040515
1:10
So x = 10
If this was right to find out the "number of molecules of water of crystallisation in Borax do I simply take the "Weight of water in the borax sample " answer and multiply it by 10 since x=10 and then multiply that by Avogadro's number? Thanks
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I think I finally see where the confusion is.
Question is a bit ambiguous. "The number of molecules of water of crystallization in Borax" can mean both x in ·xH2O (and I believe this is what they are asking for), and the number of molecules in the whole sample.
What you did is OK, that's the correct way of finding x.
To calculate total number of molecules it is enough that you have determined mass of water in the sample from the experimental data.
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I think I finally see where the confusion is.
Question is a bit ambiguous. "The number of molecules of water of crystallization in Borax" can mean both x in ·xH2O (and I believe this is what they are asking for), and the number of molecules in the whole sample.
What you did is OK, that's the correct way of finding x.
To calculate total number of molecules it is enough that you have determined mass of water in the sample from the experimental data.
If I was to find the "number of molecules of water of crystallisation in Borax" do I simply take the "Weight of water in the borax sample " answer and multiply it by 10 since x=10 and then multiply that by Avogadro's number? Or would I take the "Weight of water in the borax sample " and multiply this directly by Avogadro's number? Cheers
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Say you have 2 g of water.
Does the number of molecules depend on where the water comes from and what was the formula of the original compound that contained it?
Is 2 g of water taken from CuSO4·5H2O different from 2 g of water taken from Na2B4O7·10H2O?
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Say you have 2 g of water.
Does the number of molecules depend on where the water comes from and what was the formula of the original compound that contained it?
Is 2 g of water taken from CuSO4·5H2O different from 2 g of water taken from Na2B4O7·10H2O?
No it doesn't matter because to know the amount of molecules present when given grams of a substance you would need to:
1. Divide the grams of the substance by its molar mass which gives the moles of a substance.
2. Then you need to multiply this number by Avogadro's number to get the amount of molecules.
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Do you see how it answers your question?
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Do you see how it answers your question?
Yes I see, I just need to:
1. Divide the grams of the substance by its molar mass which gives the moles of a substance.
2. Then you need to multiply this number by Avogadro's number to get the amount of molecules.
1. 2.431463775/18.01528 = 0.1349667491 moles
2. 0.1349667491* 6.022 x 1023 = 8.1276976 x 1022
Thanks for all the help finally got there 8) Once again cheers.