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Offline Tittywahah

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Stoichiometry help asked
« on: January 22, 2015, 09:57:16 AM »
I am having so much mental agony on this so if someone has to correct my maths, please could you show me the steps.  So grateful really.  Iodometric titration done of copper nitrate with unknown hydrate, is it hexa, deca, or tri or something else?

2.5 g of Copper Nitrate added to water to make up a 100 mL solution
From this mother solution 30 mL was extracted and titrated (dilution factor of 3.33)
80 mL of 0.01 M thiosulphate titrated to the endpoint.

My reasoning:
0.08 L titrant x 0.01 M Thiosulphate = 0.0008 moles of Cu2+ in 30 mL sample
Dilution factor is 3.33 therefore this x 0.0008 = 0.00266 moles in the 100 mL original solution.
0.00266 x 187.55 g/mol (anhydrous copper nitrate) = 0.5 g of Cu(NO3)2 in that 100 mL

Therefore logic dictates that 2.5 g - 0.5 g = 2.0 g of this nitrate sample is water
This means that 2.5/0.5 = 5  which means that my sample is Cu(NO3)2.5H2O

Thankyou

Offline mjc123

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Re: Stoichiometry help asked
« Reply #1 on: January 22, 2015, 10:10:35 AM »
Quote
This means that 2.5/0.5 = 5  which means that my sample is Cu(NO3)2.5H2O
Where does this come from?
You have 2.0 g water. How many moles of water is this? How many moles per mole of copper? You will find the answer is over 40, which is clearly wrong. Your calculations look right (up to this point), so look through them and see whether you have copied anything down wrongly.

Offline Tittywahah

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Re: Stoichiometry help asked
« Reply #2 on: January 22, 2015, 10:19:48 AM »
Quote
This means that 2.5/0.5 = 5  which means that my sample is Cu(NO3)2.5H2O
Where does this come from?
You have 2.0 g water. How many moles of water is this? How many moles per mole of copper?
2 g water = 0.1 mole, but I have no idea what you mean by this: """How many moles per mole of copper"""?  I get the 2.5 g from the Original weight of the copper nitrate sample I would have taken out of the jar.  I am using fictitious numbers because all my titrations failed due to some weird thing with the corn starch.  I wanted to get the maths correct so I could get rid of this confusion then I would go back and re-do the titration in a relaxed manner.

Did you mean this: 2 g water = 0.1 mole  and  0.5 g Cu. Nitrate = 0.0027 mole
Therefore 0.1/0.0027 = 37 which hypothetically means .37H2O
« Last Edit: January 22, 2015, 10:44:25 AM by Tittywahah »

Offline mjc123

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Re: Stoichiometry help asked
« Reply #3 on: January 22, 2015, 12:09:23 PM »
Quote
Did you mean this: 2 g water = 0.1 mole  and  0.5 g Cu. Nitrate = 0.0027 mole
Therefore 0.1/0.0027 = 37 which hypothetically means .37H2O
Yes
Quote
I am using fictitious numbers
Oh well. No wonder you get a wrong answer. But your maths was correct, up to the last step.

Offline Tittywahah

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Re: Stoichiometry help asked
« Reply #4 on: January 22, 2015, 12:20:43 PM »
Quote
Did you mean this: 2 g water = 0.1 mole  and  0.5 g Cu. Nitrate = 0.0027 mole
Therefore 0.1/0.0027 = 37 which hypothetically means .37H2O
Yes
Quote
I am using fictitious numbers
Oh well. No wonder you get a wrong answer. But your maths was correct, up to the last step.
Well I am relieved. Chemistry is self taught and the stoichiometry of things is something I have had to work at triple hard to get right.  So thanks.

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