Calculation of water of crystallization

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of water of crystallization in sodium sulfate decahydrate (Na2SO4·10H2O) after an experiment where the substance was heated to evaporate water. Participants explore the calculations related to the moles of water evaporated and the resulting formula after evaporation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The initial poster calculates that 0.2 grams of water evaporated, leading to an estimate of 0.011 moles of water evaporated.
  • Some participants suggest that the assumption of starting with decahydrate may not be valid due to various factors affecting the sample's history and water content.
  • Concerns are raised about the accuracy of the calculations, particularly regarding the presentation of significant figures and rounding.
  • The initial poster acknowledges the feedback and expresses intent to improve accuracy in future calculations.

Areas of Agreement / Disagreement

Participants generally agree on the calculations being mostly correct but highlight uncertainties regarding the initial assumptions about the sample and the accuracy of the presented numbers. There is no consensus on the validity of the initial assumption regarding the decahydrate form.

Contextual Notes

Limitations include the uncertainty about the initial state of the sample and the influence of external factors such as humidity and storage conditions on the amount of water present.

lolbits
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Homework Statement


We did an experiment at school. We heated [tex]Na_2SO_4*10H_2O[/tex] so that the water evaportated. We started with 2 gram of the substance, and ended up with 1.8 grams. Unfortunately, not all the water did evaporate because of the equipment we used at school. So the question is, a) how many moles of water evaporated and b) what would the new formula be?

Homework Equations



mol = m(g) / M_m

The Attempt at a Solution



a) Moles of water evaporated:
Know that 0.2 grams of water evaporated.
0.2/18 = 0.011 mol H2O evaporated.
That would be about 2 moles for each mol Na2SO4.

b) 2grams [tex]Na_2SO_4*10H_2O[/tex] [tex]\rightarrow[/tex] [tex]Na_2SO_4*xH_2O[/tex]

[tex]M_m[/tex] of [tex]Na_2SO_4*10H_2O[/tex] is [tex]322g/mol[/tex]

[tex]\frac{2g}{322g/mol}[/tex] gives [tex]0.0062 mol[/tex]

[tex]M_m[/tex] of [tex]Na_2SO_4[/tex] = [tex]142 g/mol[/tex]

[tex]\frac{1.8g}{0.0062 mol}=290.32 g/mol[/tex]

[tex]290.32g/mol - 142g/mol = 148g/mol[/tex]

[tex]M_m[/tex] of [tex]H_2O=18[/tex]

[tex]\frac{142 g/mol}{18g/mol} = 8[/tex]

So the new formula would be:

[tex]Na_2SO_4*8H_2O[/tex]

Is this right?
 
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No one? This should not be hard..
 
Calculations are almost OK, there is one weak point - you can't be sure you started with decahydrate, as amount of water may depend on many factors (including sample history - was it kept in open/closed bottle, what was humidity at the time and so on). But you can't check it, so your assumption is not bad.

The only thing I don't like is that you ignore accuracy - that is, some numbers are presented with guard digits, while other are presented already rounded down in a rather unpredictable way.

--
 
Borek said:
Calculations are almost OK, there is one weak point - you can't be sure you started with decahydrate, as amount of water may depend on many factors (including sample history - was it kept in open/closed bottle, what was humidity at the time and so on). But you can't check it, so your assumption is not bad.

The only thing I don't like is that you ignore accuracy - that is, some numbers are presented with guard digits, while other are presented already rounded down in a rather unpredictable way.

--
www.titrations.info, www.chemistry-quizzes.info, www.ph-meter.info

Thank you for replying!

What I wrote was just quick. I will be more accurate next time. But again, thanks!
 

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