Hydrated copper (II) sulfate mass calculation

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Discussion Overview

The discussion revolves around the calculation of the percent by mass of water in a sample of hydrated copper (II) sulfate, including considerations of the chemical reaction involved in dehydration and the implications of incomplete conversion of the hydrate to its anhydrous form.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the percent by mass of water in hydrated copper (II) sulfate but is challenged by others regarding the accuracy of their approach.
  • Another participant suggests calculating the mass of copper sulfate before and after roasting to clarify the process.
  • A participant describes expected observations when heating a hydrated ionic compound, noting the presence of water droplets and steam, and discusses the implications of incomplete conversion to the anhydrous compound.
  • There is a discussion about how not fully converting the hydrate would affect the molecular formula, with some suggesting that more water molecules would be present.
  • A later reply clarifies that incomplete conversion means that the mass of the anhydrous compound would not be pure, affecting the calculated percent by mass of water.

Areas of Agreement / Disagreement

Participants express disagreement regarding the initial calculation of percent by mass and the implications of incomplete conversion of the hydrate. Multiple competing views remain on how these factors affect the results.

Contextual Notes

Participants highlight the importance of accurately measuring the mass of the anhydrous compound and the potential for varying results based on the completeness of the dehydration process. There are unresolved aspects regarding the exact calculations and assumptions made in the initial approach.

Aya
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calculate the precent by mass of water in a sample of hydrated copper (II) sulfate.

mass of beaker-165.3g
mass of beaker with bluestone-173.9g
mass of dehydrated bluestone-170.7g

Would this be a ballanced equation for this reaction?
CuSO4•5H2O+ heat= CuSO4+5H2O


so would you do this

%by mass=mass of part/mass of whole
=90.1g/249.7g/molx100
= 36%

is this correct?
 
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No. But it is so wrong I have no idea how to help you :(

Start calculating mass of the copper sulfate before roasting and after roasting.
 
No. But it is so wrong I have no idea how to help you :(

lol, its ok i figured that part out

i have different questions now

Suppose you heated a sample of hydrated ionic compound in a test tube. What might you expect to see inside the test tube, near the mount of the test tube? Explain

You would see water droplets and steam. This is because a hydrate indicates the presents of water, when you heat a hydrated ionic compound the water will separate from the compound, and you will see some of the water collecting near the mount of the test tube. The steam could be a result of the water, or the compound evaporating

would that be right?

****
Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the molecular formula you determined?

you would get more molecules of water?
 
or the compound evaporating

Ionic compounds generally are nonvolatile solids and have high boiling points. Stick to the water answer.

Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the molecular formula you determined?

you would get more molecules of water?

No, think about what the question is saying here. Some water molecules are leaving the compound, just not all of them. When all the water has gone from the compound, then you will have the anhydrous product.
 
^ok thanks
 
7. Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the calculated percent by mass of water in the compound?

If the hydrate was not completely converted to the anhydrous compound, the mass of anhydrous Cu(II)SO4 would increase because it would not be pure anhydrous Cu(II)SO4, instead it would be CuSO4•5H2O, in other words you would be weighing the compound plus water. In turn, this would change the mass of H2O, making to higher (for example instead of 3.2g it might be 4.2g.) the results of the percent composition calculation would also change to a higher number, for example %= (4.2/8.6)100 would equal 48.8% of water, compared to 37.2% of water, which is what we got by completely convert the hydrate to the anhydrous compound.
 

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