Need help finding the formula of a compound

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Discussion Overview

The discussion revolves around determining the formula of a hydrated oxalic acid compound, specifically identifying the number of water molecules (x) in the formula H2C2O4 • xH2O. Participants are exploring the experimental setup involving titration and calculations based on the data provided.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a problem involving a bottle of hydrated oxalic acid and requests assistance in determining the formula based on a titration experiment.
  • Another participant suggests calculating the moles of NaOH used in the titration to find the corresponding moles of H2C2O4 and hints at using the mass of the solution to derive further information.
  • A third participant mentions that retail oxalic acid is typically a bihydrate, which may provide a reference point for checking results.
  • A later reply indicates that the participant has calculated the moles of NaOH and the moles of the solution but is uncertain how to proceed to find the number of water molecules.
  • Another participant advises finding the molar mass of the sample, noting the mass of the solid used and the molar concentration, while also reminding that oxalic acid is diprotic.

Areas of Agreement / Disagreement

Participants are generally engaged in a collaborative effort to solve the problem, but there is no consensus on the exact method to determine the number of water molecules in the hydrated compound. Multiple approaches and calculations are being discussed without a definitive resolution.

Contextual Notes

Participants express uncertainty regarding the application of their calculations and the relationship between the moles of NaOH and the hydrated oxalic acid. There are also references to the diprotic nature of oxalic acid, which may influence the calculations.

bohemian
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I need help finding the formula for the compound in this problem:

You have been given a bottle of hydrated oxalic acid (contains x number of water molecules) with a half-torn label, and have been asked to determine the exact formula of this compound. Your best guess on the formula of this molecule is [tex]H_{2}C_{2}O_{4} \bullet xH_{2}O[/tex]. To figure out the exact formula of this compund, you make a solution by taking 5.00g of this compound, dissolve it in exactly 250mL of water. You then remove 25.0mL of this solution and titrate it with 0.500 M sodium hydroxide. You find that the solution requires 15.9mL of the base for neutralization. Based on this experiment, determine the exact formula of the compound (you need to determine x).

If anyone can walk me through this and help me get the answer, I'd really appreciate it.
 
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bohemian said:
I need help finding the formula for the compound in this problem:

You have been given a bottle of hydrated oxalic acid (contains x number of water molecules) with a half-torn label, and have been asked to determine the exact formula of this compound. Your best guess on the formula of this molecule is [tex]H_{2}C_{2}O_{4} \bullet xH_{2}O[/tex]. To figure out the exact formula of this compund, you make a solution by taking 5.00g of this compound, dissolve it in exactly 250mL of water. You then remove 25.0mL of this solution and titrate it with 0.500 M sodium hydroxide. You find that the solution requires 15.9mL of the base for neutralization. Based on this experiment, determine the exact formula of the compound (you need to determine x).

If anyone can walk me through this and help me get the answer, I'd really appreciate it.

Not going to give you the answer, but some hints. How many moles of NaOH were used in the titration? So how many moles of H2C2O4 does that make ? You know the # of grams in the solution and the fraction used. Just due the math ;)
 
DrMark pointed you in the right direction. Retail oxalic acid is usually bihydrate - that may, or may not help check your results ;)
 
Last edited by a moderator:
Thanks for your help. However, I'll still a little stuck. I know that there are .0075 mols of NaOH and .318 mols of the 25 ml of solution were needed to neutralize it. The 25 ml of solution is 1/10 of the 250 ml that dissolved the 5 g. sample. That's pretty much all I know but I don't know how to use these numbers to find out how many molecules of water there are. Thanks again for your help.
 
Try to find out molar mass of the sample - you know the mass of solid used for solution preparation and you know what was the molar concentration. Remember that oxalic acid is diprotic.
 
Last edited by a moderator:

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