Need help finding the formula of a compound

[bohemian]

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 H_{2}C_{2}O_{4} \bullet xH_{2}O. 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.

 

[DrMark]

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 H_{2}C_{2}O_{4} \bullet xH_{2}O. 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 ;)

 

[Borek]

DrMark pointed you in the right direction. Retail oxalic acid is usually bihydrate - that may, or may not help check your results ;)

 

[bohemian]

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.

 

[Borek]

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.