[Julie Smith]

hi,

i have a chemisty question that I'm kind of stuck on so any help is appreciated. Here's the question:

consider the following experimental data to determine the empirical formula of calcium hydroxide by gravimetric analysis. You add 50mL of .200M calcium nitrate solution to 50mL of .500M sodium hydroxide solution (excess reagent) and filter out the resulting calcium hydroxide precipitate at 25 degrees celcius. The calcium hydroxide precipate is dried out and found to have a mass of 590mg.

a) determine the empirical formula of calcium hydroxide based on this experimental data.

b) the solubility of calcium hydroxide in water at 25 degrees celcius is 150mg of calcium hydroxide per 100mL of water. taking this into account in the above gravimetric analysis experiment, recalculate the empirical formula of calcium hydroxide.

[Kryolith]

Please post your own ideas, according to the forum rules

[Rabn]

Show us what you've done and we will guide you towards success.

[Julie Smith]

ok, so what i've done for a) is that I converted the Molarity of calcium nitrate and sodium hydroxide into seperate moles of calcium and hydroxide. When i did that I got 2 moles for calcium and 2.5 moles for hydroxide, so I multiplied both by 2 to get whole numbers. In the original question, they give me mg but I don't know if I'm supposed to use that for part a.

For part b) I figured out the moles of calcium and hydroxide seperately from the mg given, but i'm not sure if that's right.

Any help is appreciated.

[Padfoot]

ok, so what i've done for a) is that I converted the Molarity of calcium nitrate and sodium hydroxide into seperate moles of calcium and hydroxide.

Correct, converting to moles is the first step.  For this you use the formula n=cv though (check your textbook).  Next step is a balanced reaction equation.  How many moles of Calcuim hydroxide form your given amount of reactants?

For part b) I figured out the moles of calcium and hydroxide seperately from the mg given, but i'm not sure if that's right.

You are forgetting a step.  Think about what the solubility info is telling you.  What changes do you have to make to the 590mg?

[Julie Smith]

ok, so for part b) I'm supposed to change the mg into grams and then find the moles right? However, how do I know out of the two mg choices they give me, which one to choose? Or do I use both? From there, then I would have to find the moles again right?

Also, is their any significance to the 25 degrees they give me?

Thanks

[Padfoot]

This is what you were doing before - what about the forgotten step 
 

The solubilty of Ca(OH)2 is 150mg per 100mL of water so in our case 150mg of it is in dissociated form.  From this, we can say there was really (150+590)mg of Ca(OH)2 formed.  This is the amount you continue with. 

Also, is their any significance to the 25 degrees they give me?
Thanks

Solubility changes with temp, they have only included it to show that the solubility they gave is applicable.

[Julie Smith]

haha, ok I figured out the "forgotten step"

Thanks for the help Brett!