[laddoo12]

Lab of Empirical formula of a compound ca(Oh)n2

*im not sure if you do this but with the .5 grams Cax(OH)x would you go .5/.250 x .20 because you transport the solid 3 times into 3 things giving you 0.04 grams of it as a number to do your calculations with :S


1.Weigh 0.5g calcium hydroxide into 50ml beaker
2.Pipet 15ml of 2.0M HCl into the 50ml beaker, dissolve the calcium hydroxide, then transfer contents into a 250ml volumetric flask, then dilute the flask to the mark.
3.Pipet three 20ml samples into erlenmeyer flasks, and add indicator drops
4.Titrate excess HCl in the three samples with standard NaOH solution (0.1M), and thus record the volume of NaOH used.

The prelab says the following:

The student will react a known mass of calcium hydroxide with a known (excess) amount of HCl. The excess HCl is then back titrated with a standard NaOH solution. From this information the student can calculate the moles of hydroxide in the calcium hydroxide as well as the mass of hydroxide in the calcium hydroxide sample. From the known mass of calcium hydroxide the student can thus calculate the mass of calcium in the original sample and thus the moles of calcium in the calcium hydroxide sample. Based on the moles of hydroxide and calcium in the original calcium hydroxide sample the student can thus establish the empirical formula of calcium hydroxide.

Keep in mind i can't use the periodic table to figure out the empirical formula

what i did so far was

1. Hcl+NaOH-->Nacl+h20
2. initical Hcl 0.015x2.0M/.250Lx0.02 = (0.0024moles HCL initial)
3.Excess HCL (1 to 1 mole ratio with NAOH thus (naoh= .1Mx0.00132ml) (Excess HCl=0.000132)
4. Total Hcl reacted = 0.0024-0.000132=0.002268 moles
5 for reference i wrote (2HCL+Ca(OH)2-->Cacl2+2h20)
6. so i could see that H+ ions is 1 to 1 with OH- thus
7. 0.002268moles OH- x 17 = 0.038556 grams OH Which I'm not sure is right :S


are the above calculations right :S and f not could someone point out where i went wrong
what would i do next after finding the grams of OH to find the empirical formula   I'm quite confused

[fledarmus]

I've often found that the best way to deal with the confusion of a problem of this sort is to break it down into things you know and things you don't, stepwise. So let's look at your problem. What are you specifically given and what are you trying to find out?

As I understand it, you know that you have some compound of calcium and hydroxide, and you don't know how much of each. Basically, this means you don't know the oxidation state of calcium, right? You are allowed to assume that hydroxide is OH^-? So you are trying to identify x and y for a molecule with empirical formula Ca_x(OH)_y.

Now, what did you do in the lab?

1. Weigh 0.5g calcium hydroxide into 50 mL beaker.

Using your variables x and y, how many moles of calcium and hydroxide would this represent?

2. Pipet 15mL of 2.0M HCl into the 50 mL beaker, dissolve the calcium hydroxide...

What chemical reaction(s) occurred here? Now how much calcium, hydroxide, and hydrochloric acid are in the flask? (using your variables)

...then transfer contents into a 250mL volumetric flask, then dilute the flask to the mark.
3. Pipet three 20mL samples into erlenmeyer flasks...

How much of each component is now present in each sample?

and so forth. If you work your way through forwards like this, it will make it easier to calculate your way backwards to your answer once you've collected your data.

[laddoo12]

i don't think I'm aloud to do that and I'm not quite sure how to put in variables
i must follow the prelab

which is

The student will react a known mass of calcium hydroxide with a known (excess) amount of HCl. The excess HCl is then back titrated with a standard NaOH solution. From this information the student can calculate the moles of hydroxide in the calcium hydroxide as well as the mass of hydroxide in the calcium hydroxide sample. From the known mass of calcium hydroxide the student can thus calculate the mass of calcium in the original sample and thus the moles of calcium in the calcium hydroxide sample. Based on the moles of hydroxide and calcium in the original calcium hydroxide sample the student can thus establish the empirical formula of calcium hydroxide.

which i did but I'm not sure if I'm doing it correctly, and whatever i do to my final answer is off or i don't even know how to do the calculations afterward to find the empirical formula

[laddoo12]

Or could you maybe lead me through what you mean a bit better by providing me how you do it with variables

[laddoo12]

So i got a answer from yahoo answers

I assume that the back-titration took 13.2 mL of 0.1 M NaOH. So
moles NaOH = M NaOH x L NaOH = (0.1)(0.0132) = 0.00132 moles, ten times more than you have above.

So then, total moles HCl reacted (in one 20-mL aliquot) = 0.0024 - 0.00132 = 0.00108 moles HCl
You're right, the mole ratio between H+ and OH- in the back titration is 1:1, so moles OH- neutralized (in one aliquot) = 0.00108 moles OH-.
0.00108 moles OH- x (17.0 g NaOH / 1 mole OH-) = 0.0184 g OH- in one aliquot

The amount of the original sample represented by one aliquot = 0.5000 g x (20 / 250) = 0.0400 g.

So g Ca2+ in one aliquot = g sample - g OH- = 0.0400 - 0.0184 = 0.0216 g Ca2+
0.0216 g Ca2+ x (1 mole Ca2+ / 40.1 g Ca2+) = 0.00054 moles Ca2+

So the molar ratio between OH- and Ca2+ = moles OH- / moles Ca2+ = 0.00108 / 0.00054 = 2!


is this the correct method of doing what i tried doing

[laddoo12]

Nvm iv solved it