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