[laddoo12]

consider the fallowing experimental data to determine the formula of calcium hydroxide by gravimetric analysis. you add 0.050 L of .200M calcium nitrate solution to 0.050L of 0.5M sodium hydroxide solution(excess reagent) and filter out the resulting calcium hydroxide precipitate at 25 Celsius. The calcium hydroxide precipitate is dried out and found to have a mass of 590 mg (0.59 grams)

A) determine the empirical formula of calcium hydroxide based on this expire mental data

B) the solubility of calcium hydroxide in water at 25 Celsius is .150 grams of calcium hydroxide per 100ml of water (i.e .150 grams remain dissolved in water and didn't precipitate out). Taking this into account in the above gravimetric analysis experiment, recalculate the empirical formula of calcium hydroxide.

could some please give me a step by step guide to approach this question and a solution I'm really having trouble to even start it .I'd love you for life

please and thank you

[laddoo12]

What i did was

a)Ca(NO3)2 + 2 NaOH = Ca(OH)2 + 2 NaNO3
NaOH is in excess so Ca(NO3)2 is the limiting reagent
moles of Ca(NO3)2 = .2 x .05 = .01 moles
from the equation we see the moles of Ca(OH)2 formed = .01
these have mass of .590g so 1 mole has mass of .590 / .01 = 59 g / mole
b) when we combine the 50 ml of NaOH with the 50 mL of NaNO3 we have 100 mL of solution
at the given temperature we have 150 mg of Ca(OH)2 still in solution so total actually formed
= .590 + .150 g = .740 g
mass of 1 mole is now .740 / .01 = 74 g


but I'm not sure if this is correct could someone please tell me if i went wrong somewhere

[Borek]

OK so far.

[laddoo12]

if so how would i use that data to find out the empirical formula of the compound

[Borek]

Find x in Ca(OH)_x that fits found molar masses.