To isolate the barium ion the hydrate was reacted with sodium carbonate to from

Question

To isolate the barium ion the hydrate was reacted with sodium carbonate to from a white precipitate of barium carbonate. Data from this reaction is given below:

Initial mass of hydrated barium hydroxide: 0.471g

Mass of barium carbonate precipitate collected: 0.295g

To determine the amount of water present in the hydrate another sample of the hydrate was weight and heated in a crucible until the mass no longer changed. Data from this procedure is given below:

Mass of empty crucible: 10.281g

Mass of crucible + hydrate: 12.621g

Mass of crucible + anhydrous substance after heating: 11.553g

Use the data given above to determine the empirical formula of the hydrate.

Explanation / Answer

Moles of Ba = moles of BaCO3 = mass/molar mass of BaCO3

= 0.295/197.34 = 0.001495 mol

Mass of Ba = moles x molar mass of Ba

= 0.001495 x 137.33 = 0.2053 g

Mass% of barium = mass of Ba/mass of sample 1 x 100%

= 0.2053/0.471 x 100% = 43.6%


Mass of water = 12.621 - 11.553 = 1.068 g

Mass of sample 2 = 12.621 - 10.281 = 2.340 g

Mass% of water = mass of water/mass of sample 2 x 100%

= 1.068/2.340 x 100% = 45.6%


Mass% of hydroxide = 100 - mass% of barium - mass% of water

= 100 - 43.6 - 45.6 = 10.8%


Since moles = mass/molar mass:

Moles of Ba : OH : H2O = 43.6/137.33 : 10.8/17.01 : 45.6/18.02

= 0.317 : 0.635 : 2.53

= 1 : 2 : 8


Empirical formula is: Ba(OH)2.8H2O

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