Record results in the blanks below and perform any calculations. Mass of empty d
ID: 1048629 • Letter: R
Question
Record results in the blanks below and perform any calculations. Mass of empty drying dish 41.48 g Mass of drying dish + sample 42.48 g Mass of sample 1 g Mass of drying dish + sample after first 5min heating 42.33 g Mass of drying dish + sample after second 5 min heating 42.32 g Mass of water (lost) 0.16 g Mass of Bacl_2 (anhydrous =n without water) Calculate the number of moles of Bacl_2 show work! Calculate the number of moles of water, show work! Calculate the value "X" in the chemical formula: Bacl_2.XH_2O of moles of water, show work! X = Calculate the number of moles of Bacl_2.ZH_2O with which you started, Show work! Calculate the %(Wt/Wt) water on original sample:Explanation / Answer
Initial (hydrated) mass of sample = 1.00 g
Mass of water lost = 0.16 gram
1. Mass of anhydrous sample (pure BaCl2 only) = (1.00 – 0.16) g = 0.84 g
So, mass of BaCl2 (anhydrous = without water) = 0.84 g
2. number of moles of BaCl2 = mass of anhydrous BaCl2 / molecular mass of BaCl2
= 0.84 g / 208.23 g mol-1
= 0.004034000864 moles
3. Number of moles of water = mass of water lost / molecular mass of water
= 0.16 g / 18.015 g mol-1
= 0.00888 moles
4. From 2 and 3.
0.00888 moles of H2O is released by 0.004034 moles BaCl2 during desiccation.
So, number of moles of water molecules associated per mol BaCl2
= 0.00888 / 0.004034 = 2.2
That, 1 mol BaCl2 is hydrated with 2.2 moles ware.
Thus, X = 2.2
Actual empirical formula of BaCl2 in the sample = BaCl2. 2.2 H2O
Ans. 5. Number of moles of BaCl2. 2.2 H2O in initial (hydrated sample)
= Initial mass of sample / molecular mass of BaCl2. 2.2 H2O
= 1.00 g / 247.863 g mol-1
= 0.004034486793 moles
Ans. 6. % (wt/wt) in original sample = (mass of water lost / initial mass of sample) x 100
=( 0.16 g/ 1.00 g) x 100 = 16.0 %
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