2. A chemist heated an unknown hydrate, MgSO, X HO, in a crucible. The water was
ID: 555120 • Letter: 2
Question
2. A chemist heated an unknown hydrate, MgSO, X HO, in a crucible. The water was driven off, leaving the anhydrous salt in the crucible. After cooling, the mass of the anhydrous salt and water lost were calculated. The mass of the anhydrous salt was 1.570 grams g and the mass of the water lost was 1.641 g What was the mass of the unknown? Show your calculations. (25 points) Your lab book instructed you last week to heat and cool off your hydrated sample three times. Why three times not one time? Explain. Give me a scenario with specific example where you may have to heat and cool off the sample five times instead of three. (20 points) 3.
Explanation / Answer
2.
Mass of anhydrous salt = 1.570 g
Mass of the water lost = 1.641 g
Mass of the unknown = (Mass of anhydrous salt) + (Mass of the water lost)
= 1.570 g + 1.641 g
= 3.211 g
4.
3.
Mass of potassium chlorate = (Mass of test tube + potassium chlorate) - Mass of empty test tube
= 15.7 g – 13.7 g
= 2.0 g
6. Mass of oxygen evolved
= (Mass of test tube and total contents including catalyst before heating) - (Mass of test tube and total contents including catalyst after heating) -
= 17.7 g – 17.2 g
= 0.5 g
7.
% of oxygen from the chlorate sample = [ (Mass of oxygen) / (Mass of chlorate sample) ] x 100
= (0.5 g / 2.0 g) x 100
= 25 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.