Question

1) You are given a solid hydrate. You dry and weigh a sample tube. You put in the hydrate and reweigh.
You heat the tube and contents to drive off any water then cool and reweigh.
You obtain the following fictional data:

Mass of sample tube and hydrate	37.9143 g
Mass of sample tube and hydrate after heating and cooling	36.8824 g
Mass of empty sample tube	35.6254 g
Molecular weight of the anhydrous salt	112.0 g/mol

Calculate:

Mass of hydrate weighed out	_______________ g	(a)
Mass of water lost after heating	_______________ g	(b)
Per cent water in the hydrate	_______________ %	(c)
Moles of water lost during heating	_______________ mol	(d)
Moles of anhydrous salt remaining after heating	_______________ mol	(e)
Ratio-moles of water per mole of anhydrous salt	_______________	(f)

Report this ratio (f) as a single number to two decimal places, e.g. the ratio of 2.00 mole water and 1.50 mol salt is 1.33

Answer 1

Mass of hydrate weighed out = Mass of sample tube and hydrate - Mass of empty sample tube

Mass of hydrate weighed out = 37.9143 g - 35.6254 g = 2.2889 g

Mass of water lost after heating = Mass of sample tube and hydrate - Mass of sample tube and hydrate after heating and cooling

Mass of water lost after heating = 37.9143 g - 36.8824 g = 1.0319 g

Per cent water in hydrate = ( Mass of water lost after heating / Mass of hydrated salt ) * 100

Per cent water in hydrate = ( 1.0319 g / 2.2889 g ) * 100

Per cent water in hydrate = 45.08%

Moles of water lost during heating = Mass of water lost during heating / molar mass = 1.0319 g / 18.0153 g/mol

Moles of water lost during heating = 0.0573 mol

Moles of anhydrous salt remaining after heating = Mass of anhydrous salt / molar mass of anhydrous salt

Moles of anhydrous salt remaining after heating = (36.8824 - 35.6254) g / 112 g/mol = 0.0112 mol

0.0112 mol of anhydrous salt contains 0.0573 mol of water

1 mol of anhydrous salt contains 0.0573 / 0.0112 = 5.12

Ratio of moles of water per mole of anhydrous salt = 5.12