1) You are given a solid hydrate. You dry and weigh a sample tube. You put in th

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:

Calculate:

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

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

Explanation / Answer

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

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