Question

A solid hydrate of Ca(NO3)2 is to be analyzed (Ca(NO3)2 •XH2O). A sample of the hydrate is weighed in a clean, dry and covered crucible. When empty, the crucible and cover are heated to a constant weight of 30.2016 g. The mass of the crucible and sample is 33.2255 g. The crucible with sample is heated (to drive off the water of hydration) to a final constant weight of 32.2995 g

Table 1: Mass Data for the Analysis of a Hydrate

Heating # / Crucible Mass (g) / With Hydrate (g)(no heating) / With Anhydr. (g)

0 / - / 33.2255 / -

1 / 30.2115 / - / 32.6777  

2 / 30.2018 / - / 32.5888

3 / 30.2016 / - / 32.2997

4 / - / - / 32.2995

Calculate the following from the data given above (using last value in each column = Constant Mass):

1. Mass of hydrate sample

2. Mass of water driven off by the heating

3. Mass of the anhydrous Ca(NO3)2

4. Percent water in the hydrate (from masses in #2 and #1)

5. Moles water in the sample (use 18.015 g/mol for water)

6. Moles Ca(NO3)2 in the sample

7. Ratio of moles H2O to moles Ca(NO3)2 (rounded to a whole number)

8. Formula of the hydrate

9. Percent water as determined from formula

10. % error assuming % water from formula to be the accepted value

Answer 1

1. Mass of hydrate sample = mass of crucible with sample - mass of empty crucible

= 33.2255-30.2016 = 3.0239 g

2. Mass of water driven off by the heating = mass of crucible with sample before heating - mass of crucible with sample after heating

= 33.2255-32.2995 = 0.926 g

3. Mass of anhydrous Ca(NO3)2 = Mass of hydrate sample - Mass of water driven off by the heating

= 3.0239-0.9260 = 2.0979 g

4. Percent water in the hydrate = (Mass of water driven off by the heating/Mass of hydrate sample)*100

= (0.9260/3.0239)*100 = 30.62 %

5. Mass of water driven off = 0.9260 g

Molar mass of water = 18.015 g/mol

Moles of water in the sample = mass/molar mass = 0.9260/18.015 = 0.051 mol

6. Mass of anhydrous Ca(NO3)2 = 2.0979 g

Molar mass of Ca(NO3)2 = 164.088 g/mol

Moles Ca(NO3)2 in the sample = 2.0979/164.088 = 0.013

7. Ratio of moles H2O to moles Ca(NO3)2 = moles of H2O/moles of Ca(NO3)2 = 0.051/0.013 = 3.9 = 4(approx)

8. Formula of Hydrate = Ca(NO3)2.4H2O

9. Percent water as determined from formula:

Molar mass of Ca(NO3)2.4H2O = 236.146 g/mol

Molar mass of H2O = 18.015

There are 4 water molecules, so mass of H2O in one mole of molecule = 4*18.015 = 72.06 g

Percent water = (72.06/236.146)*100 = 30.51 %

10. % error = (Theoretical value-experimental value/experimental value)*100

= [(30.51-30.62)/30.51]*100 = 0.36 %