John FloresChristina GeeJosh RojoChemistry 11GDec 17, 2010

Formula of a Hydrate

Purpose

The purpose of this lab is to determine the percentage of water in an unknown hydrate, to determine the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt, and to write the empirical formula of the hydrate.

Materials

Refer to page 46 of Essential Experiments of Chemistry

Procedure

1. Put on your lab apron and safety goggles.2. Set up the pipestem triangle, iron ring, stand, and Bunsen burner as shown in Figure 4C-1

(on page 46). The ring should be 5-6 cm above the top of the burner. Place a clean, dry crucible with lid on the pipestem triangle, and heat with the burner for approximately 3 min to make sure that the crucible is dry. This lid should be slightly off so that any moisture present may escape. Do not use a luminous flame on the burner, as this may deposit carbon on it – make sure that the flame is blue.

3. Remove the burner and allow the crucible to cool for about 3 min. Check by bringing the back of your hand close to the crucible without touching it. If you can still feel the heat, it is not yet cool enough.

4. Determine the mass of the empty crucible and lid. Record the mass in your copy of Table 1. 5. Place the hydrate that you are assigned into the crucible until it is approximately one-third

full. Determine and record the mass of the crucible, lid, and hydrate.6. Place the crucible and contents on the pipestem triangle, arrange the lid slightly off, and

begin heating. Gradually increase the heat until the bottom of the crucible is a dull red. Maintain this temperature for 5 min. To get the maximum heat, position the ring on the stand so that the crucible is just above the top of the light blue inner cone of the flame.

7. Turn off the burner and allow the crucible to cool for about 5 min. Check as before to make sure it is cool enough, then determine and record the mass of the crucible, lid, and contents.

8. To make sure that all of the water is driven off, you should reheat the crucible for another 5 min. Cool as before, then determine and record the mass. If the masses that you determine in Steps 6 and 7 do not agree within 0.03 g, check with your instructor to see if there is time for one more heating. Use the lowest mass in your calculations.

9. After being satisfied with you final mass reading and the crucible has cooled, add a few drops of water to the contents of the crucible. Note any changes that occur.

Data and Observations

Table 1 – Experimental Results

Mass of empty crucible and lid 39.60 g

Mass of crucible, lid, and hydrate 45.77 g

Mass of hydrate 6.17 g

Mass of crucible, lid, and anhydrous salt (first heating)

42.84 g

Mass of crucible, lid, and anhydrous salt (second heating)

42.81 g

Mass of anhydrous salt 3.21 g

Mass of water given off 2.96 g

Mass of one mole of anhydrous salt 164.0 g/mol

Any changes that you observed when adding water to the crucible

No visible change

Observations during heatingHotter fire than average; hydrate liquefies and starts boiling aggressively; took off the fire momentarily, but put back on fire with

Analysis

1.2.96g H 2O

6.17g salt=48.0 % H 2O

2.3.21gansalt

1×

mol164.0g

=0.0196molansalt

3.2.96g H 2O

1×mol

18.0 g=0.164mol H 2O

4.0.164mol H 2O

0.0196molansalt=8.40mol H 2O /molansalt

5. AB∙8.5H 2O or 2 AB ∙17H 2O

Discussion Questions

1. One reason why this procedure may not yield good results for some hydrates may be because some compounds can boil at a lower temperature than water. Although, the boiling points of most ionic compounds are higher relative to water. Related to this is the possible decomposition of some compounds when faced with heat. If you apply too much heat at once, the salt will start decomposing, leading to inaccurate results. Also, there could be gaseous hydrates which this would not work at all for.

2.

Assume 100 g ÷0.350 Empirical FormulaNa 16.1g

1×mol

23.0g=0.700mol 2

Na2CO3 · 10H2OC 4.2g

1×mol

12.0 g=0.350mol 1

O 16.8g1

×mol

16.0g=1.05mol 3

H2O 62.9g1

×mol

18.0g=3.49mol 10

3.

The anhydrous salt’s mass is 2.15 g.

4.

Assume 100 g ÷0.8056 × 2 Empirical FormulaNa 18.53g

1×mol

23.0g=0.8056mol 1 2

Na2S2O3 · 5H2OS 25.87g

1×mol

32.1g=0.8059mol 1 2

O 19.34 g1

×mol

16.0 g=1.209mol 1.5 3

H2O 36.26g1

×mol

18.0g=2.014mol 2.5 5

Conclusion

To find the percentage of water in an unknown hydrate, one must simply weigh the hydrate before and after a through heating, during which the salt is heated past water’s boiling point, allowing the water to evaporate out of the salt. When given the molar mass of the anhydrous salt, one can also find the number of moles of water present in each mole of the hydrate. The empirical formula of the hydrated salt was found to be AB · 8.5 H2O or 2AB · 17 H2O. One possible source of error was the size of the fire. In this experiment, the fire was considerably hotter than usual, and may have caused the experimenters to take it out too early, due to risk for decomposition. Thus, not all the water may have been boiled out.

Na 23.0 g O 16.0 g 5.82g1

×mol

286.0g=0.0203mol

C 12.0 g H2O 18.0 gNa2CO3 106.0 g Na2CO3 ·10H2O 286.0 g 0.0203mol

1×

106.0 gmol

=2.15 g