1. Empirical Formula of a Hydrate In addition to this presentation, before coming to lab or attempting the prelab quiz you must also:  Read the following pages in the “Laboratory Handbook…” byStanitski et al. before coming to lab: pages 39-40 (using a bunsen burner) page 42 (heating a solid)  Watch the prelab video for this experiment  Review the video on use of the balances if needed  Read the introduction to the lab in the coursepack

2. What’s the point? In this lab you willIn this lab you will Gain more practical experience - bunsen burners, heating apparatus, balancesGain more practical experience - bunsen burners, heating apparatus, balances Determine formula stoichiometry by experiment - find the molar ratio of water to salt in a hydrateDetermine formula stoichiometry by experiment - find the molar ratio of water to salt in a hydrate Practice empirical formula calculationsPractice empirical formula calculations

3. Hydrates Ionic compounds with water molecules incorporated into their crystal structuresIonic compounds with water molecules incorporated into their crystal structures Waters are shown in the chemical formulaWaters are shown in the chemical formula The number of moles of water (“waters of hydration”) can be anywhere from 1 to 12The number of moles of water (“waters of hydration”) can be anywhere from 1 to 12 Example formulas are:Example formulas are: ZnSO 4  7H 2 O (7 mol H 2 O for every for 1 mol of ZnSO 4 ) CuSO 4  5H 2 O (5 mol H 2 O for every for 1 mol of CuSO 4 )

4. Naming Hydrates To name a hydrate, name the ionic compound, then indicate the amount of water of hydration by adding… …a greek prefix to indicate the number of waters… 1 = mono; 2 = di; 3 = tri; 4 = tetra; 5 = penta; etc... …and then the word hydrate e.g., CuSO 4  5H 2 O = copper (II) sulfate pentahydrate Note: There is no way to predict the number of waters of hydration for a given ionic compound  This must be obtained from experiment

5. Measuring the Amount of Water The hydrate molar mass includes waters of hydration ZnSO 4  7H 2 O would be,The hydrate molar mass includes waters of hydration ZnSO 4  7H 2 O would be, 65.37 + 32.06 + (4 x 16.00) + (7 x [2 x 1.008 + 16.00]) = 287.57 g / mol The moles of water in a hydrate can be determined by measuring the mass of water lost upon heating (a) Original mass = mass of salt + water (b) Mass after heating = mass of salt (c) The difference gives the mass of water (d) To get the formula, convert masses to molesThe moles of water in a hydrate can be determined by measuring the mass of water lost upon heating (a) Original mass = mass of salt + water (b) Mass after heating = mass of salt (c) The difference gives the mass of water (d) To get the formula, convert masses to moles

6. Bunsen Burners Open flame heat source Gas goes in here Air mixes with gas here. The collar is adjustable to let more or less air in. Use this valve to adjust gas flow Don’t adjust from nozzle on bench

7. Connect rubber hose to the lab gas valve and burner, checking for holes or cracks in the rubber.Connect rubber hose to the lab gas valve and burner, checking for holes or cracks in the rubber. Turn on the main gas valve (on bench).Turn on the main gas valve (on bench). Light burner with striker.Light burner with striker. Adjust the flame as desired by turning the needle valve (under burner tube) and/or adjusting the collar height.Adjust the flame as desired by turning the needle valve (under burner tube) and/or adjusting the collar height. When finished heating, turn off the burner at the main valve (on bench).When finished heating, turn off the burner at the main valve (on bench). Never try to turn off the burner from the needle valve.Never try to turn off the burner from the needle valve. To use a Bunsen burner

8. A hot blue flame can be obtained by raising the Bunsen Burner collar to allow more air in A cooler, yellow flame can be obtained by lowering the Bunsen Burner collar to allow less air in Less Air “reducing” More Air “oxidizing” This is generally an easier type of flame to control This type of flame is good if you need a low temperature, but is harder to control

9. Experimental Details Measure mass of hydrate sampleMeasure mass of hydrate sample Heat in crucible to remove waterHeat in crucible to remove water Cool the crucible in a dry location, the desicooler keeps salt from absorbing atmospheric water vaporCool the crucible in a dry location, the desicooler keeps salt from absorbing atmospheric water vapor Measure mass of water lost You must heat the sample until no more mass is lost This is to ensure all water is removed Verify by several heating steps until constant massMeasure mass of water lost You must heat the sample until no more mass is lost This is to ensure all water is removed Verify by several heating steps until constant mass Measure mass of dry saltMeasure mass of dry salt From mass of water and mass of salt, determine molar ratio of water/saltFrom mass of water and mass of salt, determine molar ratio of water/salt

10. Experimental Set-Up Clay triangle, on iron ring, to support the crucible Iron ring, clamped to post Bunsen burner, Adjusted to hot, blue flame Post on ceramic stand

11. Use of Balances (Review) Use Analytical Balances for All Measurements In This LabUse Analytical Balances for All Measurements In This Lab precise to +/- 0.001 g (1 mg)precise to +/- 0.001 g (1 mg) RULESRULES never weigh a hot/warm object on balancenever weigh a hot/warm object on balance keep the balance area cleankeep the balance area clean

12. Necessary Calculations calculating hydrate empirical formulascalculating hydrate empirical formulas moles of salt = grams dry salt / molar massmoles of salt = grams dry salt / molar mass moles of water = grams water lost / molar massmoles of water = grams water lost / molar mass Heating removes 0.438 g of H 2 O from a hydrate, leaving behind 0.562 g dry salt. If the salt has a molar mass of 161.26 g mol -1, what is the hydrate empirical formula?

13. salt mass = 0.562 gsalt mass = 0.562 g water mass = 0.438 gwater mass = 0.438 g 0.562 g = 0.00349 mol salt 0.562 g = 0.00349 mol salt 161.26 g/mole Divide the larger number by the smaller:Divide the larger number by the smaller: (0.0243 mol H 2 O) / (0.00349 mol salt) = 6.96 0.438 g = 0.0243 mol H 2 O 0.438 g = 0.0243 mol H 2 O 18.02 g/mole

14. This means, 6.96 mol H 2 O per mol salt It would seem safe to say that there are 7 mol water for each mol salt the empirical formula, written in the appropriate form for a hydrate, is Salt  7H 2 Othe empirical formula, written in the appropriate form for a hydrate, is Salt  7H 2 O this is a heptahydratethis is a heptahydrate

15. Safety Lab goggles and coat must be onLab goggles and coat must be on Use care with the bunsen burner tie back long hair secure loose clothing keep your notebook away from the work areaUse care with the bunsen burner tie back long hair secure loose clothing keep your notebook away from the work area Do not handle hot objects by handDo not handle hot objects by hand TURN OFF THE GAS USING THE MAIN VALVE ON THE BENCH WHEN EXTINGUISHING BUNSEN BURNERSTURN OFF THE GAS USING THE MAIN VALVE ON THE BENCH WHEN EXTINGUISHING BUNSEN BURNERS