1. Lab 8 Sodium Carbonate or Sodium Bicarbonate? Objective To determine a compound to be either Na 2 CO 3 or NaHCO 3.

2. NaHCO 3 + HCl  CO 2 + H 2 O + NaCl Na 2 CO 3 + 2HCl  CO 2 + H 2 O + 2NaCl Background NaHCO 3 = sodium bicarbonate (baking soda) Na 2 CO 3 = sodium carbonate HCl = hydrochloric acid 84.007 g/mol 58.443 g/mol 105.988 g/mol 58.443 g/mol

3. Slowly add drops of HCl to the dish until all of the solid has reacted. Slowly evaporate the liquid in the dish with the watch glass on top of the dish. Avoid spattering. Let the evap. dish cool. (M1) Mass of empty evap. dish + watch glass = ____ g Experiment - (M2) Mass of dish + watch glass + 1.00 gram of solid = ____ g (M3) Mass of evap. dish + watch glass + product = ____ g Clean out dish and watch glass.

4. 1. What reactant was the excess reactant in this lab? Analysis - 4. Convert the mass of the NaCl into moles of NaCl (keep at least 3 sig figs). 1.00 g Na 2 CO 3 105.988 g Na 2 CO 3 = 0.009435 mol Na 2 CO 3 1.00 g NaHCO 3 84.007 g NaHCO 3 ÷ = 0.0119 mol NaHCO 3 6. If the solid was Na 2 CO 3, calculate how many moles reacted. 5. If the solid was NaHCO 3, calculate how many moles reacted. You started with 1.00 gram of a solid, ÷ 2. How many grams of the solid did you begin with? 3. Calculate the mass of the NaCl that was produced.

5. 9. Identify the unknown as NaHCO 3 or Na 2 CO 3. Explain how you know. Results - 7. If the unknown was NaHCO 3, how many moles of NaCl would have been produced? 8. If the unknown was Na 2 CO 3, how many moles of NaCl would have been produced? Using your answers from 5 & 6 and the balanced chemical equations,