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Mass % NaHCO 3 in Alka Seltzer

Mass % NaHCO 3 in Alka Seltzer. In addition to this presentation, before coming to lab or attempting the prelab quiz you must also: Review the video about use of the balances if needed Read the introduction to the lab exercise in the coursepack. What’s the point?.

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Mass % NaHCO 3 in Alka Seltzer

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  1. Mass % NaHCO3 in Alka Seltzer • In addition to this presentation, before coming to lab or attempting the prelab quiz you must also: • Review the video about use of the balances if needed • Read the introduction to the lab exercise in the coursepack

  2. What’s the point? • Experimentally apply reaction and formula stoichiometry • Introduce the idea of limiting reagents • Chapter 3, Brown, LeMay and Burstein • Review the concept and use of mass % • Practice interpretation of graphs

  3. Alka-Seltzer medicine used to neutralize stomach acid by reaction with a base the basic component of Alka-Seltzer is NaHCO3 (sodium bicarbonate) Background • In the presence of acid (H+), the reaction is • NaHCO3(aq) + HCl(aq)  H2O(l) + CO2(g) + NaCl(aq) • Notice that a gas (CO2) is generated. This makes the “fizz”

  4. Reaction Stoichiometry • NaHCO3(aq) + HCl(aq)  H2O(l) + CO2(g) + NaCl(aq) • For each mole of CO2 (g) produced, 1 mole of NaHCO3 reacts • We say there is a 1:1 stoichiometric ratio between CO2 and NaHCO3 • Notice also that for each mole of CO2 (g) produced, 1 mole of HCl must also have reacted • There is a 1:1 stoichiometric ratio between CO2 and HCl

  5. Example Calculation: Mass % • mass percent NaHCO3 in a tablet: % NaHCO3 = (mass of NaHCO3 / tablet mass) * 100% An Alka-Seltzer tablet was 3.234 g. Reaction of the tablet yielded 0.672 g of CO2. What was the mass % NaHCO3 in the tablet?

  6. Alka-Selter tablet mass = 3.234 g • CO2 mass = 0.627 g • Hmm, we need the NaHCO3 mass! • How is the CO2 mass related to NaHCO3? • To relate amounts of two chemicals in a reaction we use the stoichiometric ratio (i.e., the mole ratio in the chemical equation). • We can relate the moles of CO2 to moles of NaHCO3 • To get from g CO2 to moles CO2, use the molar mass. 0.672 g x 1 mole CO2 = 0.0153 mole CO2 44.01 g CO2

  7. CO2 moles = 0.0153 moles • Use stoichiometry to convert mol CO2 to mol NaHCO3: NaHCO3(aq) + HCl(aq)  H2O(l) + CO2(g) + NaCl(aq) • For each mole of CO2 (g), 1 mole of NaHCO3 reacts Convert moles CO2 into moles NaHCO3 by stoichiometry 0.0153 mol CO2 x 1 mol NaHCO3 = 0.0153 1 mol CO2 mol NaHCO3

  8. NaHCO3 moles = 0.0153 moles • To convert moles NaHCO3 to g, use the molar mass. 0.0153 mole NaHCO3 x 84.01 g NaHCO3 = 1.28 g 1 mole NaHCO3 • Given the mass of NaHCO3 (1.28 g) and the tablet mass (3.234 g), we can find the mass % NaHCO3: 1.28 g x 100% = 39.6% 3.234 g Note: 100% is considered an exact number

  9. Preparing Solutions by Dilution • In this experiment, you must prepare 1M HCl by diluting stock 6 M HCl • Remember, M = molar concentration (mol per L) The Dilution Equation M1V1 = M2V2 M1 = initial concentration (before dilution) M2 = final concentration (after dilution) V1 = volume of undiluted sample V2 = volume of diluted sample

  10. Dilution Calculation and Procedure How many mL of 6 M HCl are needed to prepare 250 mL of 1 M HCl by dilution? • We know our starting concentration, but not how much we need to use to get the final desired volume and concentration, so… • M1 = 6 M V1 = ? • M2 = 1 M V2 = 250 mL

  11. Dilution Calculation and Procedure M1V1 = M2V2 (6 mol L-1)(? L) = (1 mol L-1)(0.250 L) V1 = 0.0417 L So, if we dilute 41.7 mL of 6 M HCl to 250 mL with DI water, we will have prepared 1 M HCl. • Procedure • measure 41.7 mL of 6 M HCl into a graduated cylinder • transfer to a 250 mL volumetric flask • dilute to the mark with DI water and mix thoroughly

  12. Limiting Reagent • NaHCO3(aq) + HCl(aq)  H2O(l) + CO2(g) + NaCl(aq) • If you had 1 mol of NaHCO3 but only 0.5 mol of HCl, you’d run out of HCl before all the NaHCO3 reacted. • Then, HCl would be the “limiting reagent” (LR): the amount of CO2 produced would be limited by the amount of HCl. • The moles of product obtained is is based on the moles of LR: 0.5 mol HCl x 1 mole CO2 = 0.5 mole CO2 1 mole HCl

  13. Safety • lab goggles and coats must be on • use care in diluting the desktop 6M HCl to 1M HCl • HCl is a strong acid and is caustic • spills should be cleaned up immediately with copious amounts of water • contact with skin should be avoided • in the event of contact, flush the area with lots of water and have someone notify the instructor

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