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Volumetric Analysis

Volumetric Analysis

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Volumetric Analysis

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  1. Volumetric Analysis • Frequently, we will react 2 solutions with each other • These reactions are called titrations: • Water analysis • Environmental Sciences • Food industry • Medical laboratories • Typical titrations involve reacting acids and bases • We add an indicator (ie: phenolphthalein) that changes color when the titration is complete • The concentration of the analyte is determined by measuring how much of the standard material of known concentration was used

  2. Approach is the same, but instead of working with solids, we have solutions

  3. Remember the definition of Molarity:

  4. In a titration, 3.25g of acid (HX) requires 68.8 mL of 0.750M NaOH for complete reaction. What is the molar mass of the acid? Step 1: Balanced Chemical reaction Step 2: Calculate the # of moles of NaOH Step 3: Determine the # of moles of HX Step 4: Calculate the molar mass of HX 

  5. Suppose that 25.0mL of 0.5M K2CrO4(aq) reacts with 15.0 mL of AgNO3(aq) completely. What mass of NaCl is needed to react completely with 35.0mL of the same AgNO3 solution? Step 1: Balanced chemical equation Step 2: Determine the # of moles of K2CrO4 Step 3: Determine the # of moles of AgNO3 Step 4: Determine the Molarity of the AgNO3 solution Step 5: Reaction between AgNO3 and NaCl Step 6: Determine the # of moles of AgNO3 in 35.0 mL Step 7: Calculate the NaCl mass in that # of moles 

  6. Fundamentals M: Limiting Reagents • We have only considered reactions where the exact amounts of reactants were present • In this chapter, we’re going to look at more realistic reactions and determine how to ascertain the extent of reaction based upon the amount of reactants present • First, we’ll need to define some terms…

  7. Reaction Yield • The Theoretical Yield is the maximum quantity of products that can be obtained from a given amount of reactant • The Percentage Yield is equal to the Actual Yield divided by the Theoretical Yield multiplied by 100. • Many “real world” experiments are reported in terms of percent yield 

  8. In an automotive engine, 1.00L of octane (C8H18) (702g of octane) is burned but only 1.84 kg of carbon dioxide is produced. What is the percentage yield of carbon dioxide? Step 1: Balanced chemical reaction Step 2: Calculate the number of moles of octane Step 3: Calculate the # of moles of CO2 that should be made (theoretical yield) Step 4: Calculate the actual moles of CO2 made

  9. The Limiting Reactant • A limiting reactant is the reactant that governs the maximum yield of product obtained. • The limiting reactant is consumed before the other reactant(s) are and makes the reaction stop. • In order to determine the limiting reactant, we must know the balanced chemical equation

  10. How do we determine the limiting reactant? • Step 1: You’ll be given something (mass or # of moles) of the reactant. Convert to moles Or Calculate the # of moles of each reactant you actually have • Step 2: Look at the balanced chemical equation. Find the reactant with the highest stoichiometric coefficient and divide it by the stoichiometric coefficient of the other reactant. Now, do the same for the # of moles of reactant you calculated in Step 1.

  11. Step 3: Look at the values from Step 2. • Does the last calculation (of the actual # of moles) give you a higher number than the calculation with the stoichiometric coefficients? If so, the reactant with the lower stoichiometric coefficient is limiting, if not, then the reactant with the higher stoichiometric coefficient is limiting • Step 4: The # of moles of the limiting reactant determines the maximum amount of product that can be formed. • Alright, that was really long. Let’s look at some examples for help… 

  12. If 28g of NO2 and 18g of H2O are allowed to react to form nitric acid and nitrogen monoxide, and 22g of nitric acid is produced in the reaction, what is the percentage yield? Step 1: Determine the limiting reactant Step 2: Calculate the theoretical # of moles of product Step 3: Calculate the actual # of moles of product Step 4: Calculate the % yield 

  13. In the synthesis of ammonia, what is the percentage yield of ammonia when 100 kg of hydrogen reacts with 800 kg of nitrogen to produce 400 kg of ammonia. Step 1: Balanced chemical equation Step 2: Calculate the # of moles of reactants Step 3: Identify the limiting reactant Step 4: Calculate the theoretical yield based upon the limiting reactant Step 5: Calculate the actual # of moles of product Step 6: Calculate the % yield

  14. Combustion Analysis • One of the most important techniques in the analysis of an unknown organic compound is Combustion Analysis • In this technique, an organic sample is combusted in a furnace and the products are quantitated in a gas chromatograph

  15. Combustion Analysis CxHyOz + O2 --> CO2 + H2O • All the hydrogen in water comes from the unknown sample • All the carbon in carbon dioxide comes from the unknown sample • We will know the mass of the unknown compound and the mass of the products • We use the Law Of Conservation of Mass and these values to calculate the amount of oxygen (or other atom) in the unknown sample. • We can then use other information given to calculate the molecular formula 

  16. A combustion analysis was carried out on 1.621g of a newly synthesized compound, which was known to contain only C, H and O. The masses of water and carbon dioxide produced were 1.902g and 3.095g, respectively. What is the empirical formula of the compound? Step 1: Determine the amounts of H and C in the products Step 2: Using the Law of Conservation of Mass, determine the amount of oxygen in the compound Step 3: Determine the # of moles just like we did in the mass percentage problems you know and love 

  17. When 0.528g of sucrose (which consists of carbon, hydrogen and oxygen) is burned, 0.306g of H2O and 0.815g of CO2 are formed. What is the empirical formula of the compound? Step 1: Determine the amounts of H and C in the products Step 2: Determine the amount of oxygen in the sample Step 3: Calculate the ratios of the moles of the atoms