SUSB052
 Gravimetric Analysis of a Mixture 
1. Preliminaries In this exercise, you weigh a sample of a mixture of NaHCO_{3} and NaCl, heat it to decompose the NaHCO_{3} and weigh the product to determine the composition of the mixture. The NaCl is unaffected by the heating. Any changes in weight are due to NaHCO_{3} alone.

2. Stoichiometry: The stoichiometric relationship between the reagents is given by the overall equation: 2NaHCO_{3}(s) = Na_{2}CO_{3}(s) + H_{2}O(g) + CO_{2}(g) This equation implies that: 2 moles of NaHCO_{3} react to produce 1 mole of water and 1 mole of CO_{2} or, 1 mole of NaHCO_{3} react to produce 1/2 mole of water and 1/2 mole of CO_{2} or, 84.01 grams of NaHCO_{3} react to produce 18.01/2 grams of water and 44.01/2 grams of CO_{2} or, 84.01/84.01 = 1.000 gram of NaHCO_{3} reacts to produce 18.01/2*84.01 = 0.1072 grams of water and 44.01/2*84.01 = 0.2619 grams of CO_{2} 1.000 gram of NaHCO_{3} reacts to produce (0.1072 + 0.2619 = 0.3691) grams of gaseous products I.e., weight loss is directly proportional to the initial weight of NaHCO_{3}. weight loss = 0.3691 * weight of NaHCO_{3} or, weight of NaHCO_{3} = weight loss / 0.3691 Or, rephrasing the above, consider what happens when a 1.0000 g sample of pure NaHCO_{3} is heated to decomposition  a temperature high enough to vaporize water. ( CO_{2} is certainly gaseous as well. )
When it decomposes, it will produce 1/2 X 11.90 = 5.950 mmol of gaseous H_{2}O and 5.950 mmol of CO_{2} These will weigh, respectively,
If we know the original and final weight of a sample of NaHCO_{3}, the loss in weight will be proportional to the weight of NaHCO_{3} in the sample, with a proportionality constant of 0.3691 . I.e., again we have as above, weight loss in heating (g) = 0.3691 X weight of NaHCO_{3} (g)

3. A Sample Calculation Suppose a 0.8643 g sample of an unknown weighs 0.6098 g after heating. The weight loss is 0.8643  0.6098 = 0.2545 g. This corresponds to 0.2545 / 0.3691 = 0.6895 g of NaHCO_{3}, and the sample must have contained


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20150827
