. CONSERVATION OF MASS Revised October 2. 1999 6 The principle of conservation of mass is fundamental to all chemical engineering analysis The basic idea is relatively easy to understand since it is fact of our everyday life Let us consider a simple example. Suppose we are required to prepare one kilogram of a solution of ethanol in water such that the solution will contain 40% ethanol by weight. So, we weigh out 400 grams of ethanol and 600 grams of water and mix the two together in a large beaker. If we weigh the resulting mixture(making appropriate allowance for the weight of the aker), experience says it will weigh 1000 grams or one kilogram. And it will. This is manifestation of the conservation of mass That, in the absence of nuclear reactions, mass is conserved is a fundamental law ofnature This law is used throughout these notes and throughout all chemical engineering Suppose we happened to measure the volumes involved in making up our alcohol solution. Assuming that we do this at 20 C. we would find that we added 5989 ml of water to 315.7 ml of ethanol to obtain 935.2 ml of solution. However, the sum of the volumes of the pure components is 914.6 ml. We conclude that volume is not conserved Let us take note of one other fact about our solution. If we were to separate it back into its pure components(something we could do, for instance, by azeotropic distillation) and did this with extreme care to avoid any inadvertent losses, we would obtain 400 gm of ethanol and 600 gm of water. Thus, in this case, not only was total mass conserved but the mass of each of the components was also This is not always true. Suppose that instead of adding ethanol and water, we added ( carefully and slowly) sodium hydroxide to sulfuric acid. Suppose that the H2sO4 solution contains exactly 98.08 pounds of H2SO4 and that we add exactly 80.00 pounds of NaOH. A chemical reaction will take place as follow H2 SO4 +2 Naoh, Na,SO4 2 H,O Notice that the amount of H2so4 in the original solution is 1.0 lb-mol and that the amount of NaoH added is exactly 2.0 lb-mols. What we are left with is 1.0 lb-mol of Na2 SO4 or 142.05 lbs and 2.0 lb-mols of H2O or 36.03 lbs. No individual component is conserved; the H2SO4 and the Naoh have disappeared and in their place we have Na2 SO4 and H2O. However, if we look at the atomic species H, O, S, and Na, we will find that these are all con-served. That is exactly what the reaction equation expresses-5- I. CONSERVATION OF MASS Revised October 2, 1999 The principle of conservation of mass is fundamental to all chemical engineering analysis. The basic idea is relatively easy to understand since it is fact of our everyday life. Let us consider a simple example. Suppose we are required to prepare one kilogram of a solution of ethanol in water such that the solution will contain 40% ethanol by weight. So, we weigh out 400 grams of ethanol and 600 grams of water and mix the two together in a large beaker. If we weigh the resulting mixture (making appropriate allowance for the weight of the beaker), experience says it will weigh 1000 grams or one kilogram. And it will. This is a manifestation of the conservation of mass. That, in the absence of nuclear reactions, mass is conserved is a fundamental law of nature. This law is used throughout these notes and throughout all chemical engineering. Suppose we happened to measure the volumes involved in making up our alcohol solution. Assuming that we do this at 20 C, we would find that we added 598.9 ml of water to 315.7 ml of ethanol to obtain 935.2 ml of solution. However, the sum of the volumes of the pure components is 914.6 ml. We conclude that volume is not conserved. Let us take note of one other fact about our solution. If we were to separate it back into its pure components (something we could do, for instance, by azeotropic distillation) and did this with extreme care to avoid any inadvertent losses, we would obtain 400 gm of ethanol and 600 gm of water. Thus, in this case, not only was total mass conserved but the mass of each of the components was also. This is not always true. Suppose that instead of adding ethanol and water, we added (carefully and slowly) sodium hydroxide to sulfuric acid. Suppose that the H2SO4 solution contains exactly 98.08 pounds of H2SO4 and that we add exactly 80.00 pounds of NaOH. A chemical reaction will take place as follows: H2SO4 + 2 NaOH ‡ Na2SO4 + 2 H2O. Notice that the amount of H2SO4 in the original solution is 1.0 lb-mol and that the amount of NaOH added is exactly 2.0 lb-mols. What we are left with is 1.0 lb-mol of Na2SO4 or 142.05 lbs and 2.0 lb-mols of H2O or 36.03 lbs. No individual component is conserved; the H2SO4 and the NaOH have disappeared and in their place we have Na2SO4 and H2O. However, if we look at the atomic species H, O, S, and Na, we will find that these are all con-served. That is exactly what the reaction equation expresses