11. Conservation of Mass,       Charge, and Energy
Conservation of Mass in Chemical Reactions
 Note that we proceeded in five steps: 1.Write an unbalanced expression with the correct reactants and products. 2.Balance the equation properly, and obtain the ratio of number of moles of the reactants and products of interest. 3.Calculate molecular weights of the reactants and products of interest. 4.Calculate the number of moles of reactant used, and use the mole ratio from step 2 to find the number of moles of product. 5.Use the molecular weight of the product to obtain the weight in grams. Example. Glucose is a sugar with the chemical formula , and is a common energy source for living organisms. How many moles of oxygen are required to burn a mole of glucose, and how many grams of are needed for a kilogram of glucose? In passing, observe that burning a kilogram of glucose requires only 1066 g of oxygen, whereas burning the same weight of propane requires 3627 g of oxygen. This is because glucose already is partially oxidized. You should not be surprised later when we calculate that the combustion of glucose produces only half as much heat per gram as combustion of propane. Glucose is a poorer fuel than propane on a weight basis. In Chapter 21 we will return to the question as to why plants selected glucose for energy storage (in plant starch), whereas animals developed fats as a kind of "solid propane."
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