-50.0 = 5.0 g-10 78 Chapter Three Stoichiometry Can we count these nonidentical beans by weighing? Yes. The key piece of information we need is the average mass of the jelly beans. Let's compute the average mass for our 10-bean sample. total mass of beans Average mass = -------number of beans 5.1 g + 5.2 g + 5.0 g + 4.8 g + 4.9 g + 5.0 g + 5.0 g + 5.1 g + 4.9 g + 5.0 g 10 Figure 3.1. (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer. The average mass of a jelly bean is 5.0 g. Thus, to count out 1000 beans, we need to weigh out 5000 g of beans. This sample of beans, in which the beans have an average mass of 5.0 g, can be treated exactly like a sample where all of the beans are identical. Objects do not need to have identical masses to be counted by weighing. We simply need to know the average mass of the objects. For purposes of counting, the objects behave as though they were all identical, as though they each actually
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 12/13/2010 for the course CHEM 2301 taught by Professor Bill during the Spring '10 term at South Texas College.