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# Scan_Doc0088 - 82 Chapter Three Stoichiometry Table 3.1...

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82 Chapter Three Stoichiometry Table 3.1. Comparison of 1.-MoleSamples of Various Elements Mass of Sample (g) Element Number of Atoms Present 6.022 X 10 23 6.022 X 10 23 6.022 X 10 23 6.022 X 1023 6.022 X 1023 6.022 X 10 23 Aluminum Copper Iron Sulfur Iodine Mercury 26.98 63.55 55.85 32.07 126.9 200.6 and 6.022 X 10 23 amu = 1 g t Exact number This relationship can be used to derive the unit factor needed to convert between atomic mass units and grams. EXAMPLE 3.2 Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms. Solution From the table inside the front cover of the text, we note that one americium atom has a mass of 243 amu. Thus the mass of six atoms is

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Unformatted text preview: amu 6 atemS X 243 --= 1.46 X 10 3 amu atern Using the relationship 6.022 X 10 23 amu = 1 g we write the conversion factor for converting atomic mass units to grams: 1 g 6.022 X 10 23 amu The mass of six americium atoms in grams is l.46 X 10 3 aarn X 1 g = 2.42 X 10-21 g 6.022 X 1023 aarn Reality Check: Since this sample contains only six atoms, the mass should be very small as the amount 2.42 X 10-21 g indicates. SEE EXERCISE 3.41 To do chemical calculations, you must understand what the mole means and how to determine the number of moles in a given mass of a substance. These procedures are illustrated in Examples 3.3 and 3.4....
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