Conversions with Avogadros Number

Avogadro s number is the conversion factor that iinks the number of moles with the number of individual particles. To determine the number of atoms in a sample of an element, we multiply the number of moles by Avogadro s number — n Likewise, if a sample contains a certain number of atoms, the number of moles in the sample... 

Ans. There are 6.02 X 10 i atoms in 1.00 mol Na (Avogadro s number). There is 23.0 g of Na in LOO mol Na (equal to the atomic weight in grams). This problem requires use of two of the most important conversion factors involving moles. Note which one is used with masses and which one is used with numbers of atoms (or molecules of formula units). With numbers of atoms, molecules, or formula units, use Avogadro s number with mass or weight use the formula weight. 

You use molar masses as conversion factors in the same way you use Avogadro s number. The right side of Skills Toolkit 3 shows how the amount in moles relates to the mass in grams of a substance. Suppose you must find the mass of 3.50 mol of copper. You will use the molar mass of copper. By checking the periodic table, you find the atomic mass of copper, 63.546 amu, which you round to 63.55 amu. So, in calculations with copper, use 63.55 g/mol. 

Calculate the mass in grams of one formula unit of AICI3. Start with molar mass and use the inverse of Avogadro s number as a conversion factor. 

Did you notice that we multiplied and divided by Avogadro s number These two steps cancel each other out Perhaps you now see the beauty of Avogadro s number. Using the number of atomic mass units in one gram eliminates our need to make this conversion every time we do one of these calculations. Example 4 shows the same calculation, except it starts with the molar mass of carbon dioxide, which has the same value as the molecular mass, except with different units. 

During irradiation with monochromatic light of frequency v, quanta of known energy hv) are delivered. The photochemist often refers to a mole of quanta, which refers to Avogadro s number 6.02 X 10 of quanta, and is called an einstein of radiation. By using equation (1) and the appropriate conversion factors, it is possible to derive the expression... 

The value of Avogadro s number will also be given in a table of useful information or, sometimes it is included with the question itself. Since you use this value so often, you probably already know it. These two conversion factors are arranged so that the units cancel in making the change from grams through moles to atoms. If the conversion were desired in the reverse direction, then both factors would be inverted. ... 

You can even calculate the number of carbon atoms it takes to react with that 1.000 kilogram of rust. Basically, you use the same conversions, but instead of converting from moles of carbon to grams, you convert from moles of carbon to carbon atoms using Avogadro s number ... 

Draw a solution map, beginning with silver atoms and ending at moles. The conversion factor is Avogadro s number. 

Amount-Mass-Number Conversions Involving Elements We begin with amount-mass-number relationships of elements. As Figure 3.2 shows, convert mass (in grams) or number of entities (atoms or molecules) to amount (mol) first. For molecular elements, Avogadro s number gives molecules per mole. 

You need two conversion factors. From the periodic table, find the molar mass of copper to be 63.55 g/mol. You also need Avogadro s number, 6.022 X 10 atoms = 1 mol. Starting with the mass, first convert to moles and then to number of atoms ... 

Here a short remark on the designations atomic weight and atomic mass is necessary They are atomic mass ratio numbers, since they are scaled according to the atomic mass of They are relative atomic masses and therefore they are often given without the unit u. Ideally, these numbers should be used as molar mass of the specific element, M(E), with the units g mol . The conversion of relative atomic masses into molar masses is accomplished by using the Avogadro constant as given in Eq. (6.1). Details can be obtained elsewhere [36, 37). 

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