Conversion of Mass to Moles Help
In 1811, Italian physicist, Amedeo Avogadro, presented a theory in the Journal de Physique that the mass of one mole of a sample contains the same basic number of particles as in 12 grams of 12 C. That number of atoms is called Avogadro’s number or a mole of sample.
Avogadro’s number (N) or one mole (mol) is equal to 6.02 × 10 23 atoms or molecules.
If you wanted to find the number of molecules in 1.35 moles, you would multiply the number of moles by Avogadro’s number.
6.022 × 10 23 × 1.35 moles = 8.1297 × 10 23 molecules
That same year, Avogadro proposed that equal volumes of different gases at the same pressure and temperature all contain the same number of particles. Like Avogadro’s number, this observation was used to solve chemical bonding and molecular composition problems.
To better understand the size of Avogadro’s number, think of the size of a pea. If one mole of peas were spread over the earth’s surface, then the surface would be covered by a layer nearly 20 miles thick. A high-speed computer can count all the fish in the sea in less than a second, but would take over a million years to count a mole of fish.
The atomic mass for a given element in the Periodic Table is measured out in grams equal to one mole of atoms of that element. The molar mass (MM) of elements and compounds is the mass, in grams, equal to the atomic and formula masses of those elements and compounds. Molar mass is measured in grams/mole.
Conversion Of Mass To Moles
In the course of an experiment, it is usually necessary to figure out the mass of an element. For example, you might be asked to calculate the mass of 2 moles of potassium (K). To do this, you need a conversion factor that changes moles to mass. The conversion of moles to mass would look like this:
2.0 mol K × 39.0 g/mol K = 78.0 g
How about finding the number of moles in 124 grams of calcium? Mass to moles then would be mol/g:
124 g × 1 mol Ca/40.1 g = 3.09 mol Ca
Using these simple conversion factors, then, a chemist is able to find the molar mass of a compound. The following solution of CaCO 3 demonstrates how it works:
Ca = 1 mol Ca × 40.1 g/mol Ca = 40.1 g
C = 1 mol C × 12.0 g/mol C = 12 g
O = 3 mol O × 16 g/mol O = 48.0 g
Molar mass = 100.1 g/mol CaCO 3
To see the % of each element in CaCO 3 , you would take the mass of each element in the one mole sample and then divide by the molar mass.
% Ca = 40.1 g Ca/100.1 g CaCO 3 × 100 = 40.1%
% C = 12 g C/100.1 g CaCO 3 × 100 = 12.0%
% O = 48 g O/100.1 g CaCO 3 × 100 = 48.0%
Practice problems for these concepts can be found at – Concentration and Molarity Practice Test
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