the extreme smallness of atoms and molecules makes it necessary to define a very large “standard sample”
it is referred to as the MOLE
it’s the SI base unit for amount of chemical substance
one mole consists of 6.02 x 10^{23} atoms or molecules; it is known as Avogadro’s Number
the mass of one mole of an element is the same as the atomic mass of the element
for
example:
one mole of bromine, atomic mass 79.90 amu, has a mass of 79.90 g; this is
called the molar mass
the molar mass of chromium,
Cr, is 52.00 g/mol.
types of calculations:
(i) mass of 1.50 moles of chromium
(ii) # of moles in 5.00g of O_{2}
(iii) determining the molar mass of a compound; what is the molar mass of CuSO_{4}?
(iv) # of moles given a particular mass of a compound; how many moles is 56.00 g of NaCl?
(v) mass of a compound given a particular # of moles; determine the mass of 0.1 moles of H_{2}SO_{4}.
Examples:
1. How many grams of sulfur are present in 0.150 mol of Au?
solution:
1 mole of Au → 197.0 g
0.150 mol of Au→ x g
x = 197.0 g x 0.150 moles = 29.6 g Au
1 mole
2. How many atoms are in a sample of uranium with a mass of 1.00 x 10^{–6 }g ?
solution:
1 mol U → 238.0 g U
1 mol U → 2.03 x 1023 atoms
therefore:
238.0 g U → 2.03 x 1023 atoms
1.00 x 10–6 g U → x atoms
x = (2.03 x 1023 atoms)( 1.00 x 10–6 g) = 2.53 x 1015 atoms
238.0 g
3. How many grams is 0.115 mol of Ca_{3}(PO_{4})_{2} ?
Solution:
1 mol of Ca_{3}(PO_{4})_{2 }→ 310.18 g
0.115 mol → x g
x = 310.18 g x 0.115 mol = 35.7 g Ca_{3}(PO_{4})_{2}
1 mol
4. How many grams of Cl are needed to combine with 24.4 g of Si to make silicon tetrachloride, SiCl_{4}?
Solution:
The balanced equation for the reaction must be written
Si(s) + 2Cl_{2}(g) → SiCl_{4}(s)
From the equation we see that:
1 mol Si reacts with 2 mol of Cl_{2} 1 mol Si = 28.09 g
x mol Si = 24.4 g
0.869 mol Si reacts with x mol Cl_{2}
x = 1 mol x 24.4 g = 0.869 mol Si
x mol Cl2 = 2 mol Cl_{2} x 0.869 mol Si 24.4 g
1 mol Si use this value in your calculations
= 1.74 mol
1 mol Cl_{2} = 70.90 g
1.74 mol Cl_{2} = x g
x g Cl_{2} = 70.90 g x 1.74 mol
1 mol
= 123.366 g = 123 g Cl_{2}
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Series of Metals