Chapter 10: The Mole
Measuring atoms, molecules, or formulas is based on:
Elements
Cannot be broken down by chemical means
Weighted atomic masses of elements do not vary from place to place
Atomic mass is defined by 12C ≡ 12 amu (and therefore proportional for all elements)
Compounds
Always occur in a fixed ratio (unlike mixtures which can occur in varying ratios), therefore:
Chemical formulas are fixed regardless of the type of chemical formula, any of:
Empirical formula: smallest whole number ratio
Molecular formula: actual counts of atoms in a molecule
Structural formula: a representation showing the bonding of atoms in molecules
Elements in formulas (and compounds) occur in small, whole number ratios
Chemical reactions (and therefore balanced chemical equations) are bound by conservation of:
Atoms (and therefore mass)
Energy
Charge
Converting amu to mass in grams is based on the mole
Mole ≡ the number of atoms in exactly 12 g of 12C
Mole = 6.022 1367 x 1023 particles mole–1
Importance of the mole (Why 6.022 1367 x 1023?)
A set of wooden blocks have a length of 4.5 inches.
What is length of 12 such blocks measured in feet?
4.5 feet
What made this calculation so easy?
A steel rod has a length of 7.15 feet.
What is length of 3 of these rods measured in yards?
7.15 yards
What made this calculation so easy?
What is length of 5280 of these rods measured in miles?
7.15 miles
What made this calculation so easy?
What about 5280 aluminum rods 3.14 feet long?
3.14 miles
Why are 5280 Al rods only 3.14 miles long while 5280 steel rods are 7.15 miles long?
How were we able to calculate the total length of both rods so fast?
Scientists chose 6.022 1367 x 1023 particles in one mole so that:
1 mole of silicon (atomic mass = 28.0855 amu) = 28.0855 g of silicon while
1 mole of lead (atomic mass = 207.2 amu) = 207.2 g of lead
Converting between particles and moles
Calculate the number of water molecules in 3.15 moles of water
6.022 × 1023 molecules H2O
3.15 mol H2O ×
= 1.90 × 1024 molecules H2O
1 mol H2O
Calculate the number of moles of He in 9.75 x 1022 atoms of He
1 mol He
9.75 × 1022 atoms He×
= 0.162 mol He
6.022 × 1023 atoms He
Calculating molar mass
Find the molar mass of KClO3
K = 1 (39.0983g mol–1) = 39.0983
Cl = 1 (35.453 g mol–1)
= 35.453
–1
O = 3 (15.9994 g mol ) = 47.9982
KClO3
= 122.550
g mol–1
g mol–1
g mol–1
g mol–1
Converting between mass and moles
Find the mass of 6.25 moles of KClO3 (122.55 g mol–1)
122.55 g KClO3
6.25 mol KClO3 ×
= 766 g KClO3
1 mol KClO3
How many moles of KClO3 are represented by 6.25 g?
1 mol KClO3
6.25 g KClO3 ×
= 0.0510 mol KClO3
122.55 g KClO3
Converting between mass and particles
How many molecules of water are in 8.04 g of H2O (18.0153 g mol–1)?
1 mol H2O
6.022 × 1023 molecules H2O
8.04 g H2O ×
×
= 2.69 × 1023 molecules H2O
18.0153 g H2O
1 mol H2O
Find the mass of 8.04 x 1024 molecules of H2O
1 mol H2O
18.0153 g H2O
8.04 × 1024 molecules ×
×
= 241 g H2O
6.022 × 1023 molecules H2O
1 mol H2O
Analyzing a hydrate
Example:
Barium chloride is a known hydrate. Before heating, a BaCl2 sample has a mass of 5.00 g.
After heating, the anhydrous sample has a mass of 4.26 g. What is the formula of the
original barium chloride hydrate?
Mass of anhydrous BaCl2 = 4.26 g
Moles of anhydrous BaCl2
1 mol BaCl2
4.26 g BaCl2 ×
= 0.0205 mol BaCl2
208.23 g BaCl2
Mass of H2O is the mass lost on heating = 5.00 g – 4.26 g = 0.74 g
Moles of H2O
1 mol H2O
0.74 g H2O ×
= 0.041 mol H2O
18.0153 g H2O
Ratio of moles of H2O to moles of BaCl2
mol H2O
0.041 mol H2O
2 mol H2O
=
=
mol BaCl2
0.0205 mol BaCl2
1 mol BaCl2
Original formula of the hydrate = BaCl2 · 2 H2O