Chem. 1A Week 3 Discussion Notes Dr. Mack/S12 Isotopes and Isotope abundance: Atoms of a particular element with differing numbers of neutrons are known as "Isotopes". Each isotope has a different number of neutrons. For example, carbon has three isotopes, one with 6 neutrons, one with 7 and the third with 8 neutrons. Each isotope of a particular element occurs in nature with a specific fraction or "% abundance". The sum of these abundances is always equal to 100%. A
Z
X X = elemental symbol Z = atomic number (# of protons) A = mass number (# of protons + neutrons) C−12, C−13 & C−14 (shorthand notation) Example: Boron B−10 (10.0129u) = 19.91% B−11 (11.0093u) = 80.09% The average weighted atomic mass for an element that is listed on the periodic table is determined by the following: ⎛ % abundance isotope 1 ⎞
⎛ % abundance isotope 2 ⎞
Atomic weight = ⎜
⎟ × (mass isotope 1 ) + ⎜
⎟ × (mass isotope 2 ) + ...
100
100
⎝
⎠
⎝
⎠
Example Boron: ⎛ 19.91 ⎞
⎛ 80.09 ⎞
Atomic weight = ⎜
⎟ × (10.0129u) + ⎜
⎟ × (11.0093 ) = 10.81u ⎝ 100 ⎠
⎝ 100 ⎠
Dimensional analysis and mole/mass calculations: Moles and mass are related by molar mass. Mass/Mass calculations always go through moles: Example: How many grams of hydrogen are in 10.5g of water: g
moles
moles ×
= g g ×
= moles 1 mol H2 O 2 mols H 1.0079g H
moles
g
10.5g H2 O ×
×
×
= 1.17g H
18.02g H2 O 1 mol H2 O 1 mole H
Empirical formula calculations • The Empirical Formula of a compound represents the simplest ratio of elements in the compound. • Empirical formulas are often determined from the % composition of a compound. • In this calculation, the relative percentages of each element are equated to grams, i.e. 100% = 100g. • The grams are converted to moles and the mole ratios of the elements in the compound are found by dividing each mole value by the smallest mole value. Page 1 of 2 Chem. 1A Week 3 Discussion Notes Dr. Mack/S12 Chemical formulas from mass data: Another method of determining the formula of a compound is to use mass data from chemical reactions. There are many different approaches to these problems, however they all are base on a common principle; matter is conserved in any chemical process. 62.69 g
Example: An ionic compound has the formula MCl2. Molar mass (MCl2 ) =
= 208.2 g/mol
The mass of 0.3011 mol of the compound is 62.69 0.3011 mol
grams. What is the identity of the metal? Solution: The molar mass of the compound is found 208.2 g / mol − 2 × (35.45 g / mol) = 137.3 g / mol from the data. Subtracting of the molar mass of chlorine (×2) yields M = Barium
the molar mass of M. Hydrate formula determination from mass data: A hydrate is an inorganic salt that has water physically bound as part of the crystal structure. The general formula for hydrate is given by MX⋅nH2O where n = the number of water molecules associates with one mole of the salt. When a hydrate is heated, the water is liberated leaving behind the anhydrous salt. Δ
MX ⋅ nH2 O ⎯⎯
→ MX(s) + H2 O(g) The formula of the salt can be determined from mass data. Example: A hydrated salt of copper (II) sulfate was Mass of empty Test Tube: 25.1051 g heated to liberate the water. From the data given, Mass of Test Tube + hydrate: 25.5846 g determine the chemical formula of the hydrate. Mass of TT + hydrate after heating: 25.4116 g From the mass data, the mass of water lost and the mass of anhydrous salt that remains may be determined. These masses yield there corresponding moles which in turn gives the formula for the salt. Mass & moles water lost
25.5846g − 25.4116g = 0.1730 g H2 O ×
1 mole H2 O
= 0.009600 moles H2 O
18.02g
Mass & moles CuSO4 remaining
25.4116g − 25.1051g = 0.3065 g CuSO4 ×
n=
1 mole CuSO4
= 0.001920 moles H2 O
159.61g
mols H2 O from hydrate 0.009600
=
= 5 the formula is CuSO4 ⋅ 5H2 O
mols anhydrous salt
0.001920
Page 2 of 2