Tutorial #2

Atomic Structure, Mole Calculations

I. Atomic Structure

Energy levels in a simple hydrogen atom can be calculated using the Balmer-Rydberg equation:

1/l = R[1/m² - 1/n²] where R is the Rydberg constant, 0.01097 nm^-1, n is the upper energy level and m is the lower one.

Note that three series of lines have been identified, differing only in the value of m. If m=2, we have the Balmer series, in the visible range. If m=1, we have the Lyman series, in the UV range. And if m=3, we have the Paschen series in the IR.

Note also that if n=infinity, this is the largest energy transition in the series, and 1/l = R[1/m²].

1. (question 5.51 fron text, 2nd ed.)

Lines in the Brackett series have m=4. Find the l of the first line in the series. Find the photon energy in kJ/mol.

1/l = R[1/m² - 1/n²]

The first line has n=5.

Thus, 1/l = 0.01097 nm^-1 [1/4² - 1/5²] = 0.000247 nm^-1. Thus l = 4051 nm (IR)
E = hn = hc/l = 6.63 x 10^-34 J s (3.00 x 10⁸ m s^-1) / 4051 x 10^-9 m = 4.90 x 10^-20 J/photon
x 6.02 x 10²³ mol^-1 = 29,600 J/mol photons = 29.6 kJ/mol

Quantum Numbers

The Schroedinger equation yields three quantum numbers: n, l and m_l. n is the principle quantum number. It signifies the energy of the electrons in it, and their distance from the nucleus. The value of n must be >0. We speak of the n=1 shell, the n=2 shell and so on.

Within a shell there are subshells. These are numbered l = 0, 1,...n-1. Thus, the number of subshells depends on the value of n. In the n=1 shell, l=0 only, and thus the only type of orbital is a so-called s-orbital. In the n=2 shell, l can have values of 0 or 1, which signify s or p orbitals. In the n=3 shell, there can be s, p or d orbitals, and in the n=4 shell, there can be s, p, d and f orbitals. s, p, d or f orbitals are often referred to as subshells.

Orbitals can have different orientations. The magnetic quantum number m_l specifies which p, d or f orbital we are talking about. In general, -l <= m_l <= l. There are thus three different p-orbitals, 5 d-orbitals and 7 f-orbitals, but only one s-orbital.

AUFBAU

Use the nmemonic to remember the orbital filling order:

1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d 6f
7s 7p 7d 7f

(and draw in the diagonal lines.)

2. What is the ground electronic state of Br (element 35)?

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
or, [Ar] 4s² 3d¹⁰ 4p⁵

3. Draw an orbital filling diagram for Br. Place the orbitals in ascending order of energy.

II. Use of the Mole

4. How many molecules are there in 341 mL of beer? Assume beer is mostly water, whose density = 1.00 g/mL

Thus, g water = 341 mL x 1.00 g/mL = 341 g water
MW(H₂O) = 2 x 1.01 + 16.0 = 18.02 g/mol
Thus, 341 g water = 341 g / 18.02 g/mol = 18.9 mol water
= 18.9 mol x 6.02 x 10²³ molecules mol^-1 = 1.14 x 10²⁵ molecules

For a reaction, we use the balanced chemical reaction as a guide to calculating the amount of product possible from a certain amount of reactant(s). For example,

5. For the reaction 2 Na_(s) + 2  H₂O_(l) ® 2 Na⁺_(aq) + 2 OH^-_(aq) + H_2(g)

calculate the mass of hydrogen gas liberated by the reaction of 1 g Na with excess water.

The method for any such calculation is:

(i)convert mass of reactant to moles of reactant
(ii) use the balanced reaction to determine the moles of product
(iii) convert moles of product to mass of product

For this problem,

(i) 1 g Na = 1 g / 23.0 g/mol = 0.0434 mol Na
(ii) from the reaction, 2 mol Na gives 1 mole H_2(g).
Thus, moles H_2(g) = 0.0434 mol Na x (1 mol H_2(g) / 2 mol Na) = 0.0217 mol H_2(g)
(iii) g H_2(g) = 0.0217 mol H_2(g) x 2.02 g/mol = 0.0878 g H_2(g).

Limiting Reactants

Often (usually, in fact), reactions are performed with one of the reactants in excess of the amonut required to completely react with the other. We need to determine how much of the products are formed under these circumstances.

6. Suppose we react 1 g of chlorine gas with 1 gram of ethene according to the reaction:

Cl_2(g) + C₂H_4(g) ® C₂H₄Cl_2(g)

     Which reactant gets used up first, i.e. which is the limiting reactant? Also, how much product (in g) is produced?

For limiting reactant problems, the approach is similar to the above problem:

(i) calculate the moles of each reactant available
(ii) calculate which one is in excess - the other is the limiting reactant, and will determine how much product is formed.

1 g Cl_2(g) = 1 g / 70.9 g/mol = 0.0141 mol Cl_2(g)
1 g C₂H_4(g) = 1 g / 28.0 g/mol = 0.0357 mol C₂H_4(g)

Thus, since they react in a 1:1 mol ratio, the C₂H_4(g)  is in excess, and so the Cl_2(g) is the limiting reactant.

Thus, moles C₂H₄Cl_2(g) = moles Cl_2(g) = 0.0141 mol
= 0.0141 mol x 98.9 g/mol = 1.39 g  C₂H₄Cl_2(g) produced

7. For the reaction HI + KHCO₃ à KI + H₂O + CO₂
(a) how much KI is produced from reacting 48 g HI with 318 g KHCO₃?

481 g HI x (1 mol / 127.9 g) = 3.76 mol HI
318 g KHCO₃ x (1 mol / 100.1 g) = 3.18 mol KHCO₃

Thus, HI is in excess, since the two reactants react in a 1:1 mole ratio. Thus, the amount of KI formed is limited by the amount of KHCO₃.

? g KI = 3.18 mol KHCO₃ x (1 mol KI / 1 mol KHCO₃) x (166.0 g KI / 1 mol KI)
= 528 g KI
The term in red comes from inspection of the reaction, showing the 1:1 mole ratio of KI to KHCO₃

(b) How much HI is consumed and how much is left over?

HI consumed = 3.18 mol KHCO₃ x (1 mol HI / 1 mol KHCO₃) x (127.9 g HI / 1 mol HI)
= 407 g HI
Thus, HI left over = 481 - 407 = 74 g HI left over.