1. Moles and Masses

The number of moles of a substance is defined as:

number of moles = mass/RMM

- where the mass is in grammes (denoted by g) and RMM denotes the Relative Molecular Mass of the substance.

Example 1: Calculate the number of moles of sodium chloride (NaCl) in 5.85 g given the Relative Atomic Mass (RAM) data Na = 23, Cl = 35.5.

Solution: From the question, the RAMs for Na and Cl are, respectively, 23 and 35.5. The RMM for NaCl is thus 23 + 35.5 = 58.5. The number of moles is thus:

number of moles = mass/RMM = 5.85 /58.5 = 0.1

Exercise 1: Given the following additional RAM data: Mg = 24, C = 12, H = 1, calculate the following :

(a)The number of moles in 0.95 g of magnesium chloride (MgCl2)

Solution: Since the RAMs for Mg and Cl are, respectively, 24 and 35.5, the RMM for MgCl2 is:

24 + (2 × 35.5) = 24 + 71 = 95

Then:

number of moles = mass/RMM = 0.95 /95 = 0.01
(b)The mass of 0.05 moles of Cl2

Solution: The RMM for Cl2 is 2 × 35.5 = 71. Rearranging the equation:

number of moles = mass/RMM

We obtain:

mass = number of moles × RMM = 0.05 × 71 = 3.55g
(c)The mass of 0.35moles of benzene (C6H6)

Solution: The RMM for C6H6 is:

(6 × 12) + (6 × 1) = 78

Thus:

mass = number of moles × RMM = 0.35 × 78 = 27.3g

Click on questions to reveal solutions

Now try this quiz:

Quiz 1: Given that the RAMs for S and O are 32 and 16 respectively, which of the following is the mass of 0.125moles of MgSO4?

(a)15gCorrect - well done!
(b)36gIncorrect - please try again!
(c)960gIncorrect - please try again!
(d)1.04 × 100-3g ???????Incorrect - please try again!

Explanation: The RMM for MgSO4 is:

24 + 32 + (4 × 16) = 120

Since:

number of moles = mass/RMM

we have:

mass = number of moles × RMM
  = 0.125 × 120
  = 15g

2. Density

The density of a substance is defined to be its mass per unit of volume. Symbolically:

density = mass/volume

To see how this is used look at the following example:

Example 2: Find the density of water if 20cm3 has a mass of 20.4g.

Solution: Using the definition above:

density = mass/volume = 20.4/20 = 1.02g cm−3

Now for some exercises and a short quiz for practice:

Exercise 2: Use the formula for the density of a substance to calculate the following:

(a)The volume of 5g of ethanol if its density is 0.8g cm−3

Since:

density = mass/volume

Rearranging the equation gives:

volume = mass/density = 5/0.8 = 6.25cm3
(b)The mass of 25 cm3 of mercury if its density is 13.5g cm−3

Rearranging the equation for the density, we have:

mass = density × volume = 13.5 × 25 = 337.5g
(c)The volume of 0.1moles of acetone (C3H6O) if its density is 0.83g cm−3

The solution to this part of the exercise involves two calculations. First, calculate the mass of 0.1moles. of acetone, and then use this to find the volume of the substance.

The RMM of acetone (C3H6O) is given by:

RMM = 3 × 12 + 6 × 1 + 16 = 5.8g

As we have seen earlier:

volume = mass/density = 5.8/0.83 = 7cm3

Click on questions to reveal solutions

Now try this quiz:

Quiz 2: If the density of cyclohexane (C6H12) is 0.78g cm−3, which of the following is the number of moles in 100cm3 of the substance?

(a)15 gCorrect - well done!
(b)36 gIncorrect - please try again!
(c)960 gIncorrect - please try again!
(d)1.04 × 100-3 gIncorrect - please try again!

Explanation: The solution to this problem involves two steps; first use the density to find the mass of 100cm3 of the substance, then use this to find the number of moles. The mass is:

mass = density × volume = 0.78 × 100 = 78g

The number of moles is then:

number of moles = mass/RMM = 78/84 = 0.93

so we have:

mass = number of moles × RMM
  = 0.125 × 120
  = 15g

3. Concentrations of Chemicals in Solution

The number of moles of a chemical substance contained in a solution is defined by:

number of moles = volume × molarity/1000

In this expression, the concentration term is represented by molarity. This can also be represented by M and is equivalent to the number of moles of substance in 1000 cm3, or 1dm3, so the units are also in mol dm−3.

Example 3: Calculate the number of moles of Cu2+ ions in 25cm3 of a 0.1M solution.

Solution: Using the definition above, we have:

number of moles = volume × molarity/1000 = 25 × 0.1/1000 = 2.5 × 10−3

Exercise 3: Calculate the following:

(a)A solution of Cl− ions (0.35M) was titrated and found to contain 7 × 10−4moles of chloride. What volume was titrated?

Solution: Since:

number of moles = volume × molarity/1000

Rearranging, this gives:

volume = number of moles × 1000/molarity = 7 × 10-4 × 103/0.35 = 7 × 10-1/0.35 = 2cm3
(b)An acidic solution (50cm3) is titrated and found to contain 5 × 10−2moles of H+. What is the molarity of H+?

Solution: Since:

number of moles = volume × molarity/1000

we have, on rearranging the equation:

molarity = number of moles × 1000/volume = 5 × 10-2 × 1000/50
= 5 × 10-2 × 103/50 = 5 × 10/50 = 10M
(c)Given that the pH of a substance is minus the log10 of its molarity, what is the pH of the solution in part (b)?

Solution: The pH is given by:

pH = − log10[H+]

Since the concentration is [H+] = 1.0, its pH is:

pH = − log10[10] = 0
i.e. pH = 0

Click on questions to reveal solutions.

And now a quiz:

Quiz 3: If 11.1 g of CaCl2, with a Relative Molecular Mass of 111 g mol−1 is dissolved in 2500 cm3 of water, which of the following pairs represents the concentrations of Ca2+ and Cl− ions?

(a)0.04 and 0.04 MIncorrect - please try again!
(b)0.04 and 0.08 MCorrect - well done!
(c) 0.25 and 0.05 MIncorrect - please try again!
(d)4 and 8 MIncorrect - please try again!

Explanation: From Section 1:

no of moles = mass/RMM

So, 11.1g of CaCl2 (RMM = 111g mol−1) corresponds to 11.1 ⁄ 111 = 0.1moles of CaCl2.

The number of moles of a chemical substance contained in a solution is related to the molarity by:

no of moles = volume × molarity/1000

So that:

molarity = no of moles × 1000/volume

Now, 1mole CaCl2 is composed of 1mole Ca2+ and 2moles Cl−, thus we obtain:

molarity of Ca2+ = (0.1×100) ⁄ 2500 = 0.04 ; molarity of Cl− = (0.2×100) ⁄ 2500 = 0.08

4. Thermodynamics

The Gibbs free energy equation is defined as:

ΔG = ΔH − T ΔS (1)

- where ΔG, ΔH and ΔS correspond to the Gibbs free energy, the enthalpy and the entropy changes associated with a chemical process.

Example 4: Rearrange equation (1) to express it in terms of the enthalpy, ΔH.

Solution: Adding T ΔS to both sides of (1), we obtain:

ΔG + T ΔS = ΔH

Exercise 4: Calculate the following:

(a)Rearrange (1) to obtain ΔS as the subject of the equation.

Solution: From Example 4 we have:

ΔH=ΔG + T ΔS
ΔH − ΔG = T ΔS
ΔH − ΔG/T = ΔS

Click on question to reveal solution.

Exercise 5: The standard equilibrium isotherm is defined as:

ΔG0 = −RT ln K (2)

- where K corresponds to the equilibrium constant for a chemical reaction, and the superscript 0 denotes the isotherm.

(a)Rearrange (2) to obtain ln K as the subject of the equation.

Solution: From (2) we have:

ΔG0 = − RT ln K

Dividing both sides by −RT:

ΔG0/−RT = ln K

or:

ln K = − ΔG0/RT
(b)Given that ln(10) ≈ 2.3, determine the expression for the equilibrium isotherm in terms of log10 K.

Solution: From the module on logarithms we have the formula for changing the bases of logarithms:

loga(c) = loga(b) × logb(c)

With a = e, b = 10, c = x, we have:

ln(x) = (ln 10) × (log10 x)

so that, from (2), since ln 10 ≅ 2.3:

ΔG0 = −2.3 RT log10(K)

Click on questions to reveal solutions.

Now try this quiz:

Quiz 4: Combining (1) and (2), one can generate a further equation, known as the van’t Hoff equation, which relates ln K to ΔH, ΔS and T. Which of those below is this equation?

(a)ln K = ΔH0/RT − ΔS0/R Incorrect - please try again!
(b)ln K = ΔH0 − T ΔS0 + RTIncorrect - please try again!
(c)ln K = − ΔH0/RT + ΔS0/RCorrect - well done!
(d)ln K = RT − ΔH0 − T ΔS0Incorrect - please try again!

Explanation: From (1) and (2), we obtain:

ΔG0 = −RT ln K = ΔH0 − T ΔS0
so that:   RT ln K = −ΔH0 + T ΔS0
    ln K = −ΔH0 + T ΔS0 /RT
      = −ΔH0 /RT + ΔS0 /R

5. Kinetics

When two chemical species A and B, with concentrations [A] and [B] respectively, react together, the general rate equation for the reaction is:

Rate=k [A]x [B]y (3)

- where k is the rate constant and x and y are the appropriate stoichiometric coefficients.

Quiz 5: Using (3), which of the following is true when x = 0?

(a)Rate = 0Incorrect - please try again!
(b)Rate = kIncorrect - please try again!
(c)Rate = k [B]yCorrect - well done!
(d)∞Incorrect - please try again!

Explanation: The rate equation is:

Rate=k [A]x [B]y

When x = 0 we have [A]0 = 1. In this case the expression simplifies to:

Rate=k [B]y

Quiz 6: Which of the following is an alternative expression?

(a)log Rate= log k[A][B]xyIncorrect - please try again!
(b)log Rate= log k + [A] + x + [B] + yIncorrect - please try again!
(c)log Rate= log k + x[A] + y[B]Incorrect - please try again!
(d)log Rate= log k + x log[A] + y log[B]Correct - well done!

Explanation: Since:

Rate=k [A]x [B]y

We have, on taking logs:

log Rate = log k + x log[A] + y log[B]

Note: Here we have used the following laws of logarithms:

log (A × B) = log A + log B
log (Ak) = k log A

(See the Logarithms module for details)

PPLATO material © copyright 2002, Plymouth University