The number of moles of a substance is deﬁned as:
- 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:
Exercise 1: Given the following additional RAM data: Mg = 24, C = 12, H = 1, calculate the following :
Solution: Since the RAMs for Mg and Cl are, respectively, 24 and 35.5, the RMM for MgCl_{2} is:
Then:
Solution: The RMM for Cl_{2} is 2 × 35.5 = 71. Rearranging the equation:
We obtain:
Solution: The RMM for C_{6}H_{6} is:
Thus:
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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 MgSO_{4}?
Explanation: The RMM for MgSO_{4} is:
24 + 32 + (4 × 16) = 120
Since:
we have:
mass | = | number of moles × RMM |
= | 0.125 × 120 | |
= | 15g |
The density of a substance is deﬁned to be its mass per unit of volume. Symbolically:
To see how this is used look at the following example:
Example 2: Find the density of water if 20cm^{3} has a mass of 20.4g.
Solution: Using the deﬁnition above:
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:
Since:
Rearranging the equation gives:
Rearranging the equation for the density, we have:
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 (C_{3}H_{6}O) is given by:
As we have seen earlier:
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Now try this quiz:
Quiz 2: If the density of cyclohexane (C_{6}H_{12}) is 0.78g cm^{−3}, which of the following is the number of moles in 100cm^{3} of the substance?
Explanation: The solution to this problem involves two steps; first use the density to find the mass of 100cm^{3} 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:
so we have:
mass | = | number of moles × RMM |
= | 0.125 × 120 | |
= | 15g |
The number of moles of a chemical substance contained in a solution is deﬁned by:
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 cm^{3}, or 1dm^{3}, so the units are also in mol dm^{−3}.
Example 3: Calculate the number of moles of Cu^{2+} ions in 25cm^{3} of a 0.1M solution.
Solution: Using the deﬁnition above, we have:
Exercise 3: Calculate the following:
Solution: Since:
Rearranging, this gives:
Solution: Since:
we have, on rearranging the equation:
Solution: The pH is given by:
Since the concentration is [H^{+}] = 1.0, its pH is:
Click on questions to reveal solutions.
And now a quiz:
Quiz 3: If 11.1 g of CaCl_{2}, with a Relative Molecular Mass of 111 g mol^{−1} is dissolved in 2500 cm^{3} of water, which of the following pairs represents the concentrations of Ca_{2}^{+} and Cl^{−} ions?
Explanation: From Section 1:
So, 11.1g of CaCl_{2} (RMM = 111g mol^{−1}) corresponds to 11.1 ⁄ 111 = 0.1moles of CaCl_{2}.
The number of moles of a chemical substance contained in a solution is related to the molarity by:
So that:
Now, 1mole CaCl_{2} is composed of 1mole Ca^{2+} and 2moles Cl^{−}, thus we obtain:
molarity of Ca^{2+} = (0.1×100) ⁄ 2500 = 0.04 ; molarity of Cl^{−} = (0.2×100) ⁄ 2500 = 0.08
The Gibbs free energy equation is deﬁned as:
ΔG = ΔH − T ΔS | (1) |
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- 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:
Exercise 4: Calculate the following:
Solution: From Example 4 we have:
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Exercise 5: The standard equilibrium isotherm is defined as:
ΔG^{0} = −RT ln K | (2) |
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- where K corresponds to the equilibrium constant for a chemical reaction, and the superscript ^{0} denotes the isotherm.
Solution: From (2) we have:
Dividing both sides by −RT:
or:
Solution: From the module on logarithms we have the formula for changing the bases of logarithms:
With a = e, b = 10, c = x, we have:
so that, from (2), since ln 10 ≅ 2.3:
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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?
Explanation: From (1) and (2), we obtain:
ΔG^{0} | = | −RT ln K | = | ΔH^{0} − T ΔS^{0}_{} |
---|---|---|---|---|
so that: | RT ln K | = | −ΔH^{0} + T ΔS^{0}_{} | |
ln K | = | −ΔH^{0} + T ΔS^{0}_{} RT | ||
= | −ΔH^{0} RT + ΔS^{0} R |
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) |
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- 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?
Explanation: The rate equation is:
When x = 0 we have [A]^{0} = 1. In this case the expression simplifies to:
Quiz 6: Which of the following is an alternative expression?
Explanation: Since:
We have, on taking logs:
Note: Here we have used the following laws of logarithms:
(See the Logarithms module for details)