Introduction
- Quantitative analysis in chemistry focuses on determining and calculating the quantities of substances involved in chemical reactions.
- It enables chemists to identify the strength of solutions, estimate the amount of products formed, and evaluate how effective a reaction is.
- Through the use of experiments and mathematical calculations, it ensures reactions are performed precisely, minimizes unnecessary waste, and enhances safety in laboratory and industrial processes.
Applications of Quantitative Analysis:
- Medicine – calculating accurate drug dosages for safe treatments.
- Food industry – testing sugar, salt, or nutrient levels in foods.
- Water treatment – measuring concentrations of harmful ions or chemicals in drinking water.
- Industrial chemistry – ensuring the right amounts of reactants are used to maximise yield.
What is Concentration?
- Concentration tells us the amount of solute (the substance being dissolved) present in a given volume of solution.
Example
When sugar is added to water, the sweetness increases as more sugar dissolves. This shows that the solution becomes more concentrated with each spoon of sugar added.
Sugar dissolving in water
- Concentration is usually measured in $mol/dm^3$ (moles per cubic decimetre).
- In chemistry, moles are represented by $n$, concentration by $C$, and volume by $V$.
- Concentration can be calculated using the triangle formula, where moles ($n$) are at the top and concentration ($C$) and volume ($V$) are at the bottom.
Concentration Formula Triangle
Note: How do we convert between $g/dm^3$ and $mol/dm^3$?
- To change from $g/dm^3$ to $mol/dm^3$ divide by the molar mass of the solute.
- To change from $mol/dm^3$ to $g/dm^3$ multiply by the molar mass.
Solved Example
A solution contains $10~g$ of sodium hydroxide (NaOH) dissolved in $1~dm^3$ of solution. Calculate its concentration in $mol/dm^3$.
SOLUTION
Given:
- Mass of NaOH $= 10~g$
- Volume of solution $= 1~dm^3$
- Molar mass of NaOH $= 40~g/mol$
1
Convert mass into moles
$$\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar Mass}}$$
$$\text{Moles of NaOH} = \frac{10}{40} = 0.25~mol$$
2
Find concentration
Using formula:
$$C = \frac{n}{V}$$
$$C = \frac{0.25}{1}$$
Final Answer: $C = 0.25~mol/dm^3$
What is Titration?
- The experimental technique used to find the exact concentration of an unknown acid or alkali is called titration.
- It is very important in chemistry because it gives accurate results.
How is titration carried out?
- Place the acid into a burette and note the starting volume.
Acid in a burette
- Use a pipette to transfer a known volume of alkali into a conical flask.
Transferring alkali using a pipette
- Add a few drops of a suitable indicator, such as phenolphthalein or methyl orange.
- Slowly release the acid from the burette while gently swirling the flask, until the indicator changes colour. This point is called the end point.
- Record the volume of acid used – this is known as the titre.
- Repeat the experiment several times until you obtain results that are very close together, known as concordant results.
Calculations from Titration Results
- We can find the concentration of an unknown solution or the volume required to neutralise another solution.
How do we calculate it?
- Use the titre volume and the concentration of the known solution to determine the number of moles.
- Apply the balanced chemical equation to identify the mole ratio between the acid and the alkali.
- Using this ratio, calculate the moles of the unknown solution.
- Calculate the unknown concentration or volume using the formula for concentration.
Solved Example
If $30.0~cm^3$ of $0.250~mol/dm^3~H_2SO_4$ completely neutralises $25.0~cm^3$ of NaOH solution, what is the concentration of the NaOH?
SOLUTION
1
To calculate moles:
Moles of $H_2SO_4 = \text{concentration} \times \text{volume (in } dm^3 \text{)}$
$$\text{Moles of } H_2SO_4 = 0.250 \times \left(\frac{30}{1000}\right)$$
$$\text{Moles of } H_2SO_4 = 0.250 \times 0.0300$$
$$\text{Moles of } H_2SO_4 = 0.00750~mol$$
2
Apply the balanced chemical equation:
Balanced equation:
$$H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$$
Ratio $H_2SO_4 : NaOH = 1:2$
3
Using this ratio, calculate the moles of the unknown solution.
Moles of NaOH:
$$= 2 \times 0.00750 = 0.0150~mol$$
4
Calculate the unknown concentration or volume using the formula for concentration.
$$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}$$
First calculate the volume:
$$\text{Volume} = \frac{25}{1000}$$
$$\text{Volume} = 0.0250$$
Now,
$$\text{Concentration} = \frac{0.0150}{0.0250}$$
$$\text{Concentration} = 0.600~mol/dm^3$$
Final Answer: The concentration of the NaOH solution is $0.600~mol/dm^3$.
What is molar volume?
- At room temperature and pressure (RTP, $20^\circ\text{C}$, $1\text{ atm}$), 1 mole of any gas occupies $24~dm^3$. This is called the molar volume of a gas.
- At room temperature and pressure (RTP), one mole of any gas occupies about $24~dm^3$ ($24,000~cm^3$).
- At standard temperature and pressure (STP), one mole of any gas occupies about $22.4~dm^3$.
Formulas:
- Volume $(dm^3) = \text{moles} \times 24$
- Volume $(cm^3) = \text{moles} \times 24,000$
Solved Example
Calculate the volume of carbon dioxide gas produced at RTP when $10~g$ of calcium carbonate ($CaCO_3$) reacts with hydrochloric acid.
SOLUTION
1
Write the balanced equation
$$CaCO_3 + 2HCl \longrightarrow CaCl_2 + H_2O + CO_2$$
2
Calculate moles of Calcium Carbonate
Molar mass of Calcium Carbonate ($CaCO_3$) $= 100~g/mol$
$$\text{Moles of } CaCO_3 = \frac{10}{100}$$
$$\text{Moles of } CaCO_3 = 0.1~mol$$
3
Calculate moles of Calcium Carbonate
From the balanced equation:
- 1 mole of $CaCO_3$ produces 1 mole of $CO_2$.
- Therefore, 0.1 mole of $CaCO_3$ will produce 0.1 mole of $CO_2$.
4
Convert moles of $CO_2$ to volume
$$\text{Volume of } CO_2 = 0.1 \times 24 = 2.4~dm^3$$
Final Answer: So, $2.4~dm^3$ of $CO_2$ gas is produced at RTP.
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Book a Consultation →What is Avogadro’s law?
- Avogadro’s law states that 1 mole of any substance contains $6.02 \times 10^{23}$ particles, such as atoms, ions, or molecules.
It is useful because:
- Connects moles to the number of particles.
- Equal moles of gases have the same volume in the same conditions. Makes it easy to compare and calculate gas volumes.
Solved Example
How many molecules are present in 2 moles of hydrogen gas ($H_2$)? (Take Avogadro’s number $= 6.02 \times 10^{23}$).
SOLUTION
$\text{Number of molecules} = \text{moles} \times \text{Avogadro’s number}$
$$\text{Number of molecules} = 2 \times 6.02 \times 10^{23}$$
$$\text{Number of molecules} = 1.204 \times 10^{24} \text{ molecules}$$
Final Answer: 2 moles of $H_2$ contain $1.204 \times 10^{24}$ molecules.
What is percentage yield?
- Percentage yield tells us the portion of product we actually get compared to the amount we would get if the reaction worked perfectly. It is a measure of efficiency.
Formula:
$$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$
Solved Example
In a chemical reaction, the theoretical yield of a product is $50~g$, but only $40~g$ is actually obtained. Calculate the percentage yield.
SOLUTION
Given:
- Theoretical yield of a product $= 50~g$
- Actual yield of a product $= 40~g$
Now using formula:
$$\text{Percentage yield} = \frac{40}{50} \times 100$$
$$\text{Percentage yield} = 0.8 \times 100$$
$$\text{Percentage yield} = 80\%$$
Final Answer: The percentage yield is $80\%$.
What is atom economy?
- Atom economy tells us what fraction of the reactants is actually turned into the product we want.
- It tells us how efficient a process is.
- It is calculated by using formula:
$$\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100$$
Importance of atom economy:
- High atom economy $\longrightarrow$ produces less waste and is more sustainable.
- Helps protect the environment.
- Saves money in industrial processes.
Solved Example
Calculate the atom economy for the reaction:
$$N_2 + 3H_2 \longrightarrow 2NH_3$$
(Molar masses: $N_2 = 28~g/mol$, $H_2 = 2~g/mol$, $NH_3 = 17~g/mol$)
SOLUTION
1
Find the total mass of desired product:
2 moles of $NH_3: 2 \times 17 = 34~g$
2
Find the total mass of reactants:
$N_2 + 3H_2$
$$= 28 + (3 \times 2) = 28 + 6 = 34~g$$
3
Calculate atom economy:
Using formula:
$$\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100$$
Now plug the values into formula,
$$\text{Atom economy} = \frac{34}{34} \times 100$$
Final Answer: Atom economy $= 100\%$
Solved Example
Calculate the atom economy for the reaction:
$$CH_4 + 2Cl_2 \longrightarrow CH_2Cl_2 + 2HCl$$
(Molar masses: $CH_4 = 16~g/mol$, $Cl_2 = 71~g/mol$, $CH_2Cl_2 = 85~g/mol$, $HCl = 36.5~g/mol$)
SOLUTION
1
Identify the desired product
- The desired product is $CH_2Cl_2$ (dichloromethane).
- The by-product is $HCl$.
2
Calculate total mass of products
$$\text{Total Mass of all reactants} = (1 \times 85) + (2 \times 36.5)$$
$$= 85 + 73 = 158$$
3
Calculate atom economy
$$\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100$$
Now plug the values into formula,
$$\text{Atom economy} = \frac{85}{158} \times 100$$
Final Answer: Atom economy $= 53.8\%$
