Quantitative Chemistry | GCSE - Wyatt's Notes

Quantitative Chemistry

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1. Relative Formula Mass

1.1 Calculating Relative Formula Mass

The relative formula mass ( $M_r$ ) of a compound is the sum of the relative atomic masses of all Atoms in the formula. For a molecule, this is also called the relative molecular mass.

Worked Example. Calculate $M_r$ for CaCO $_3$ .

$M_r(\mathrm{CaCO_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$

Worked Example. Calculate $M_r$ for (NH $_4$ ) $_2$ SO $_4$ .

$M_r = (2 \times 14.0) + (8 \times 1.0) + 32.1 + (4 \times 16.0) = 28 + 8 + 32.1 + 64 = 132.1$

The formula, including atoms inside brackets. The subscript outside the bracket applies to every Atom inside.

Worked Example. Calculate $M_r$ for Cu(NO $_3$ ) $_2$ .

$M_r = 63.5 + 2(14.0 + 3 \times 16.0) = 63.5 + 2(14.0 + 48.0) = 63.5 + 2(62.0) = 63.5 + 124.0 = 187.5$

Worked Example. Calculate $M_r$ for Al $_2$ (SO $_4$ ) $_3$ .

$M_r = 2(27.0) + 3(32.1 + 4 \times 16.0) = 54.0 + 3(32.1 + 64.0) = 54.0 + 3(96.1) = 54.0 + 288.3 = 342.3$

Worked Example. Calculate $M_r$ for Fe $_2$ O $_3$ .

$M_r = 2(55.8) + 3(16.0) = 111.6 + 48.0 = 159.6$

1.2 Conservation of Mass and Balancing Equations

In a balanced chemical equation, the total $M_r$ of reactants equals the total $M_r$ of products. This is a direct consequence of the conservation of mass.

Worked Example. Verify that the equation $\mathrm{CaCO_3 \to \mathrm{CaO + \mathrm{CO_2$ obeys the Law of conservation of mass.

$M_r(\mathrm{CaCO_3) = 100.1$$M_r(\mathrm{CaO) = 56.1$$M_r(\mathrm{CO_2) = 44.0$ .

$100.1 = 56.1 + 44.0 = 100.1$ . The equation is balanced.

Worked Example. Verify the conservation of mass for: $2\mathrm{Mg + \mathrm{O_2 \to 2\mathrm{MgO$ .

$2(24.3) + 2(16.0) = 48.6 + 32.0 = 80.6$ $2(24.3 + 16.0) = 2(40.3) = 80.6$

Both sides equal 80.6.

2. The Mole Concept

2.1 Definition

The mole is the unit for amount of substance. One mole contains exactly $6.02 \times 10^{23}$ Particles (this number is Avogadro”s constant, $N_A$ ).

The mole bridges the gap between the atomic scale and the macroscopic scale. One mole of carbon-12 Has a mass of exactly 12 g. One mole of any substance contains the same number of particles as there Are atoms in 12 g of carbon-12.

2.2 Why the Mole Is Useful

Without the mole, chemists would have to work with individual atoms — far too small to weigh. The Mole provides a convenient scale: instead of saying “react $6.02 \times 10^{23}$ molecules”, we say “react 1 mole”. The balanced equation then directly gives the mole ratio, making calculations Straightforward.

2.3 Key Relationships

$n = \frac{m}{M_r}$

Where $n$ is the number of moles, $m$ is the mass in grams, and $M_r$ is the relative formula mass.

$m = n \times M_r$

This equation is the single most useful relationship in quantitative chemistry. Every stoichiometric Calculation ultimately relies on converting between mass and moles.

2.4 Calculating Moles, Mass, and $M_r$

Worked Example. Calculate the number of moles in 15 g of NaOH.

$M_r(\mathrm{NaOH) = 23 + 16 + 1 = 40$ $n = \frac{15}{40} = 0.375 \mathrm{ mol$

Worked Example. Calculate the mass of 0.5 mol of CO $_2$ .

$M_r(\mathrm{CO_2) = 12 + (2 \times 16) = 44$ $m = 0.5 \times 44 = 22 \mathrm{ g$

Worked Example. Calculate the mass of 2.5 mol of H $_2$ SO $_4$ .

$M_r(\mathrm{H_2\mathrm{SO_4) = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98$ $m = 2.5 \times 98 = 245 \mathrm{ g$

Worked Example. Calculate the number of moles in 50 g of CaCO $_3$ .

$n = \frac{50}{100.1} = 0.500 \mathrm{ mol$

Worked Example. What is the mass of 0.25 mol of nitrogen gas (N $_2$ )?

$M_r(\mathrm{N_2) = 2 \times 14 = 28$ $m = 0.25 \times 28 = 7.0 \mathrm{ g$

2.5 Number of Particles

$\mathrm{Number of particles = n \times N_A = n \times 6.02 \times 10^{23}$

Worked Example. How many molecules are there in 0.25 mol of H $_2$ O?

$\mathrm{Molecules = 0.25 \times 6.02 \times 10^{23} = 1.505 \times 10^{23}$

Worked Example. How many atoms are there in 0.5 mol of O $_2$ ?

Each O $_2$ molecule contains 2 atoms, so:

$\mathrm{Atoms = 0.5 \times 6.02 \times 10^{23} \times 2 = 6.02 \times 10^{23}$

Worked Example. How many molecules are in 88 g of CO $_2$ ?

$n = \frac{88}{44} = 2 \mathrm{ mol$ $\mathrm{Molecules = 2 \times 6.02 \times 10^{23} = 1.204 \times 10^{24}$

2.6 Molar Volume of Gas

At room temperature and pressure (RTP, approximately $25^{\circ}\mathrm{C$ and 1 atm), one mole of Any gas occupies approximately 24 dm $^3$ .

$n = \frac{V}{24}$

Where $V$ is the volume in dm $^3$ .

The fact that one mole of any gas occupies the same volume is a consequence of Avogadro’s law: equal Volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Worked Example. What volume does 3 mol of CO $_2$ occupy at RTP?

$V = 3 \times 24 = 72 \mathrm{ dm^3$

Worked Example. 480 cm $^3$ of a gas is collected at RTP. How many moles is this?

$n = \frac{0.480}{24} = 0.020 \mathrm{ mol$

Note: 480 cm $^3$ = 0.480 dm $^3$ (divide by 1000).

Worked Example. What volume does 4 g of methane (CH $_4$ ) occupy at RTP?

$M_r(\mathrm{CH_4) = 12 + 4(1) = 16$ $n = \frac{4}{16} = 0.25 \mathrm{ mol$ $V = 0.25 \times 24 = 6.0 \mathrm{ dm^3$

Worked Example. What mass of nitrogen gas (N $_2$ ) occupies 6 dm $^3$ at RTP?

$n = \frac{6}{24} = 0.25 \mathrm{ mol$ $m = 0.25 \times 28 = 7.0 \mathrm{ g$

Worked Example. What volume does 0.125 mol of oxygen gas occupy at RTP?

$V = 0.125 \times 24 = 3.0 \mathrm{ dm^3$

2.7 Derivation: Avogadro’s Law

Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain Equal numbers of molecules. This arises because gas molecules are very far apart compared to their Size, so the volume depends on the number of molecules, not on their identity. At RTP, one mole of Any ideal gas occupies 24 dm $^3$ .

3. Conservation of Mass and Balanced Equations

3.1 The Law of Conservation of Mass

In a closed system, the total mass of the reactants equals the total mass of the products. No atoms Are created or destroyed in a chemical reaction.

This law holds exactly in a closed system. In an open system, mass may appear to be lost (e.g., gas Escaping) or gained (e.g., gas from the atmosphere dissolving), but the atoms are still conserved.

3.2 Using Equations to Calculate Masses

Step-by-step method:

Write the balanced equation
Write the $M_r$ values under each substance
Write the known mass (or moles) and find the moles of the known substance
Use the mole ratio to find the moles of the unknown substance
Convert moles to mass

Worked Example. Calculate the mass of iron produced when 16 g of iron(III) oxide is reduced by Carbon.

$\mathrm{Fe_2\mathrm{O_3 + 3\mathrm{C \to 2\mathrm{Fe + 3\mathrm{CO$

$M_r(\mathrm{Fe_2\mathrm{O_3) = 2 \times 56 + 3 \times 16 = 160$ $M_r(\mathrm{Fe) = 56$

$n(\mathrm{Fe_2\mathrm{O_3) = \frac{16}{160} = 0.1 \mathrm{ mol$

From the equation: $n(\mathrm{Fe) = 2 \times n(\mathrm{Fe_2\mathrm{O_3) = 0.2 \mathrm{ mol$

$m(\mathrm{Fe) = 0.2 \times 56 = 11.2 \mathrm{ g$

Worked Example. What mass of magnesium oxide is formed when 12 g of magnesium burns in oxygen?

$2\mathrm{Mg + \mathrm{O_2 \to 2\mathrm{MgO$

$n(\mathrm{Mg) = \frac{12}{24} = 0.5 \mathrm{ mol$

Mole ratio: $n(\mathrm{MgO) = n(\mathrm{Mg) = 0.5 \mathrm{ mol$

$m(\mathrm{MgO) = 0.5 \times (24 + 16) = 0.5 \times 40 = 20 \mathrm{ g$

Worked Example. What mass of zinc chloride is produced when 6.5 g of zinc reacts with excess Hydrochloric acid?

$\mathrm{Zn + 2\mathrm{HCl \to \mathrm{ZnCl_2 + \mathrm{H_2$

$n(\mathrm{Zn) = \frac{6.5}{65} = 0.10 \mathrm{ mol$

Mole ratio: 1:1, so $n(\mathrm{ZnCl_2) = 0.10 \mathrm{ mol$ .

$m(\mathrm{ZnCl_2) = 0.10 \times (65 + 2 \times 35.5) = 0.10 \times 136 = 13.6 \mathrm{ g$

Worked Example. What volume of CO $_2$ is produced when 25 g of CaCO $_3$ reacts with excess HCl?

$\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2$

$n(\mathrm{CaCO_3) = \frac{25}{100.1} = 0.250 \mathrm{ mol$

Mole ratio 1:1, so $n(\mathrm{CO_2) = 0.250 \mathrm{ mol$ .

$V(\mathrm{CO_2) = 0.250 \times 24 = 6.0 \mathrm{ dm^3$

4. Concentration

4.1 Concentration in mol/dm $^3$

$c = \frac{n}{V}$

Where $c$ is concentration (mol/dm $^3$ ), $n$ is amount of substance (mol), and $V$ is volume (dm $^3$ ).

Rearrangements:

$n = c \times V, \qquad V = \frac{n}{c}$

Worked Example. What is the concentration of a solution containing 0.5 mol of NaCl in 250 Cm $^3$ ?

$V = 0.250 \mathrm{ dm^3$ $c = \frac{0.5}{0.250} = 2.0 \mathrm{ mol/dm^3$

Worked Example. How many moles of solute are present in 30 cm $^3$ of 0.5 mol/dm $^3$ HCl?

$V = 0.030 \mathrm{ dm^3$ $n = 0.5 \times 0.030 = 0.015 \mathrm{ mol$

Worked Example. What volume of 0.200 mol/dm $^3$ NaOH contains 0.040 mol?

$V = \frac{0.040}{0.200} = 0.200 \mathrm{ dm^3 = 200 \mathrm{ cm^3$

4.2 Concentration in g/dm $^3$

$\mathrm{Concentration (g/dm^3) = \frac{\mathrm{mass (g)}{\mathrm{volume (dm^3)}$

Converting between units:

$\mathrm{g/dm^3 = \mathrm{mol/dm^3 \times M_r$

Worked Example. What is the concentration of 5.85 g of NaCl dissolved in 200 cm $^3$ ?

$\mathrm{Concentration = \frac{5.85}{0.200} = 29.25 \mathrm{ g/dm^3$

In mol/dm $^3$ : $\frac{29.25}{58.5} = 0.5 \mathrm{ mol/dm^3$

Worked Example. What is the concentration in g/dm $^3$ of a 0.100 mol/dm $^3$ solution of H $_2$ SO $_4$ ?

$\mathrm{Concentration = 0.100 \times 98 = 9.8 \mathrm{ g/dm^3$

4.3 Dilution

When a solution is diluted, the number of moles of solute remains constant:

$c_1 V_1 = c_2 V_2$

Worked Example. 25 cm $^3$ of 2.0 mol/dm $^3$ HCl is diluted to 250 cm $^3$ . What is the new Concentration?

$c_2 = \frac{c_1 V_1}{V_2} = \frac{2.0 \times 25}{250} = 0.20 \mathrm{ mol/dm^3$

Worked Example. How much water must be added to 100 cm $^3$ of 6.0 mol/dm $^3$ NaOH to make a 2.0 Mol/dm $^3$ solution?

$V_2 = \frac{c_1 V_1}{c_2} = \frac{6.0 \times 100}{2.0} = 300 \mathrm{ cm^3$

Volume of water added = $300 - 100 = 200 \mathrm{ cm^3$ .

Worked Example. What is the concentration after diluting 10 cm $^3$ of 1.0 mol/dm $^3$ CuSO $_4$ to 100 cm $^3$ ?

$c_2 = \frac{1.0 \times 10}{100} = 0.10 \mathrm{ mol/dm^3$

4.4 Unit Conversion Reminder

$1000 \mathrm{ cm^3 = 1 \mathrm{ dm^3$

To convert cm $^3$ to dm $^3$ : divide by 1000.

To convert dm $^3$ to cm $^3$ : multiply by 1000.

5. Limiting Reactants and Yield

5.1 Limiting Reactant

The limiting reactant is the reactant that is completely used up first. It determines the Maximum amount of product that can be formed. The other reactant is in excess.

To identify the limiting reactant: convert all reactant masses to moles, then compare the actual Mole ratio with the stoichiometric ratio. The reactant that has fewer moles relative to its Coefficient in the balanced equation is the limiting reactant.

Systematic method: Divide the number of moles of each reactant by its stoichiometric Coefficient. The smallest result is the limiting reactant.

Worked Example. 8 g of hydrogen reacts with 32 g of oxygen to form water. Which is the limiting Reactant? What mass of water is formed?

$2\mathrm{H_2 + \mathrm{O_2 \to 2\mathrm{H_2\mathrm{O$

$n(\mathrm{H_2) = \frac{8}{2} = 4 \mathrm{ mol$ $n(\mathrm{O_2) = \frac{32}{32} = 1 \mathrm{ mol$

Divide by coefficients: H $_2$ : $4/2 = 2$ O $_2$ : $1/1 = 1$ . O $_2$ gives the smaller value, so it is Limiting.

From the equation: 1 mol O $_2$ produces 2 mol H $_2$ O.

$m(\mathrm{H_2\mathrm{O) = 2 \times 18 = 36 \mathrm{ g$

Worked Example. 5.6 g of iron reacts with 3.2 g of sulfur. Find the limiting reactant and the Mass of iron(II) sulfide formed.

$\mathrm{Fe + \mathrm{S \to \mathrm{FeS$

$n(\mathrm{Fe) = \frac{5.6}{56} = 0.10 \mathrm{ mol$ $n(\mathrm{S) = \frac{3.2}{32} = 0.10 \mathrm{ mol$

The mole ratio is 1:1, and both reactants have 0.10 mol. Neither is in excess; both are limiting.

$m(\mathrm{FeS) = 0.10 \times (56 + 32) = 0.10 \times 88 = 8.8 \mathrm{ g$

Worked Example. 10 g of calcium carbonate reacts with 100 cm $^3$ of 2.0 mol/dm $^3$ hydrochloric Acid. Identify the limiting reactant.

$\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2$

$n(\mathrm{CaCO_3) = \frac{10}{100.1} = 0.0999 \mathrm{ mol$ $n(\mathrm{HCl) = 2.0 \times 0.100 = 0.200 \mathrm{ mol$

Divide by coefficients: CaCO $_3$ : $0.0999/1 = 0.0999$ HCl: $0.200/2 = 0.100$ . CaCO $_3$ gives the Smaller value, so it is the limiting reactant.

5.2 Percentage Yield

The theoretical yield is the maximum amount of product possible, based on the limiting reactant.

The actual yield is the amount actually obtained in the experiment.

$\mathrm{Percentage yield = \frac{\mathrm{actual yield}{\mathrm{theoretical yield} \times 100\%$

Worked Example. The theoretical yield of ammonia is 17 g. The actual yield is 12.75 g. Calculate The percentage yield.

$\mathrm{Percentage yield = \frac{12.75}{17} \times 100\% = 75\%$

Reasons for yield being less than 100%:

Incomplete reaction (reaction has not gone to completion)
Side reactions producing unwanted products
Product lost during separation and purification (filtration, transfer between vessels)
Reversible reactions not reaching completion

5.3 Atom Economy

Atom economy measures the efficiency of a reaction in terms of how many atoms from the reactants End up in the desired product.

$\mathrm{Atom economy = \frac{M_r \mathrm{ of desired product}{\sum M_r \mathrm{ of all products} \times 100\%$

Worked Example. For the reaction $\mathrm{CaCO_3 \to \mathrm{CaO + \mathrm{CO_2$ Calculate the Atom economy if CaO is the desired product.

$\mathrm{Atom economy = \frac{56}{56 + 44} \times 100\% = \frac{56}{100} \times 100\% = 56\%$

Note: Addition reactions have 100% atom economy. Reactions that produce waste products have Lower atom economy.

Worked Example. For the reaction $\mathrm{CH_4 + \mathrm{H_2\mathrm{O \to \mathrm{CO + 3\mathrm{H_2$ Calculate the atom economy if H $_2$ is the desired product.

$\mathrm{Atom economy = \frac{3 \times 2}{28 + 3 \times 2} = \frac{6}{34} \times 100\% = 17.6\%$

This is a very low atom economy, meaning most of the mass of the reactants ends up in the by-product (CO) rather than the desired product (H $_2$ ).

Worked Example. Calculate the atom economy for the reaction of iron with copper(II) sulfate: $\mathrm{Fe + \mathrm{CuSO_4 \to \mathrm{FeSO_4 + \mathrm{Cu$ Where copper is the desired product.

$\mathrm{Atom economy = \frac{63.5}{151.9 + 63.5} = \frac{63.5}{215.4} \times 100\% = 29.5\%$

5.4 Choosing Reactions: Yield vs. Atom Economy

In industry, both yield and atom economy are important. A reaction with high atom economy but low Yield may waste less material per mole of reactant but produce little product overall. A reaction With high yield but low atom economy produces a lot of waste. The best processes maximise both.

Reaction	Atom Economy	Comment
Addition (e.g. H $_2$ + Cl $_2$ $\to$ 2HCl)	100%	No waste products
Decomposition (CaCO $_3$ $\to$ CaO + CO $_2$ )	56%	CO $_2$ is waste
Displacement (Fe + CuSO $_4$ $\to$ FeSO $_4$ + Cu)	100%	All atoms end up in products

6. The Mole and Gas Volumes in Calculations

6.1 Comprehensive Example

Worked Example. 5.0 g of calcium carbonate is reacted with excess hydrochloric acid. Calculate: (a) the volume of CO $_2$ produced at RTP (b) the mass of CaCl $_2$ produced

$\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2$

(a) $n(\mathrm{CaCO_3) = \frac{5.0}{100} = 0.050 \mathrm{ mol$

Mole ratio: 1:1 for CaCO $_3$ and CO $_2$ .

$V(\mathrm{CO_2) = 0.050 \times 24 = 1.2 \mathrm{ dm^3$

(b) Mole ratio: 1:1 for CaCO $_3$ and CaCl $_2$ .

$m(\mathrm{CaCl_2) = 0.050 \times 111 = 5.55 \mathrm{ g$

6.2 Titration Calculations

Worked Example. 25.0 cm $^3$ of NaOH solution is titrated with 0.100 mol/dm $^3$ HCl. The average Titre is 20.0 cm $^3$ . Calculate the concentration of the NaOH solution.

$\mathrm{NaOH + \mathrm{HCl \to \mathrm{NaCl + \mathrm{H_2\mathrm{O$

$n(\mathrm{HCl) = 0.100 \times 0.0200 = 0.00200 \mathrm{ mol$

Mole ratio: 1:1, so $n(\mathrm{NaOH) = 0.00200 \mathrm{ mol$ .

$c(\mathrm{NaOH) = \frac{0.00200}{0.0250} = 0.0800 \mathrm{ mol/dm^3$

Worked Example. 20.0 cm $^3$ of 0.500 mol/dm $^3$ H $_2$ SO $_4$ is titrated with NaOH. The average Titre is 25.0 cm $^3$ . Calculate the concentration of the NaOH solution.

$\mathrm{H_2\mathrm{SO_4 + 2\mathrm{NaOH \to \mathrm{Na_2\mathrm{SO_4 + 2\mathrm{H_2\mathrm{O$

$n(\mathrm{H_2\mathrm{SO_4) = 0.500 \times 0.0200 = 0.0100 \mathrm{ mol$

Mole ratio: 1:2, so $n(\mathrm{NaOH) = 0.0200 \mathrm{ mol$ .

$c(\mathrm{NaOH) = \frac{0.0200}{0.0250} = 0.800 \mathrm{ mol/dm^3$

Worked Example. 15.0 cm $^3$ of HCl of unknown concentration reacts with 25.0 cm $^3$ of 0.200 Mol/dm $^3$ NaOH. The average titre is 18.0 cm $^3$ . Calculate the concentration of the HCl.

$n(\mathrm{NaOH) = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol$

Mole ratio 1:1, so $n(\mathrm{HCl) = 0.00500 \mathrm{ mol$ .

$c(\mathrm{HCl) = \frac{0.00500}{0.0180} = 0.278 \mathrm{ mol/dm^3$

6.3 Gas Volume Calculations

Worked Example. 2.4 g of magnesium reacts with excess hydrochloric acid. Calculate the volume of Hydrogen produced at RTP.

$\mathrm{Mg + 2\mathrm{HCl \to \mathrm{MgCl_2 + \mathrm{H_2$

$n(\mathrm{Mg) = \frac{2.4}{24} = 0.10 \mathrm{ mol$

Mole ratio: 1:1, so $n(\mathrm{H_2) = 0.10 \mathrm{ mol$ .

$V(\mathrm{H_2) = 0.10 \times 24 = 2.4 \mathrm{ dm^3$

Worked Example. What mass of magnesium is needed to produce 1.2 dm $^3$ of hydrogen at RTP?

$n(\mathrm{H_2) = \frac{1.2}{24} = 0.050 \mathrm{ mol$

Mole ratio 1:1, so $n(\mathrm{Mg) = 0.050 \mathrm{ mol$ .

$m(\mathrm{Mg) = 0.050 \times 24 = 1.2 \mathrm{ g$

7. Water of Crystallisation

Some ionic compounds contain water molecules as part of their crystal structure. These water Molecules are called water of crystallisation.

Example: Hydrated copper(II) sulfate has the formula CuSO $_4 \cdot 5$ H $_2$ O. The dot indicates That five water molecules are associated with each formula unit of copper(II) sulfate.

When hydrated salts are heated, they lose their water of crystallisation and become anhydrous.

$\mathrm{CuSO_4 \cdot 5\mathrm{H_2\mathrm{O(s) \xrightarrow{\Delta} \mathrm{CuSO_4\mathrm{(s) + 5\mathrm{H_2\mathrm{O(g)$

Worked Example. 12.5 g of hydrated CuSO $_4 \cdot x$ H $_2$ O is heated until all water is removed, Leaving 8.0 g of anhydrous CuSO $_4$ . Find $x$ .

$m(\mathrm{H_2\mathrm{O) = 12.5 - 8.0 = 4.5 \mathrm{ g$ $n(\mathrm{CuSO_4) = \frac{8.0}{159.6} = 0.0501 \mathrm{ mol$ $n(\mathrm{H_2\mathrm{O) = \frac{4.5}{18} = 0.250 \mathrm{ mol$ $x = \frac{0.250}{0.0501} \approx 5$

The formula is CuSO $_4 \cdot 5$ H $_2$ O.

Worked Example. 6.30 g of hydrated Na $_2$ CO $_3 \cdot x$ H $_2$ O gives 2.65 g of anhydrous Na $_2$ CO $_3$ on heating. Find $x$ .

$m(\mathrm{H_2\mathrm{O) = 6.30 - 2.65 = 3.65 \mathrm{ g$ $n(\mathrm{Na_2\mathrm{CO_3) = \frac{2.65}{106} = 0.0250 \mathrm{ mol$ $n(\mathrm{H_2\mathrm{O) = \frac{3.65}{18} = 0.203 \mathrm{ mol$ $x = \frac{0.203}{0.0250} \approx 8$

The formula is Na $_2$ CO $_3 \cdot 8$ H $_2$ O.

7.1 Percentage of Water of Crystallisation

Worked Example. Calculate the percentage of water of crystallisation in CuSO $_4 \cdot 5$ H $_2$ O.

$M_r(\mathrm{CuSO_4 \cdot 5\mathrm{H_2\mathrm{O) = 159.6 + 5(18) = 159.6 + 90 = 249.6$ $\%\mathrm{ H_2\mathrm{O = \frac{90}{249.6} \times 100 = 36.1\%$

8. Formula Triangle Summary

The mole triangle helps you remember the relationships:

$n = \frac{m}{M_r}, \quad m = n \times M_r, \quad M_r = \frac{m}{n}$

For concentration:

$c = \frac{n}{V}, \quad n = c \times V, \quad V = \frac{n}{c}$

For gas volume:

$n = \frac{V}{24}, \quad V = n \times 24$

8.1 Comprehensive Worked Example

Problem: 20.0 g of CaCO $_3$ is reacted with 150 cm $^3$ of 1.0 mol/dm $^3$ HCl. (a) Identify the Limiting reactant. (b) Calculate the mass of CaCl $_2$ produced. (c) Calculate the volume of CO $_2$ At RTP. (d) Calculate the percentage yield if 15.0 g of CaCl $_2$ was obtained.

$\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2$

(a) $n(\mathrm{CaCO_3) = \frac{20.0}{100.1} = 0.200 \mathrm{ mol$

$n(\mathrm{HCl) = 1.0 \times 0.150 = 0.150 \mathrm{ mol$

Stoichiometry requires $2 \times 0.200 = 0.400$ mol HCl, but only 0.150 mol is available. HCl is the Limiting reactant.

$n(\mathrm{CaCl_2) = n(\mathrm{HCl)/2 = 0.0750 \mathrm{ mol$

(b) $m(\mathrm{CaCl_2) = 0.0750 \times 111 = 8.33 \mathrm{ g$

(d) $\mathrm{Percentage yield = \frac{15.0}{8.33} \times 100\% = 180\%$

Since the yield exceeds 100%, this indicates experimental error — likely the product was wet.

Common Pitfalls

Forgetting to convert between cm $^3$ and dm $^3$ . 1000 cm $^3$ = 1 dm $^3$ . Divide cm $^3$ by 1000 to get dm $^3$ . This is the single most common arithmetic error in quantitative chemistry.
Using the wrong mole ratio. Always check the balanced equation carefully. The coefficients directly give the mole ratio.
Identifying the wrong limiting reactant. Compare the actual mole ratio with the stoichiometric ratio. Divide each number of moles by its coefficient: the smallest result is the limiting reactant.
Confusing percentage yield with atom economy. Yield is about practical losses; atom economy is about theoretical efficiency. A reaction can have 100% yield but 50% atom economy (or vice versa).
Forgetting that the molar volume of gas (24 dm $^3$ ) applies at RTP, not at standard temperature and pressure (STP uses 22.4 dm $^3$ ). Check which conditions are specified.
Arithmetic errors in calculating $M_r$ . Always double-check, especially with brackets and subscripts.
Forgetting to include water of crystallisation in $M_r$ calculations. When calculating the $M_r$ of a hydrated salt, include the mass of the water molecules.
Using the molar volume for solids or liquids. The molar volume of 24 dm $^3$ /mol applies only to gases at RTP.
Not reading the question carefully for limiting reactant problems. If a reactant is “in excess”, it is not the limiting reactant.

Practice Questions

Calculate the relative formula mass of (NH $_4$ ) $_2$ SO $_4$ .
How many moles are there in 22 g of CO $_2$ ?
6 g of magnesium reacts with excess hydrochloric acid. Calculate the volume of hydrogen produced at RTP.
25 cm $^3$ of 0.5 mol/dm $^3$ H $_2$ SO $_4$ reacts with NaOH. Calculate the mass of Na $_2$ SO $_4$ produced.
10 g of iron is heated with 6.4 g of sulfur to form iron(II) sulfide. Identify the limiting reactant and calculate the maximum mass of FeS that can be produced.
The theoretical yield of a reaction is 25 g, but the actual yield is 20 g. Calculate the percentage yield.
Calculate the atom economy for the reaction: $\mathrm{CH_4 + \mathrm{H_2\mathrm{O \to \mathrm{CO + 3\mathrm{H_2$ Where H $_2$ is the desired product.
A solution is made by dissolving 12 g of NaOH in water to make 500 cm $^3$ of solution. Calculate the concentration in mol/dm $^3$ .
In a titration, 20 cm $^3$ of 0.1 mol/dm $^3$ HCl neutralises 25 cm $^3$ of NaOH solution. Calculate the concentration of the NaOH.
2.4 g of magnesium carbonate is heated until it decomposes completely. Calculate the mass of magnesium oxide produced and the volume of CO $_2$ released at RTP.
A student dissolves 4.0 g of NaOH in 250 cm $^3$ of water, then dilutes 25 cm $^3$ of this solution to 100 cm $^3$ . Calculate the concentration of the diluted solution.
Calculate the percentage of water of crystallisation in CuSO $_4 \cdot 5$ H $_2$ O.
15.0 g of CaCO $_3$ is reacted with 100 cm $^3$ of 2.0 mol/dm $^3$ HCl. Calculate the mass of CaCl $_2$ produced and the volume of CO $_2$ released at RTP. Identify the limiting reactant.
Explain the difference between percentage yield and atom economy, and why both are important in the chemical industry.
3.36 g of iron(III) oxide is reduced by carbon monoxide according to the equation: $\mathrm{Fe_2\mathrm{O_3 + 3\mathrm{CO \to 2\mathrm{Fe + 3\mathrm{CO_2$ . Calculate the mass of iron produced and the volume of CO $_2$ released at RTP.
What volume of 0.5 mol/dm $^3$ sulfuric acid is needed to react completely with 10 g of sodium hydroxide?
Calculate the percentage yield if 4.0 g of CaCO $_3$ produces 1.8 g of CaO. The equation is $\mathrm{CaCO_3 \to \mathrm{CaO + \mathrm{CO_2$ .
What mass of aluminium is produced when 51 g of Al $_2$ O $_3$ is electrolysed? (Assume 100% yield.)
A student wants to make 250 cm $^3$ of 0.10 mol/dm $^3$ HCl from 2.0 mol/dm $^3$ stock solution. What volume of stock solution is needed, and how much water must be added?
8.0 g of CuSO $_4 \cdot x$ H $_2$ O is heated, leaving 5.1 g of anhydrous CuSO $_4$ . Calculate $x$ and state the full formula of the hydrated salt.

9. Multi-Step Calculations

9.1 Combined Mole, Mass, and Volume Problems

Worked Example. 10.0 g of limestone (CaCO $_3$ ) is reacted with excess hydrochloric acid. Calculate the mass of CaCl $_2$ formed, the volume of CO $_2$ at RTP, and the mass of water produced.

$\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2$

$n(\mathrm{CaCO_3) = \frac{10.0}{100.1} = 0.0999 \mathrm{ mol$

Since HCl is in excess, CaCO $_3$ is limiting. Mole ratio 1:1 for all products.

$m(\mathrm{CaCl_2) = 0.0999 \times 111 = 11.1 \mathrm{ g$ $V(\mathrm{CO_2) = 0.0999 \times 24 = 2.40 \mathrm{ dm^3$ $m(\mathrm{H_2\mathrm{O) = 0.0999 \times 18 = 1.80 \mathrm{ g$

9.2 Back-Titration Style Problems

Worked Example. 1.20 g of impure magnesium is reacted with excess dilute sulfuric acid. The Hydrogen produced occupies 1.22 dm $^3$ at RTP. Calculate the percentage purity of the magnesium.

$n(\mathrm{H_2) = \frac{1.22}{24} = 0.0508 \mathrm{ mol$

From $\mathrm{Mg + \mathrm{H_2\mathrm{SO_4 \to \mathrm{MgSO_4 + \mathrm{H_2$ : mole ratio 1:1.

$m(\mathrm{Mg pure) = 0.0508 \times 24.3 = 1.23 \mathrm{ g$

$\mathrm{Percentage purity = \frac{1.23}{1.20} \times 100\% = 102.5\%$

Since this exceeds 100%, there must be experimental error (perhaps the gas was not perfectly dry, or Measurements were slightly off).

9.3 Converting Between Units

Worked Example. Convert 0.05 mol/dm $^3$ to g/dm $^3$ for a solution of NaCl.

$\mathrm{g/dm^3 = 0.05 \times 58.5 = 2.925 \mathrm{ g/dm^3$

Worked Example. A solution of H $_2$ SO $_4$ has concentration 4.9 g/dm $^3$ . What is its molar Concentration?

$\mathrm{mol/dm^3 = \frac{4.9}{98} = 0.050 \mathrm{ mol/dm^3$

10. Errors in Quantitative Chemistry

10.1 Systematic Errors

Systematic errors affect all measurements in the same direction. Common sources include:

Heat loss to the surroundings in calorimetry (always underestimates the energy change)
Not rinsing the burette with the solution it will contain (always affects titre values)
Reading from the top of the meniscus instead of the bottom (always gives high readings)

10.2 Random Errors

Random errors cause measurements to scatter around the true value. Common sources include:

Difficulty judging the exact end-point of a titration
Variation in room temperature affecting gas volumes
Uncertainty in reading a balance

10.3 Reducing Errors

Repeat measurements and take a mean
Use appropriate equipment (burette, not measuring cylinder, for titrations)
Control variables (temperature, concentration)
Ensure all apparatus is clean and dry

10.4 Calculating Uncertainty

If a measurement has an uncertainty of $\pm \delta$ The percentage uncertainty is:

$\mathrm{Percentage uncertainty = \frac{\delta}{\mathrm{measured value} \times 100\%$

Worked Example. A burette reading is 22.50 $\pm$ 0.05 cm $^3$ . Calculate the percentage Uncertainty.

$\mathrm{Percentage uncertainty = \frac{0.05}{22.50} \times 100\% = 0.22\%$

11. Summary Table: Key Equations

Quantity	Equation	Units
Moles from mass	$n = m / M_r$	mol
Mass from moles	$m = n \times M_r$	g
Moles from gas volume	$n = V / 24$	mol
Gas volume from moles	$V = n \times 24$	dm $^3$
Concentration	$c = n / V$	mol/dm $^3$
Moles from concentration	$n = c \times V$	mol
Dilution	$c_1 V_1 = c_2 V_2$	various
Percentage yield	$\frac{\mathrm{actual}{\mathrm{theoretical} \times 100\%$	%
Atom economy	$\frac{M_r\mathrm{(desired)}{\sum M_r\mathrm{(products)} \times 100\%$	%

Practice Problems

Question 1: Moles and mass calculation

Calculate the number of moles in $12.5 \mathrm{ g$ of calcium carbonate ( $\mathrm{CaCO_3$ ). Then calculate the number of molecules.

Answer

$M_r(\mathrm{CaCO_3) = 40 + 12 + 3(16) = 100$ .

Moles $= \frac{12.5}{100} = 0.125 \mathrm{ mol$ .

Number of molecules $= 0.125 \times 6.02 \times 10^{23} = 7.53 \times 10^{22}$ .

Question 2: Titration calculation

$25.0 \mathrm{ cm^3$ of sodium hydroxide solution is titrated with $0.100 \mathrm{ mol/dm^3$ hydrochloric acid. The average titre is $20.0 \mathrm{ cm^3$ . Calculate the concentration of the sodium hydroxide solution.

Answer

Moles of $\mathrm{HCl = 0.100 \times 20.0/1000 = 0.00200 \mathrm{ mol$ .

$\mathrm{NaOH + \mathrm{HCl \to \mathrm{NaCl + \mathrm{H_2\mathrm{O$ (1:1 ratio)

Moles of $\mathrm{NaOH = 0.00200 \mathrm{ mol$ .

Concentration of $\mathrm{NaOH = \frac{0.00200}{25.0/1000} = \frac{0.00200}{0.0250} = 0.0800 \mathrm{ mol/dm^3$ .

Question 3: Limiting reactant

$6.0 \mathrm{ g$ of magnesium reacts with $10.0 \mathrm{ g$ of hydrochloric acid ( $\mathrm{HCl$ ): $\mathrm{Mg + 2\mathrm{HCl \to \mathrm{MgCl_2 + \mathrm{H_2$ . Identify the limiting reactant and calculate the mass of hydrogen produced.

Answer

Moles of $\mathrm{Mg = 6.0/24.3 = 0.247 \mathrm{ mol$ . Moles of $\mathrm{HCl = 10.0/36.5 = 0.274 \mathrm{ mol$ .

Required $\mathrm{HCl = 2 \times 0.247 = 0.494 \mathrm{ mol$ . Only $0.274 \mathrm{ mol$ available, so $\mathrm{HCl$ is limiting.

Moles of $\mathrm{H_2 = 0.274/2 = 0.137 \mathrm{ mol$ . Mass of $\mathrm{H_2 = 0.137 \times 2.0 = 0.274 \mathrm{ g$ .

Question 4: Percentage yield

When $10.0 \mathrm{ g$ of calcium carbonate is heated, $4.20 \mathrm{ g$ of calcium oxide is produced: $\mathrm{CaCO_3 \to \mathrm{CaO + \mathrm{CO_2$ . Calculate the percentage yield.

Answer

Moles of $\mathrm{CaCO_3 = 10.0/100 = 0.100 \mathrm{ mol$ .

Theoretical moles of $\mathrm{CaO = 0.100 \mathrm{ mol$ (1:1 ratio).

Theoretical mass of $\mathrm{CaO = 0.100 \times 56 = 5.60 \mathrm{ g$ .

Percentage yield $= (4.20/5.60) \times 100 = 75.0\%$ .

Question 5: Molar gas volume

Calculate the volume of carbon dioxide produced (at room temperature and pressure) when $5.0 \mathrm{ g$ of calcium carbonate reacts with excess hydrochloric acid. One mole of any gas occupies $24.0 \mathrm{ dm^3$ at RTP.

Answer

$\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2$ .

Moles of $\mathrm{CaCO_3 = 5.0/100 = 0.050 \mathrm{ mol$ .

Moles of $\mathrm{CO_2 = 0.050 \mathrm{ mol$ (1:1 ratio).

Volume of $\mathrm{CO_2 = 0.050 \times 24.0 = 1.2 \mathrm{ dm^3$ .

Worked Examples

Example 1: Mole calculation

Calculate the number of moles in $12.0\,\text{g}$ of $\text{NaOH}$ ( $M_r = 40.0$ ).

Solution:

$n = \frac{m}{M_r} = \frac{12.0}{40.0} = 0.300\,\text{mol}$

Example 2: Reacting masses

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

What mass of $\text{CaCl}_2$ is produced from $10.0\,\text{g}$ of $\text{CaCO}_3$ ? ( $M_r[\text{CaCO}_3] = 100$ , $M_r[\text{CaCl}_2] = 111$ )

Solution:

$n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\,\text{mol}$

From the equation, ratio is $1:1$ , so $n(\text{CaCl}_2) = 0.100\,\text{mol}$ .

$m(\text{CaCl}_2) = 0.100 \times 111 = 11.1\,\text{g}$

flowchart TD
    A[3_Quantitative Chemistry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the key concepts of Quantitative Chemistry for GCSE Chemistry. Focus on understanding the fundamental principles, practising with exam-style questions, and applying your knowledge to unfamiliar contexts.

Intuition

Chemistry explains how atoms combine to form the substances that make up everything around you. Bonding is about how electrons are shared or transferred between atoms, determining properties like melting point and conductivity. Chemical reactions are rearrangements of atoms, where old bonds break and new ones form. The mole concept bridges the atomic world and the laboratory, letting you predict exactly how much product a reaction will produce. Understanding these principles lets you predict the behavior of matter before you even enter the lab.

Cross-References

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