Chemistry - Mole Concept and Stoichiometry

Relative Atomic and Molecular Masses

Relative Atomic Mass ( $A_r$ )

The relative atomic mass of an element is the weighted average mass of one atom of the element Relative to $1/12$ the mass of one atom of carbon-12.

$A_r = \frac{\mathrm{average mass of one atom of the element}}{\frac{1}{12} \times \mathrm{mass of one atom of }^{12}\mathrm{C}}$

It is a dimensionless quantity. For elements with isotopes:

$A_r = \sum (\mathrm{isotope abundance} \times \mathrm{isotope mass})$

Worked example: Chlorine has two isotopes: $^{35}\mathrm{Cl}$ (75.77%) and $^{37}\mathrm{Cl}$ (24.23%).

$A_r(\mathrm{Cl}) = 0.7577 \times 35 + 0.2423 \times 37 = 26.520 + 8.965 = 35.49$

Relative Molecular Mass ( $M_r$ )

The relative molecular mass of a compound is the sum of the relative atomic masses of all atoms in The molecule.

$M_r(\mathrm{H}_2\mathrm{SO}_4) = 2(1.0) + 32.1 + 4(16.0) = 98.1$

For ionic compounds, the term relative formula mass is used, calculated the same way.

$M_r(\mathrm{NaCl}) = 23.0 + 35.5 = 58.5$

The Mole

Definition

The mole is the SI unit for amount of substance. One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities (atoms, molecules, ions, etc.).

$n = \frac{N}{N_A}$

Where $n$ = amount in moles, $N$ = number of particles, $N_A = 6.02 \times 10^{23} \mathrm{ mol}^{-1}$ .

Molar Mass ( $M$ )

The molar mass is the mass of one mole of a substance, numerically equal to $A_r$ or $M_r$ but with Unit g/mol.

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

Substance	Formula	$M_r$	Molar Mass
Water	$\mathrm{H}_2\mathrm{O}$	18.0	18.0 g/mol
Carbon dioxide	$\mathrm{CO}_2$	44.0	44.0 g/mol
Sodium chloride	$\mathrm{NaCl}$	58.5	58.5 g/mol
Sulfuric acid	$\mathrm{H}_2\mathrm{SO}_4$	98.1	98.1 g/mol
Calcium carbonate	$\mathrm{CaCO}_3$	100.1	100.1 g/mol

Molar Volume of Gas at STP

At standard temperature and pressure (STP: $0^\circ\mathrm{C}$ 1 atm), one mole of any ideal gas Occupies **22.4 dm $^3$ **.

At room temperature and pressure (RTP: $25^\circ\mathrm{C}$ 1 atm), one mole occupies **24.0 Dm $^3$ **.

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

Where $V_m$ = 22.4 dm $^3$ /mol (STP) or 24.0 dm $^3$ /mol (RTP).

Formulae of Compounds

Empirical Formula

The empirical formula gives the simplest whole-number ratio of atoms in a compound.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in one molecule.

$\mathrm{Molecular formula} = (\mathrm{Empirical formula})_n$

Where $n = \dfrac{M_r \mathrm{ (molecular)}}{M_r \mathrm{ (empirical)}}$ .

Procedure for Finding Empirical Formula

Write down the mass (or percentage) of each element.
Divide each by its relative atomic mass to get moles.
Divide all mole values by the smallest mole value.
Round to the nearest whole number (or multiply to clear fractions).

Worked example 1: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

Answer

Element	Mass (%)	$A_r$	Moles	Ratio
C	40.0	12.0	40.0 / 12.0 = 3.33	3.33 / 3.33 = 1
H	6.7	1.0	6.7 / 1.0 = 6.7	6.7 / 3.33 = 2
O	53.3	16.0	53.3 / 16.0 = 3.33	3.33 / 3.33 = 1

Empirical formula = $\mathrm{CH}_2\mathrm{O}$

Worked example 2: The molar mass of the compound in the previous example is 180 g/mol. Find its Molecular formula.

Answer

$M_r(\mathrm{CH}_2\mathrm{O}) = 12.0 + 2(1.0) + 16.0 = 30.0$

$n = \dfrac{180}{30.0} = 6$

Molecular formula = $(\mathrm{CH}_2\mathrm{O})_6 = \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ (glucose)

Percentage Composition

$\%\mathrm{ element} = \frac{n \times A_r}{M_r} \times 100\%$

Worked example 3: Calculate the percentage by mass of nitrogen in ammonium nitrate, $\mathrm{NH}_4\mathrm{NO}_3$ .

Answer

$M_r(\mathrm{NH}_4\mathrm{NO}_3) = 14.0 + 4(1.0) + 14.0 + 3(16.0) = 80.0$

Total mass of N = $2 \times 14.0 = 28.0$

$\%\mathrm{N} = \dfrac{28.0}{80.0} \times 100\% = 35.0\%$

Hydrated Salts

A hydrated salt contains water of crystallisation. The formula is written as $\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}$ .

Worked example 4: 12.5 g of hydrated copper(II) sulfate, $\mathrm{CuSO}_4 \cdot x\mathrm{H}_2\mathrm{O}$ Is heated until constant mass is 8.0 g. Find $x$ .

Answer

Mass of water lost = $12.5 - 8.0 = 4.5 \mathrm{ g}$

Moles of $\mathrm{H}_2\mathrm{O}$ = $4.5 / 18.0 = 0.250 \mathrm{ mol}$

Moles of $\mathrm{CuSO}_4$ = $8.0 / 159.6 = 0.0501 \mathrm{ mol}$

( $M_r(\mathrm{CuSO}_4) = 63.5 + 32.1 + 4(16.0) = 159.6$ )

$x = 0.250 / 0.0501 = 4.99 \approx 5$

Formula = $\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}$

Concentration

Molarity

Molarity (molar concentration) is the number of moles of solute per unit volume of solution.

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

Units: mol/dm $^3$ (also written as M).

Note: $1 \mathrm{ dm}^3 = 1000 \mathrm{ cm}^3$ .

$c \mathrm{ (mol/dm}^3) = \frac{n \mathrm{ (mol)}}{V \mathrm{ (dm}^3)} = \frac{n \times 1000}{V \mathrm{ (cm}^3)}$

Mass Concentration

$\mathrm{Concentration (g/dm}^3) = \frac{m \mathrm{ (g)}}{V \mathrm{ (dm}^3)}$

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

Worked example 5: What is the concentration in g/dm $^3$ of a 0.50 mol/dm $^3$ solution of $\mathrm{NaOH}$ ?

Answer

$M_r(\mathrm{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0 \mathrm{ g/mol}$

Concentration = $0.50 \times 40.0 = 20.0 \mathrm{ g/dm}^3$

Molality

Molality is the number of moles of solute per kilogram of solvent (not solution).

$b = \frac{n}{m_{\mathrm{solvent}}}$

Units: mol/kg.

Molality is temperature-independent (unlike molarity), since mass does not change with temperature.

Worked example 6: Calculate the molality of a solution prepared by dissolving 5.85 g of NaCl in 500 g of water.

Answer

Moles of $\mathrm{NaCl} = 5.85 / 58.5 = 0.100 \mathrm{ mol}$

Mass of solvent = $500 \mathrm{ g} = 0.500 \mathrm{ kg}$

Molality = $0.100 / 0.500 = 0.200 \mathrm{ mol/kg}$

Dilution

When diluting a solution, the number of moles of solute remains constant.

$c_1 V_1 = c_2 V_2$

Worked example 7: How would you prepare 250 cm $^3$ of 0.10 mol/dm $^3$ HCl from a 2.0 mol/dm $^3$ Stock solution?

Answer

$c_1 V_1 = c_2 V_2$

$2.0 \times V_1 = 0.10 \times 250$

$V_1 = \dfrac{0.10 \times 250}{2.0} = 12.5 \mathrm{ cm}^3$

Measure 12.5 cm $^3$ of the 2.0 mol/dm $^3$ stock solution using a pipette, transfer to a 250 cm $^3$ Volumetric flask, and add distilled water up to the graduation mark.

Chemical Equations and Stoichiometry

Balancing Equations

The law of conservation of mass requires that the number of atoms of each element is the same on Both sides of a chemical equation.

Steps:

Write the unbalanced equation with correct formulae.
Balance the most complex substance first.
Balance polyatomic ions as a unit if they appear unchanged on both sides.
Balance O and H last.
Check all atoms balance.

Worked example 8: Balance: $\mathrm{Fe}_2\mathrm{O}_3 + \mathrm{CO} \to \mathrm{Fe} + \mathrm{CO}_2$

Answer

$\mathrm{Fe}_2\mathrm{O}_3 + 3\mathrm{CO} \to 2\mathrm{Fe} + 3\mathrm{CO}_2$

Check: Fe: 2 = 2, O: 3 + 3 = 6 = 6, C: 3 = 3.

Molar Ratios in Equations

The stoichiometric coefficients in a balanced equation give the molar ratio of reactants and Products.

$\mathrm{N}_2 + 3\mathrm{H}_2 \to 2\mathrm{NH}_3$

1 mol $\mathrm{N}_2$ reacts with 3 mol $\mathrm{H}_2$ to produce 2 mol $\mathrm{NH}_3$ .

Worked example 9: What mass of $\mathrm{NH}_3$ is produced when 28.0 g of $\mathrm{N}_2$ reacts With excess $\mathrm{H}_2$ ?

Answer

Moles of $\mathrm{N}_2 = 28.0 / 28.0 = 1.00 \mathrm{ mol}$

From the equation $\mathrm{N}_2 + 3\mathrm{H}_2 \to 2\mathrm{NH}_3$ : 1 mol $\mathrm{N}_2$ produces 2 Mol $\mathrm{NH}_3$

Moles of $\mathrm{NH}_3 = 1.00 \times 2 = 2.00 \mathrm{ mol}$

Mass of $\mathrm{NH}_3 = 2.00 \times 17.0 = 34.0 \mathrm{ g}$

Limiting Reagent

The limiting reagent is the reactant that is completely consumed first and thus determines the Maximum amount of product that can be formed.

Procedure

Calculate the moles of each reactant.
Divide each by its stoichiometric coefficient.
The smallest value corresponds to the limiting reagent.
Use the limiting reagent to calculate the amount of product.

Worked example 10: 10.0 g of $\mathrm{Al}$ is reacted with 30.0 g of $\mathrm{HCl}$ according To:

$2\mathrm{Al} + 6\mathrm{HCl} \to 2\mathrm{AlCl}_3 + 3\mathrm{H}_2$

Find the limiting reagent and the volume of $\mathrm{H}_2$ produced at RTP.

Answer

Moles of $\mathrm{Al} = 10.0 / 27.0 = 0.370 \mathrm{ mol}$

Moles of $\mathrm{HCl} = 30.0 / 36.5 = 0.822 \mathrm{ mol}$

Divide by coefficients: $\mathrm{Al}$ : $0.370 / 2 = 0.185$ , $\mathrm{HCl}$ : $0.822 / 6 = 0.137$

Since $0.137 \lt 0.185$ , $\mathrm{HCl}$ is the limiting reagent.

From the equation: 6 mol $\mathrm{HCl}$ produces 3 mol $\mathrm{H}_2$

Moles of $\mathrm{H}_2 = 0.822 \times \dfrac{3}{6} = 0.411 \mathrm{ mol}$

Volume at RTP = $0.411 \times 24.0 = 9.86 \mathrm{ dm}^3$

Percentage Yield

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

Worked example 11: 10.0 g of $\mathrm{CaCO}_3$ is heated and 4.20 g of $\mathrm{CaO}$ is Collected. Calculate the percentage yield.

$\mathrm{CaCO}_3 \to \mathrm{CaO} + \mathrm{CO}_2$

Answer

Moles of $\mathrm{CaCO}_3 = 10.0 / 100.1 = 0.0999 \mathrm{ mol}$

Theoretical moles of $\mathrm{CaO} = 0.0999 \mathrm{ mol}$

Theoretical mass of $\mathrm{CaO} = 0.0999 \times 56.1 = 5.60 \mathrm{ g}$

$\%\mathrm{ yield} = \dfrac{4.20}{5.60} \times 100\% = 75.0\%$

Gas Calculations

Using the Ideal Gas Equation

$PV = nRT$

Symbol	Meaning	Units
P	Pressure	Pa (or atm, kPa)
V	Volume	m $^3$ (or dm $^3$ )
n	Moles	mol
R	Gas constant	8.314 J/(mol K) or 0.0821 atm dm $^3$ /(mol K)
T	Temperature	K

$V_m = \dfrac{nRT}{P} = \dfrac{1 \times 0.0821 \times 273}{1} = 22.4 \mathrm{ dm}^3$

This confirms the molar volume at STP.

Worked example 12: Calculate the volume occupied by 5.00 g of $\mathrm{O}_2$ at 2.0 atm and $27^\circ\mathrm{C}$ .

Answer

$n = 5.00 / 32.0 = 0.156 \mathrm{ mol}$

$T = 27 + 273 = 300 \mathrm{ K}$

$V = \dfrac{nRT}{P} = \dfrac{0.156 \times 0.0821 \times 300}{2.0} = \dfrac{3.842}{2.0} = 1.92 \mathrm{ dm}^3$

Converting Between Mass, Moles, and Volume of Gas

The “mole triangle” connects $n$ , $m$ And $M$ :

$n = \frac{m}{M} \quad ; \quad m = n \times M \quad ; \quad M = \frac{m}{n}$

For gases, also connect $n$ , $V$ And $V_m$ :

$n = \frac{V}{V_m} \quad ; \quad V = n \times V_m$

Worked example 13: What volume of $\mathrm{CO}_2$ at RTP is produced when 25.0 g of $\mathrm{CaCO}_3$ reacts with excess HCl?

$\mathrm{CaCO}_3 + 2\mathrm{HCl} \to \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

Answer

Moles of $\mathrm{CaCO}_3 = 25.0 / 100.1 = 0.250 \mathrm{ mol}$

1 mol $\mathrm{CaCO}_3$ produces 1 mol $\mathrm{CO}_2$ So moles of $\mathrm{CO}_2 = 0.250 \mathrm{ mol}$

Volume at RTP = $0.250 \times 24.0 = 6.00 \mathrm{ dm}^3$

Volumetric Analysis (Titration)

Volumetric analysis is a quantitative technique for determining the concentration of a solution by Reaction with a standard solution of known concentration.

Key Apparatus

Apparatus	Purpose
Burette	Delivers variable volumes, graduated to 0.05 cm $^3$
Pipette	Delivers a fixed volume accurately
Volumetric flask	Prepares solutions of precise concentration
Conical flask	Contains the analyte for titration

Titration Calculations

General approach:

Write the balanced equation.
Calculate moles of the standard solution used ( $n = c \times V$ ).
Use the mole ratio to find moles of the unknown.
Calculate concentration or mass of the unknown.

Worked example 14: 25.0 cm $^3$ of $\mathrm{NaOH}$ solution is titrated against 0.100 mol/dm $^3$ $\mathrm{HCl}$ . The average titre is 20.0 cm $^3$ . Calculate the concentration of $\mathrm{NaOH}$ .

$\mathrm{NaOH} + \mathrm{HCl} \to \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}$

Answer

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

Mole ratio $\mathrm{NaOH}:\mathrm{HCl} = 1:1$ So moles of $\mathrm{NaOH} = 0.00200 \mathrm{ mol}$

$c(\mathrm{NaOH}) = \dfrac{0.00200}{25.0 / 1000} = \dfrac{0.00200}{0.0250} = 0.0800 \mathrm{ mol/dm}^3$

Worked example 15: 1.00 g of impure $\mathrm{Na}_2\mathrm{CO}_3$ is dissolved in water and made Up to 250 cm $^3$ . 25.0 cm $^3$ of this solution requires 21.5 cm $^3$ of 0.100 mol/dm $^3$ $\mathrm{HCl}$ for complete reaction. Calculate the percentage purity of the $\mathrm{Na}_2\mathrm{CO}_3$ .

$\mathrm{Na}_2\mathrm{CO}_3 + 2\mathrm{HCl} \to 2\mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

Answer

Moles of $\mathrm{HCl}$ used = $0.100 \times 21.5 / 1000 = 0.00215 \mathrm{ mol}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ in 25.0 cm $^3$ = $0.00215 / 2 = 0.001075 \mathrm{ mol}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ in 250 cm $^3$ = $0.001075 \times 10 = 0.01075 \mathrm{ mol}$

Mass of pure $\mathrm{Na}_2\mathrm{CO}_3$ = $0.01075 \times 106.0 = 1.140 \mathrm{ g}$

Wait — this exceeds 1.00 g. Let me recalculate with correct titre.

Actually, let me use a more reasonable titre of 18.9 cm $^3$ :

Moles of $\mathrm{HCl}$ used = $0.100 \times 18.9 / 1000 = 0.00189 \mathrm{ mol}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ in 25.0 cm $^3$ = $0.00189 / 2 = 0.000945 \mathrm{ mol}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ in 250 cm $^3$ = $0.000945 \times 10 = 0.00945 \mathrm{ mol}$

Mass of pure $\mathrm{Na}_2\mathrm{CO}_3$ = $0.00945 \times 106.0 = 1.002 \mathrm{ g}$

This is close to 100%, so let me use a different titre. With 15.0 cm $^3$ :

Moles of $\mathrm{HCl}$ used = $0.100 \times 15.0 / 1000 = 0.00150 \mathrm{ mol}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ in 25.0 cm $^3$ = $0.00150 / 2 = 0.000750 \mathrm{ mol}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ in 250 cm $^3$ = $0.000750 \times 10 = 0.00750 \mathrm{ mol}$

Mass of pure $\mathrm{Na}_2\mathrm{CO}_3$ = $0.00750 \times 106.0 = 0.795 \mathrm{ g}$

$\%\mathrm{ purity} = \dfrac{0.795}{1.00} \times 100\% = 79.5\%$

Back Titration

Back titration is used when the substance being analysed reacts too slowly, is insoluble, or the Endpoint is hard to detect directly.

React the analyte with excess standard reagent.
Titrate the remaining excess reagent with another standard solution.

Worked example 16: 2.00 g of impure $\mathrm{CaCO}_3$ is reacted with 50.0 cm $^3$ of 1.00 Mol/dm $^3$ $\mathrm{HCl}$ (excess). The remaining $\mathrm{HCl}$ requires 30.0 cm $^3$ of 0.500 Mol/dm $^3$ $\mathrm{NaOH}$ for neutralisation. Calculate the percentage of $\mathrm{CaCO}_3$ in the Sample.

$\mathrm{CaCO}_3 + 2\mathrm{HCl} \to \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$ $\mathrm{HCl} + \mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}$

Answer

Total moles of $\mathrm{HCl}$ added = $1.00 \times 50.0 / 1000 = 0.0500 \mathrm{ mol}$

Moles of $\mathrm{NaOH}$ used = $0.500 \times 30.0 / 1000 = 0.0150 \mathrm{ mol}$

Moles of $\mathrm{HCl}$ remaining = $0.0150 \mathrm{ mol}$ (1:1 ratio)

Moles of $\mathrm{HCl}$ reacted with $\mathrm{CaCO}_3$ = $0.0500 - 0.0150 = 0.0350 \mathrm{ mol}$

Moles of $\mathrm{CaCO}_3$ = $0.0350 / 2 = 0.0175 \mathrm{ mol}$

Mass of $\mathrm{CaCO}_3$ = $0.0175 \times 100.1 = 1.752 \mathrm{ g}$

$\%\mathrm{ CaCO}_3 = \dfrac{1.752}{2.00} \times 100\% = 87.6\%$

Solutions Involving Both Mass and Volume

Converting Between Concentration Units

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

$\mathrm{ppm} = \frac{\mathrm{mass of solute (g)}}{\mathrm{mass of solution (g)}} \times 10^6$

Worked example 17: Convert 20.0 g/dm $^3$ of $\mathrm{NaOH}$ to mol/dm $^3$ .

Answer

$c = 20.0 / 40.0 = 0.500 \mathrm{ mol/dm}^3$

Water of Crystallisation and Stoichiometry Combined

Worked example 18: 5.00 g of hydrated magnesium sulfate, $\mathrm{MgSO}_4 \cdot x\mathrm{H}_2\mathrm{O}$ Is dissolved in water and made up to 250 cm $^3$ . 25.0 cm $^3$ of this solution requires 20.0 cm $^3$ of 0.100 mol/dm $^3$ $\mathrm{NaOH}$ to precipitate All the magnesium as $\mathrm{Mg(OH)}_2$ . Find $x$ .

$\mathrm{MgSO}_4 + 2\mathrm{NaOH} \to \mathrm{Mg(OH)}_2 + \mathrm{Na}_2\mathrm{SO}_4$

Answer

Moles of $\mathrm{NaOH}$ = $0.100 \times 20.0 / 1000 = 0.00200 \mathrm{ mol}$

Moles of $\mathrm{MgSO}_4$ in 25.0 cm $^3$ = $0.00200 / 2 = 0.00100 \mathrm{ mol}$

Moles of $\mathrm{MgSO}_4$ in 250 cm $^3$ = $0.00100 \times 10 = 0.0100 \mathrm{ mol}$

$M_r(\mathrm{MgSO}_4) = 24.3 + 32.1 + 4(16.0) = 120.4$

Mass of $\mathrm{MgSO}_4$ (anhydrous) = $0.0100 \times 120.4 = 1.204 \mathrm{ g}$

Mass of water = $5.00 - 1.204 = 3.796 \mathrm{ g}$

Moles of water = $3.796 / 18.0 = 0.211 \mathrm{ mol}$

$x = 0.211 / 0.0100 = 21.1$

This is unrealistic for magnesium sulfate. The expected value is $x = 7$ for Epsom salt. The data Likely has an issue. For $\mathrm{MgSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}$ : $M_r = 120.4 + 7(18.0) = 246.4$ . 5.00 g would give $5.00/246.4 = 0.0203 \mathrm{ mol}$ And 25 Cm $^3$ aliquot would need $0.00203 \times 2 = 0.00406 \mathrm{ mol}$ $\mathrm{NaOH}$ I.e. 40.6 Cm $^3$ of 0.100 M NaOH.

Revised problem: using titre of 40.6 cm $^3$ :

Moles of $\mathrm{NaOH}$ = $0.100 \times 40.6 / 1000 = 0.00406 \mathrm{ mol}$

Moles of $\mathrm{MgSO}_4$ in 25.0 cm $^3$ = $0.00406 / 2 = 0.00203 \mathrm{ mol}$

Moles in 250 cm $^3$ = $0.0203 \mathrm{ mol}$

Mass of $\mathrm{MgSO}_4$ = $0.0203 \times 120.4 = 2.444 \mathrm{ g}$

Mass of water = $5.00 - 2.444 = 2.556 \mathrm{ g}$

Moles of water = $2.556 / 18.0 = 0.142 \mathrm{ mol}$

$x = 0.142 / 0.0203 = 7.00$

Therefore $x = 7$ Confirming the formula $\mathrm{MgSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}$ .

Intuition

Chemical accounting: Stoichiometry is like a recipe — the balanced equation tells you exactly how much of each ingredient you need and how much product you’ll get. The mole is the chemist’s dozen.

Why it matters: From manufacturing drugs to calculating fuel efficiency, stoichiometry lets you predict quantities. Errors in stoichiometric calculations can be dangerous or wasteful.

The key insight: The mole concept bridges the gap between atoms (invisible) and grams (measurable) — Avogadro’s number is the conversion factor.

Common Pitfalls

Confusing STP and RTP: STP molar volume is 22.4 dm $^3$ /mol; RTP is 24.0 dm $^3$ /mol. Always check the conditions stated in the question.
Forgetting to convert cm $^3$ to dm $^3$ : Divide cm $^3$ by 1000 before using in $c = n/V$ .
Using total mass instead of solute mass: When calculating molarity, use the mass of the solute only, not the total mass of the solution.
Incorrect mole ratios: Always use the balanced equation to determine mole ratios. Never assume a 1:1 ratio.
Choosing the wrong limiting reagent: Always calculate both reactants and compare. The one that produces less product is the limiting reagent.
Mixing up empirical and molecular formulae: The empirical formula is the simplest ratio; the molecular formula is a multiple of it.
Ignoring significant figures in titration: Burette readings should be recorded to 2 decimal places (e.g., 24.50 cm $^3$ Not 24.5 cm $^3$ ).
Forgetting the factor of 2 in back titrations: Account for the stoichiometry of both reactions involved.

Practice Problems

Problem 1

A compound contains 36.5% sodium, 25.4% sulfur, and 38.1% oxygen by mass. Determine the empirical Formula and, given that the molar mass is 126 g/mol, the molecular formula.

Answer

Element	Mass (%)	$A_r$	Moles	Ratio
Na	36.5	23.0	36.5 / 23.0 = 1.587	1.587 / 0.793 = 2.00
S	25.4	32.1	25.4 / 32.1 = 0.791	0.791 / 0.793 = 0.997
O	38.1	16.0	38.1 / 16.0 = 2.381	2.381 / 0.793 = 3.00

Empirical formula = $\mathrm{Na}_2\mathrm{SO}_3$

$M_r(\mathrm{Na}_2\mathrm{SO}_3) = 2(23.0) + 32.1 + 3(16.0) = 126.1$

$n = 126 / 126.1 = 1.00$

Molecular formula = $\mathrm{Na}_2\mathrm{SO}_3$ (sodium sulfite)

Problem 2

25.0 cm $^3$ of 0.200 mol/dm $^3$ $\mathrm{H}_2\mathrm{SO}_4$ is completely neutralised by $\mathrm{KOH}$ solution. If 40.0 cm $^3$ of $\mathrm{KOH}$ is required, what is the concentration of The $\mathrm{KOH}$ ?

Answer

$\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{KOH} \to \mathrm{K}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}$

Moles of $\mathrm{H}_2\mathrm{SO}_4$ = $0.200 \times 25.0 / 1000 = 0.00500 \mathrm{ mol}$

Moles of $\mathrm{KOH}$ = $0.00500 \times 2 = 0.0100 \mathrm{ mol}$

$c(\mathrm{KOH}) = 0.0100 / (40.0 / 1000) = 0.0100 / 0.0400 = 0.250 \mathrm{ mol/dm}^3$

Problem 3

When 8.40 g of hydrated sodium carbonate, $\mathrm{Na}_2\mathrm{CO}_3 \cdot x\mathrm{H}_2\mathrm{O}$ Is heated to constant mass, 3.10 g of Anhydrous $\mathrm{Na}_2\mathrm{CO}_3$ remains. Find the value of $x$ .

Answer

Mass of water lost = $8.40 - 3.10 = 5.30 \mathrm{ g}$

Moles of $\mathrm{Na}_2\mathrm{CO}_3$ = $3.10 / 106.0 = 0.02925 \mathrm{ mol}$

Moles of $\mathrm{H}_2\mathrm{O}$ = $5.30 / 18.0 = 0.2944 \mathrm{ mol}$

$x = 0.2944 / 0.02925 = 10.07 \approx 10$

Formula = $\mathrm{Na}_2\mathrm{CO}_3 \cdot 10\mathrm{H}_2\mathrm{O}$ (washing soda)

Problem 4

2.43 g of magnesium is added to 100 cm $^3$ of 2.00 mol/dm $^3$ hydrochloric acid. Calculate the Volume of hydrogen gas produced at RTP and identify the limiting reagent.

$\mathrm{Mg} + 2\mathrm{HCl} \to \mathrm{MgCl}_2 + \mathrm{H}_2$

Answer

Moles of $\mathrm{Mg}$ = $2.43 / 24.3 = 0.100 \mathrm{ mol}$

Moles of $\mathrm{HCl}$ = $2.00 \times 100 / 1000 = 0.200 \mathrm{ mol}$

Divide by coefficients: $\mathrm{Mg}$ : $0.100 / 1 = 0.100$ , $\mathrm{HCl}$ : $0.200 / 2 = 0.100$

Both are in exactly the stoichiometric ratio. Neither is in strict excess; both are fully consumed.

Moles of $\mathrm{H}_2$ = $0.100 \mathrm{ mol}$ (1:1 ratio with Mg)

Volume at RTP = $0.100 \times 24.0 = 2.40 \mathrm{ dm}^3$

Problem 5

In a back titration, 0.500 g of limestone ( $\mathrm{CaCO}_3$ ) is reacted with 40.0 cm $^3$ of 0.500 Mol/dm $^3$ $\mathrm{HCl}$ . The excess $\mathrm{HCl}$ requires 28.0 cm $^3$ of 0.200 mol/dm $^3$ $\mathrm{NaOH}$ for neutralisation. Calculate the percentage of $\mathrm{CaCO}_3$ in the limestone.

Answer

Total moles of $\mathrm{HCl}$ = $0.500 \times 40.0 / 1000 = 0.0200 \mathrm{ mol}$

Moles of $\mathrm{NaOH}$ = $0.200 \times 28.0 / 1000 = 0.00560 \mathrm{ mol}$

Moles of $\mathrm{HCl}$ remaining = $0.00560 \mathrm{ mol}$

Moles of $\mathrm{HCl}$ reacted with $\mathrm{CaCO}_3$ = $0.0200 - 0.00560 = 0.01440 \mathrm{ mol}$

Moles of $\mathrm{CaCO}_3$ = $0.01440 / 2 = 0.00720 \mathrm{ mol}$

Mass of $\mathrm{CaCO}_3$ = $0.00720 \times 100.1 = 0.721 \mathrm{ g}$

$\%\mathrm{ CaCO}_3 = \dfrac{0.721}{0.500} \times 100\% = 144\%$

This is impossible. The issue is that with only 0.500 g of limestone, 40 cm $^3$ of 0.500 M HCl is Far more than needed. Let me use more appropriate values.

Revised: 0.500 g limestone + 50.0 cm $^3$ of 0.200 mol/dm $^3$ HCl, excess requires 20.0 cm $^3$ of 0.100 mol/dm $^3$ NaOH.

Total moles of $\mathrm{HCl}$ = $0.200 \times 50.0 / 1000 = 0.01000 \mathrm{ mol}$

Moles of $\mathrm{NaOH}$ = $0.100 \times 20.0 / 1000 = 0.00200 \mathrm{ mol}$

Moles of $\mathrm{HCl}$ remaining = $0.00200 \mathrm{ mol}$

Moles of $\mathrm{HCl}$ reacted = $0.01000 - 0.00200 = 0.00800 \mathrm{ mol}$

Moles of $\mathrm{CaCO}_3$ = $0.00800 / 2 = 0.00400 \mathrm{ mol}$

Mass of $\mathrm{CaCO}_3$ = $0.00400 \times 100.1 = 0.400 \mathrm{ g}$

$\%\mathrm{ CaCO}_3 = \dfrac{0.400}{0.500} \times 100\% = 80.0\%$

Problem 6

A student prepares 500 cm $^3$ of 0.500 mol/dm $^3$ $\mathrm{H}_2\mathrm{SO}_4$ by diluting Concentrated acid of density 1.84 g/cm $^3$ containing 98.0% $\mathrm{H}_2\mathrm{SO}_4$ by mass. What volume of the concentrated acid is required?

Answer

Moles of $\mathrm{H}_2\mathrm{SO}_4$ needed = $0.500 \times 500 / 1000 = 0.250 \mathrm{ mol}$

Mass of $\mathrm{H}_2\mathrm{SO}_4$ needed = $0.250 \times 98.1 = 24.525 \mathrm{ g}$

Mass of concentrated acid = $24.525 / 0.980 = 25.03 \mathrm{ g}$

Volume of concentrated acid = $25.03 / 1.84 = 13.6 \mathrm{ cm}^3$

Problem 7

An organic compound contains only carbon, hydrogen, and oxygen. On complete combustion, 0.120 g of The compound produces 0.264 g of $\mathrm{CO}_2$ and 0.108 g of $\mathrm{H}_2\mathrm{O}$ . Determine The empirical formula of the compound.

Answer

Mass of C in $\mathrm{CO}_2$ = $0.264 \times \dfrac{12.0}{44.0} = 0.0720 \mathrm{ g}$

Mass of H in $\mathrm{H}_2\mathrm{O}$ = $0.108 \times \dfrac{2.0}{18.0} = 0.0120 \mathrm{ g}$

Mass of O = $0.120 - 0.0720 - 0.0120 = 0.0360 \mathrm{ g}$

Element	Mass (g)	$A_r$	Moles	Ratio
C	0.0720	12.0	0.00600	1
H	0.0120	1.0	0.0120	2
O	0.0360	16.0	0.00225	0.375

Multiply all by 8 to clear the fraction: C : H : O = 8 : 16 : 3

Wait, $0.00225 / 0.00600 = 0.375 = 3/8$ So multiply by 8:

C : H : O = $1 \times 8 : 2 \times 8 : 0.375 \times 8$ = 8 : 16 : 3

Empirical formula = $\mathrm{C}_8\mathrm{H}_{16}\mathrm{O}_3$

Hmm, let me re-check. $0.375 \times 8 = 3$ . So empirical formula = $\mathrm{C}_8\mathrm{H}_{16}\mathrm{O}_3$ .

But this seems large. Let me double check: $0.00600 / 0.00600 = 1$ , $0.0120 / 0.00600 = 2$ $0.00225 / 0.00600 = 0.375$ . Multiply by 8: 8, 16, 3. Yes, $\mathrm{C}_8\mathrm{H}_{16}\mathrm{O}_3$ .

Alternative approach with different ratios: multiply by $8/3$ : $8/3 : 16/3 : 1$ . No, that gives Fractions. The correct simplest integer ratio is indeed 8:16:3, giving $\mathrm{C}_8\mathrm{H}_{16}\mathrm{O}_3$ .

flowchart TD
    A[1_Mole Concept And Stoichiometry] --> 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

including key reactions, underlying theories, and practical applications.

Stashed changes:docs/docs_dse/Chemistry/mole-concept-and-stoichiometry.md

Key concepts include:

key chemical principles and theories
mathematical relationships in chemistry
practical techniques and apparatus
applications of chemistry in industry
environmental and ethical considerations

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.

Stashed changes:docs/docs_dse/Chemistry/mole-concept-and-stoichiometry.md

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

Atomic Structure: Atomic structure determines bonding
Equilibrium: Equilibrium is a core topic
Organic Chemistry: Organic chemistry covers carbon compounds

Mark this page as reviewed

Chemistry - Mole Concept and Stoichiometry

Relative Atomic and Molecular Masses

Relative Atomic Mass (ArA_rAr​)

Relative Molecular Mass (MrM_rMr​)

The Mole

Definition

Molar Mass (MMM)

Molar Volume of Gas at STP

Formulae of Compounds

Empirical Formula

Molecular Formula

Procedure for Finding Empirical Formula

Percentage Composition

Hydrated Salts

Concentration

Molarity

Mass Concentration

Molality

Dilution

Chemical Equations and Stoichiometry

Balancing Equations

Molar Ratios in Equations

Limiting Reagent

Procedure

Percentage Yield

Gas Calculations

Using the Ideal Gas Equation

Cross-References

Relative Atomic Mass ( $A_r$ )

Relative Molecular Mass ( $M_r$ )

Molar Mass ( $M$ )