Comprehensive Stoichiometric Calculations and Quantitative Chemical Analysis

The Mole Concept and Fundamental Molar Quantities

  • Definition of the Mole:

    • The number of atoms that make up 12 g12\,\text{g} of carbon is termed one mole of carbon atoms.

    • Formally, one mole is defined as that quantity of matter containing exactly the same number of elementary particles as there are carbon atoms in 12 g12\,\text{g} of carbon-12 (i.e., 0612C^{12}_{\phantom{0}6}\text{C}).

    • The mole is the official SI unit for the quantity of matter.

  • Avogadro's Number (NAN_A):

    • The number of particles contained in one mole of a substance is known as Avogadro's number (NAN_A).

    • Standard numerical value:     NA=6,02×1023 mol−1N_A = 6,02 \times 10^{23}\,\text{mol}^{-1}     (also written as 6.02×1023 mol−16.02 \times 10^{23}\,\text{mol}^{-1}).

  • Molar Mass (MM):

    • Defined as the mass, in grams, of one mole of that substance.

    • Standard unit: g⋅mol−1\text{g}\cdot\text{mol}^{-1}.

  • Mathematical Relationship Between Molar Mass and Moles:

    • Expressed by the formula:     n=mMn = \frac{m}{M}

    • Variable definitions:

    • nn = moles (mol\text{mol} or quantity of matter)

  • mm = mass (in g\text{g} )

  • MM = molar mass (in g⋅mol−1\text{g}\cdot\text{mol}^{-1} )

  • Molar Mass Calculation Examples:

    • Example 1: Calculate the number of moles in 6,075 g6,075\,\text{g} of magnesium (Mg\text{Mg}).

    • Example 2: Calculate the mass of 0,15 moles0,15\,\text{moles} of copper (II) carbonate (CuCO3\text{CuCO}_3).

  • Elementary Units and Moles Relationship:

    • The relationship connecting elementary particles and moles is defined as:     mole=Number of elementary unitsAvogadro’s number\text{mole} = \frac{\text{Number of elementary units}}{\text{Avogadro's number}}     n=NNAn = \frac{N}{N_A}     elementary units=n×NA\text{elementary units} = n \times N_A

    • Where NN represents the total count of elementary units, and NAN_A is the Avogadro constant (6.02×1023 mol−16.02 \times 10^{23}\,\text{mol}^{-1}).

    • Example: Calculate the number of oxygen atoms in 0,25 moles0,25\,\text{moles} of oxygen gas (O2\text{O}_2).

Molar Gas Volume and Gas Stoichiometry

  • Definition of Molar Gas Volume (VmV_m):

    • The molar gas volume is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

  • Standard Temperature and Pressure (STP):

    • STP conditions are precisely defined as:

    • Temperature: 0 ∘C0\,^\circ\text{C} (or 273 K273\,\text{K} )

  • Pressure: 101,3 kPa101,3\,\text{kPa} (equivalent to 1 atmosphere1\,\text{atmosphere} )

    • At STP, the molar gas volume (VmV_m) is equal to 22,4 dm322,4\,\text{dm}^3.

    • Consequently, one mole of any ideal gas occupies 22,4 dm322,4\,\text{dm}^3 at 0 ∘C (273 K)0\,^\circ\text{C}\,(273\,\text{K}) and 1 atmosphere (101,3 kPa)1\,\text{atmosphere}\,(101,3\,\text{kPa}).

  • Gas Volume Formulas:

    • Equation for amount of substance:     n=VVmn = \frac{V}{V_m}

    • Rearranged equation for volume at STP:     V=22,4×nV = 22,4 \times n     (or V=22,4 nV = 22,4\,n )

  • Metric Volume Equivalencies and Conversions:

    • 1 dm3=1 L1\,\text{dm}^3 = 1\,\text{L}

    • 1 cm3=1 mℓ1\,\text{cm}^3 = 1\,\text{m}\ell

    • 1000 mℓ=1 L1000\,\text{m}\ell = 1\,\text{L}

    • 1000 cm3=1 dm31000\,\text{cm}^3 = 1\,\text{dm}^3

  • Gas Stoichiometry Example:

    • Calculate the volume of carbon dioxide gas occupied by 0,5 moles0,5\,\text{moles} of the gas at STP.

Percentage Composition, Empirical Formulas, and Molecular Formulas

  • Percentage Composition:

    • Calculated by determining the proportion of the mass contributed by each element relative to the total molecular or formula mass:     % composition=mass of element (g)mass of compound (g)×100\%\,\text{composition} = \frac{\text{mass of element (g)}}{\text{mass of compound (g)}} \times 100

  • Empirical Formula versus Molecular Formula:

    • Calculating the mole ratio of atoms in a chemical combination indicates the actual atomic ratio in the compound.

    • When this ratio is expressed in its simplest whole-number atomic ratio, it defines the empirical formula.

    • Representative Example: Benzene has the molecular formula C6H6\text{C}_6\text{H}_6; its simplest atomic ratio yields the empirical formula CH\text{CH}.

  • Formula Determination Worked Example:

    • Problem: A compound is composed of 26,67%26,67\% carbon, 71,11%71,11\% oxygen, and 2,22%2,22\% hydrogen by mass. If the relative formula mass of the compound is equal to 90 g⋅mol−190\,\text{g}\cdot\text{mol}^{-1}, calculate the molecular formula of the compound.

  • Formula Determination Exercises:

    • Exercise 1: Vinegar is a dilute form of acetic acid. A sample of acetic acid possesses the following percentage composition: 39,9%39,9\% carbon, 6,7%6,7\% hydrogen, and 53,4%53,4\% oxygen.

    • 1.1 Determine the empirical formula of acetic acid.

    • 1.2 Determine the molecular formula of acetic acid if its molar mass is 60 g⋅mol−160\,\text{g}\cdot\text{mol}^{-1}.

    • Exercise 2: A compound contains 30,4%30,4\% nitrogen and 69,6%69,6\% oxygen by mass. Show using a calculation what its empirical formula will be.

    • Exercise 3: A compound consists of 36,5%36,5\% sodium, 25,4%25,4\% sulphur, and 38,1%38,1\% oxygen by mass. Find its empirical formula.

    • Exercise 4: An organic compound contains 54,545%54,545\% carbon, 9,091%9,091\% hydrogen, and 36,364%36,364\% oxygen. The molar mass of the compound is 88 g⋅mol−188\,\text{g}\cdot\text{mol}^{-1}.

    • a) Determine the empirical formula of this organic compound.

    • b) Determine the molecular formula of this organic compound.

Solution Chemistry, Concentration, Dilution, and Mixing

  • Fundamental Solution Terminology:

    • Solution: A homogeneous mixture of solute and solvent.

    • Solute: The substance that is dissolved within the solution.

    • Solvent: The substance in which another substance dissolves to form a solution.

    • Concentration (cc): The amount of solute per unit volume of solution; indicates the amount of solute (in moles) dissolved in the volume of solvent (in dm3\text{dm}^3 ).

    • Standard Solution: A solution of known concentration.

  • Formulas and Units of Concentration:

    • Standard formula:     concentration=number of moles of solutevolume of solution in dm3\text{concentration} = \frac{\text{number of moles of solute}}{\text{volume of solution in dm}^3}     c=nVc = \frac{n}{V}

    • Combined mass-volume formula:     c=mM×Vc = \frac{m}{M \times V}     (or c=mM Vc = \frac{m}{M\,V} )

    • The SI unit for molar concentration is mol⋅dm−3\text{mol}\cdot\text{dm}^{-3}.

  • Concentration Examples and Exercises:

    • Example: Calculate the concentration of a solution formed when 19,6 g19,6\,\text{g} of hydrogen chloride (HCℓ\text{HC}\ell) is dissolved in water and the volume is made up to 250 mℓ250\,\text{m}\ell.

    • Exercise 1: Calculate the concentration of a solution formed when 19,6 g19,6\,\text{g} of hydrogen chloride (HCℓ\text{HC}\ell) is dissolved in water and the volume is made up to 250 cm3 (mℓ)250\,\text{cm}^3\,(\text{m}\ell).

    • Exercise 2: 1,97 g1,97\,\text{g} of sulphuric acid (H2SO4\text{H}_2\text{SO}_4) is dissolved in 100 cm3100\,\text{cm}^3 of solution. Calculate the concentration of the solution.

    • Exercise 3: Calculate the mass of sodium chloride (NaCℓ\text{NaC}\ell) dissolved in 80 cm380\,\text{cm}^3 of solution of concentration 0,5 mol⋅dm−30,5\,\text{mol}\cdot\text{dm}^{-3}.

    • Exercise 4: Calculate the mass of Aℓ(OH)3\text{A}\ell(\text{OH})_3 dissolved in 750 cm3750\,\text{cm}^3 of a solution of concentration 0,6 mol⋅dm−30,6\,\text{mol}\cdot\text{dm}^{-3}.

    • Exercise 5: Calculate the mass of sodium hydroxide (NaOH\text{NaOH}) that must be dissolved in 200 mℓ200\,\text{m}\ell of solution to produce a concentration of 6,4 mol⋅dm−36,4\,\text{mol}\cdot\text{dm}^{-3}.

  • Dilution of Solutions:

    • A solution is diluted by adding water, which increases the total volume of the solution.

    • The number of moles of solute remains constant throughout dilution:     n=c×Vn = c \times V

    • Conservation equation for dilution:     (c1×V1)original solution=(c2×V2)dilute solution(c_1 \times V_1)_{\text{original solution}} = (c_2 \times V_2)_{\text{dilute solution}}

    • Dilution Example: 20 cm320\,\text{cm}^3 of NaOH\text{NaOH} solution of concentration 0,4 mol⋅dm−30,4\,\text{mol}\cdot\text{dm}^{-3} is diluted by adding 30 cm330\,\text{cm}^3 of water to the original solution. Calculate the concentration of the diluted solution.

    • Dilution Exercise 1: 30 g30\,\text{g} of Mg(NO3)2\text{Mg}(\text{NO}_3)_2 powder is dissolved in a small quantity of water and the resulting solution is made up to a volume of 500 cm3500\,\text{cm}^3.

    • a) Give the balanced equation to illustrate the dissolution of Mg(NO3)2\text{Mg}(\text{NO}_3)_2 in water.

    • b) Calculate the concentration of the solution.

    • c) 100 cm3100\,\text{cm}^3 of this solution is poured into an empty beaker. Calculate the volume of water that must be added to the 100 cm3100\,\text{cm}^3 of solution to dilute it to a concentration of 0,1 mol⋅dm−30,1\,\text{mol}\cdot\text{dm}^{-3}.

    • Dilution Exercise 2: In a laboratory, 500 cm3500\,\text{cm}^3 of KNO3\text{KNO}_3 solution with a concentration of 0,2 mol⋅dm−30,2\,\text{mol}\cdot\text{dm}^{-3} is provided. This solution must be diluted to obtain a concentration of 0,18 mol⋅dm−30,18\,\text{mol}\cdot\text{dm}^{-3}. Calculate the volume of water in cm3\text{cm}^3 that must be added.

  • Mixing Solutions and Ion Concentrations:

    • Exercise 1: A 500 cm3500\,\text{cm}^3 volume of Na2CO3\text{Na}_2\text{CO}_3 solution at 0,2 mol⋅dm−30,2\,\text{mol}\cdot\text{dm}^{-3} is mixed with 250 cm3250\,\text{cm}^3 of NaCℓ\text{NaC}\ell solution at 0,4 mol⋅dm−30,4\,\text{mol}\cdot\text{dm}^{-3}.

    • 1.1 Write chemical equations displaying the dissolution of each of the two salts.

    • 1.2 Calculate the concentration of Na+\text{Na}^+ ions in this combined mixture.

    • Exercise 2: A 500 cm3500\,\text{cm}^3 volume of CaCℓ2\text{CaC}\ell_2 solution at 0,2 mol⋅dm−30,2\,\text{mol}\cdot\text{dm}^{-3} is mixed with 300 cm3300\,\text{cm}^3 of NaCℓ\text{NaC}\ell solution at 0,3 mol⋅dm−30,3\,\text{mol}\cdot\text{dm}^{-3}.

    • 2.1 Write chemical equations displaying the dissolution of each of the two salts.

    • 2.2 Calculate the concentration of Cℓ−\text{C}\ell^- ions in this combined mixture.

    • Examination Question 3: A standard aqueous solution of magnesium nitrate (Mg(NO3)2\text{Mg}(\text{NO}_3)_2) is prepared with a volume of 200 cm3200\,\text{cm}^3 and a concentration of 0,15 mol⋅dm−30,15\,\text{mol}\cdot\text{dm}^{-3}.

    • 3.1 Identify the solvent used in this aqueous solution of magnesium nitrate.

    • 3.2 Write down the balanced chemical equation for the dissociation of Mg(NO3)2\text{Mg}(\text{NO}_3)_2 in water, including state/phase symbols.

    • 3.3 Calculate the number of moles of nitrate ions (NO3−\text{NO}_3^-) in the solution.

    • 3.4 Solid sodium nitrate (NaNO3\text{NaNO}_3) is subsequently dissolved into this 200 cm3200\,\text{cm}^3 of 0,15 mol⋅dm−3 Mg(NO3)20,15\,\text{mol}\cdot\text{dm}^{-3}\,\text{Mg}(\text{NO}_3)_2 solution. The resulting concentration of nitrate ions reaches 0,5 mol⋅dm−30,5\,\text{mol}\cdot\text{dm}^{-3}. Assuming no change in the volume of the solution, calculate the mass of sodium nitrate dissolved.

Consolidation Exercises on Fundamental Chemical Concepts

  • Question 1 — Molar Mass Determinations:

    • Determine the molar mass for each of the following substances:

    • 1.1 Na2SO3\text{Na}_2\text{SO}_3

    • 1.2 Aluminium nitrite

    • 1.3 CuSO4⋅5H2O\text{CuSO}_4\cdot 5\text{H}_2\text{O}

    • 1.4 Ammonium dichromate

    • 1.5 MgSO4⋅7H2O\text{MgSO}_4\cdot 7\text{H}_2\text{O}

  • Question 2 — Balancing Chemical Equations:

    • Balance each of the following equations, indicating the balancing coefficients:

    • 2.1 Ca(OH)2+HNO3→Ca(NO3)2+H2O\text{Ca}(\text{OH})_2 + \text{HNO}_3 \rightarrow \text{Ca}(\text{NO}_3)_2 + \text{H}_2\text{O}

    • 2.2 KCℓO3→KCℓ+O2\text{KC}\ell\text{O}_3 \rightarrow \text{KC}\ell + \text{O}_2

    • 2.3 Li+O2→Li2O\text{Li} + \text{O}_2 \rightarrow \text{Li}_2\text{O}

    • 2.4 N2+H2→NH3\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3

    • 2.5 C3H8+O2→CO2+H2O\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

    • 2.6 C5H10+O2→CO2+H2O\text{C}_5\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

    • 2.7 C2H5OH+O2→CO2+H2O\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

    • 2.8 Reaction between sodium carbonate and hydrochloric acid / hydrogen chloride to produce sodium chloride, carbon dioxide, and water.

    • 2.9 Reaction between hydrogen sulphate and copper (II) oxide to form copper (II) sulphate and water.

  • Question 3 — Calculations on Magnesium Nitrate:

    • A mass of 1 kg1\,\text{kg} of magnesium nitrate (Mg(NO3)2\text{Mg}(\text{NO}_3)_2) is provided.

    • 3.1 Define the mole.

    • 3.2 Calculate the number of moles of magnesium nitrate present in 1 kg1\,\text{kg} of the substance.

    • 3.3 Calculate the number of particles of magnesium nitrate present.

    • 3.4 Magnesium nitrate is dissolved in water; write down a balanced chemical equation representing the dissociation.

  • Question 4 — Ammonium Nitrate Quantities:

    • 4.1 Define molar mass.

    • 4.2 Write down the chemical formula for ammonium nitrate.

    • 4.3 Calculate how many grams of nitrogen are present in 10 g10\,\text{g} of ammonium nitrate.

  • Question 5 — Mixed Hydroxide Solutions:

    • A mass of 20 g20\,\text{g} of barium hydroxide is dissolved in 400 cm3400\,\text{cm}^3 of water to prepare a standard solution. In a separate vessel, there is a 200 cm3200\,\text{cm}^3 solution of sodium hydroxide with a concentration of 0,1 mol⋅dm−30,1\,\text{mol}\cdot\text{dm}^{-3}. Both solutions are mixed together.

    • 5.1 Define standard solution.

    • 5.2 Calculate the concentration of the barium hydroxide solution.

    • 5.3 Calculate the total number of hydroxide ions contained in the combined solution, showing all working steps.

  • Question 6 — Empirical and Molecular Formula Determinations:

    • 6.1 A compound comprises 26,67%26,67\% carbon, 71,11%71,11\% oxygen, and 2,22%2,22\% hydrogen by mass. If its molar mass equals 90 g⋅mol−190\,\text{g}\cdot\text{mol}^{-1}, calculate its molecular formula.

    • 6.2 Analysis of a liquid sample yields 0,15 g0,15\,\text{g} of hydrogen, 2,1 g2,1\,\text{g} of nitrogen, and 7,2 g7,2\,\text{g} of oxygen. Determine its empirical formula.

    • 6.3 A compound contains 92,2%92,2\% carbon and 7,76%7,76\% hydrogen. Its molar mass is 78,1 g⋅mol−178,1\,\text{g}\cdot\text{mol}^{-1}. Calculate both the empirical formula and the molecular formula.

    • 6.4 When elemental antimony (Sb\text{Sb}) is heated in air with excess sulphur, a reaction occurs forming a binary compound containing only antimony and sulphur. In this procedure, 2,435 g2,435\,\text{g} of antimony was reacted, yielding 3,397 g3,397\,\text{g} of pure antimony-sulphur compound. Using appropriate calculations, determine its empirical formula.

  • Question 7 — Gas Calculations at STP:

    • 7.1 Calculate the volume occupied by 0,8 g0,8\,\text{g} of SO2\text{SO}_2 at STP.

    • 7.2 Determine the mass of 60 cm360\,\text{cm}^3 of ammonia gas (NH3\text{NH}_3) at STP.

    • 7.3 Calculate the volume of methane (CH4\text{CH}_4) gas that contains 5,2×10265,2 \times 10^{26} molecules of CH4\text{CH}_4 at STP.

    • 7.4 An unknown diatomic gas X2\text{X}_2 with a mass of 0,8 g0,8\,\text{g} occupies 560 cm3560\,\text{cm}^3 at STP. Calculate the molar mass of the gas and determine its identity.

  • Question 8 — Solution Concentration Computations:

    • 8.1 A mass of 5 g5\,\text{g} of sodium chloride crystals is dissolved in 500 cm3500\,\text{cm}^3 of water to produce a standard solution. Calculate its concentration.

    • 8.2 A mass of 50 g50\,\text{g} of Ca(OH)2\text{Ca}(\text{OH})_2 is dissolved in 200 mℓ200\,\text{m}\ell of water. Calculate the concentration of the solution.

    • 8.3 Calculate the mass of sodium carbonate required to prepare a 250 cm3250\,\text{cm}^3 solution of concentration 0,25 mol⋅dm−30,25\,\text{mol}\cdot\text{dm}^{-3}.

Stoichiometric Relationships Across Reaction Phases

  • Stoichiometric Flow Architecture:

    • Quantitative chemical reactions link physical quantities of reactants to products via the central bridge of balanced chemical equations:

Reactants mole conversions flowchartProducts mole conversions flowchart

  • Interconversions between macroscopic quantities and moles:

    • Mass to Moles: n=mMn = \frac{m}{M}

    • Concentration to Moles: c=nV  ⟹  n=c×Vc = \frac{n}{V} \implies n = c \times V

    • Gas Volume at STP to Moles: V=22,4 n  ⟹  n=V22,4V = 22,4\,n \implies n = \frac{V}{22,4}

    • Elementary Particles to Moles: elementary units=n×NA  ⟹  n=elementary unitsNA\text{elementary units} = n \times N_A \implies n = \frac{\text{elementary units}}{N_A}

  • Moles of reactants convert to moles of products through the mole ratio established by the balanced chemical equation.

    • Mass–Mass Calculations:

  • Worked Example: Calculate the mass of magnesium oxide produced when 6,0 g6,0\,\text{g} of magnesium reacts completely with excess oxygen:     2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)

  • Exercise 1: Calculate how much mercury (in grams) is formed when 10 g10\,\text{g} of mercury (II) oxide decomposes upon heating.

  • Exercise 2: Calculate the mass of sodium oxide produced if 5,52 g5,52\,\text{g} of Na\text{Na} reacts completely with oxygen according to the following equation:     4Na(s)+O2(g)→2Na2O(s)4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)

  • Exercise 3: Calculate the mass of aluminium oxide required to produce 16,2 kg16,2\,\text{kg} of pure aluminium according to the following equation:     2Aℓ2O3(s)→4Aℓ(s)+3O2(g)2\text{A}\ell_2\text{O}_3(s) \rightarrow 4\text{A}\ell(s) + 3\text{O}_2(g)

    • Mass–Volume Calculations:

  • Worked Example: Calculate the volume of oxygen used at STP when 12,16 g12,16\,\text{g} of magnesium reacts completely with excess oxygen:     2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)

  • Exercise 1: If 18,6 g18,6\,\text{g} of phosphorus reacts with oxygen to produce phosphorus (V) oxide, calculate the volume of oxygen at STP required:     4P(s)+5O2(g)→2P2O5(s)4\text{P}(s) + 5\text{O}_2(g) \rightarrow 2\text{P}_2\text{O}_5(s)

  • Exercise 2: Sodium reacts with water to produce sodium hydroxide and hydrogen gas according to:     2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)     Calculate the mass of sodium required to produce 0,896 dm30,896\,\text{dm}^3 of hydrogen at STP.

    • Volume–Volume Calculations:

  • Worked Example: Calculate the volume of sulphur trioxide produced when 11,2 dm311,2\,\text{dm}^3 of oxygen reacts completely with excess sulphur dioxide:     2SO2(g)+O2(g)→2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)

  • Exercise 1: Octane burns in an engine according to the balanced equation:     2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g)2\text{C}_8\text{H}_{18}(l) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)     If 1257,54 dm31257,54\,\text{dm}^3 of carbon dioxide is produced at STP, calculate:

    • 1.1 The volume of oxygen required at STP.

    • 1.2 The mass of octane required to produce this volume of carbon dioxide at STP.

    • Mass–Concentration Calculations:

  • Worked Example: Calculate the mass of zinc metal that will react with 40 cm340\,\text{cm}^3 of hydrochloric acid of concentration 0,05 mol⋅dm−30,05\,\text{mol}\cdot\text{dm}^{-3} according to:     Zn(s)+2HCℓ(aq)→ZnCℓ2(aq)+H2(g)\text{Zn}(s) + 2\text{HC}\ell(aq) \rightarrow \text{ZnC}\ell_2(aq) + \text{H}_2(g)

  • Exercise 1: Given the balanced chemical equation:     2NaHCO3(s)+H2SO4(aq)→Na2SO4(aq)+2H2O(ℓ)+2CO2(g)2\text{NaHCO}_3(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(\ell) + 2\text{CO}_2(g)     If 500 cm3500\,\text{cm}^3 of H2SO4\text{H}_2\text{SO}_4 at a concentration of 0,15 mol⋅dm−30,15\,\text{mol}\cdot\text{dm}^{-3} reacts, calculate:

    • a) The mass of sodium bicarbonate required for this reaction.

    • b) The volume of carbon dioxide produced at STP.

  • Exercise 2: The compound As2S3\text{As}_2\text{S}_3 reacts with air when heated strongly, producing a mixture of products:     2As2S3(s)+9O2(g)→As4O6(s)+6SO2(g)2\text{As}_2\text{S}_3(s) + 9\text{O}_2(g) \rightarrow \text{As}_4\text{O}_6(s) + 6\text{SO}_2(g)     A sample containing 0,198 g0,198\,\text{g} of As2S3\text{As}_2\text{S}_3 is placed in 0,1 dm30,1\,\text{dm}^3 of pure excess oxygen in a reaction chamber. The reactants are heated until all As2S3\text{As}_2\text{S}_3 is completely consumed. The molar gas volume (VmV_m) under these specific conditions is 24,0 dm3⋅mol−124,0\,\text{dm}^3\cdot\text{mol}^{-1}.

    • 2.1 Determine the volume of SO2\text{SO}_2 that will be present at this temperature and pressure.

    • 2.2 Calculate the total volume of gas present in the reaction chamber at this temperature after the reaction reaches completion.

    • Comprehensive Stoichiometry Worksheet:

  • 1. Calculate the mass of sodium sulphate formed when 200 g200\,\text{g} of sodium hydroxide reacts with excess sulphuric acid:     2NaOH+H2SO4→2H2O+Na2SO42\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4

  • 2. Calculate the mass of lithium nitrate required to react with excess lead (IV) sulphate to produce 250 g250\,\text{g} of lithium sulphate:     Pb(SO4)2+4LiNO3→Pb(NO3)4+2Li2SO4\text{Pb}(\text{SO}_4)_2 + 4\text{LiNO}_3 \rightarrow \text{Pb}(\text{NO}_3)_4 + 2\text{Li}_2\text{SO}_4

  • 3. Calculate the volume of hydrochloric acid of concentration 0,100 mol⋅dm−30,100\,\text{mol}\cdot\text{dm}^{-3} required to react completely with 5,00 g5,00\,\text{g} of calcium hydroxide:     Ca(OH)2(s)+2HCℓ(aq)→CaCℓ2(aq)+2H2O(ℓ)\text{Ca}(\text{OH})_2(s) + 2\text{HC}\ell(aq) \rightarrow \text{CaC}\ell_2(aq) + 2\text{H}_2\text{O}(\ell)

  • 4. If calcium hydroxide reacts with 75 cm375\,\text{cm}^3 of hydrochloric acid at a concentration of 0,5 mol⋅dm−30,5\,\text{mol}\cdot\text{dm}^{-3}, calculate the mass of calcium chloride formed:     Ca(OH)2(s)+2HCℓ(aq)→CaCℓ2(aq)+2H2O(ℓ)\text{Ca}(\text{OH})_2(s) + 2\text{HC}\ell(aq) \rightarrow \text{CaC}\ell_2(aq) + 2\text{H}_2\text{O}(\ell)

  • 5. Calcium carbonate decomposes at elevated temperatures to form carbon dioxide and calcium oxide:     CaCO3(s)→CO2(g)+CaO(s)\text{CaCO}_3(s) \rightarrow \text{CO}_2(g) + \text{CaO}(s)     Calculate the mass of calcium carbonate required to produce 3,45 dm33,45\,\text{dm}^3 of carbon dioxide at STP.

  • 6. Ethylene burns in oxygen to form carbon dioxide and water vapour:     C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(g)     Calculate the volume of water vapour formed if 1,25 dm31,25\,\text{dm}^3 of ethylene is consumed at STP.

  • 7. Chlorine reacts with acetylene to yield 1,1,2,2-tetrachloroethane:     2Cℓ2(g)+C2H2(g)→C2H2Cℓ4(ℓ)2\text{C}\ell_2(g) + \text{C}_2\text{H}_2(g) \rightarrow \text{C}_2\text{H}_2\text{C}\ell_4(\ell)     Calculate the volume of chlorine gas required to produce 75,0 g75,0\,\text{g} of C2H2Cℓ4\text{C}_2\text{H}_2\text{C}\ell_4 at STP.

Limiting Reagents, Percentage Yield, and Percentage Purity

  • Limiting Reagent Fundamentals:

    • Definition: A limiting agent (or limiting reactant) is the reactant that is completely consumed during a reaction and determines the theoretical maximum amount of product formed.

    • Worked Example: Hydrogen sulphide gas is synthesized by reacting 3,52 g3,52\,\text{g} of iron (II) sulphide with 60 cm360\,\text{cm}^3 of dilute hydrochloric acid of concentration 2 mol⋅dm−32\,\text{mol}\cdot\text{dm}^{-3}. Calculate the volume of hydrogen sulphide gas produced at STP according to:     FeS+2HCℓ→H2S+FeCℓ2\text{FeS} + 2\text{HC}\ell \rightarrow \text{H}_2\text{S} + \text{FeC}\ell_2

    • Exercise 1: A mass of 4 g4\,\text{g} of calcium carbonate is added to 250 cm3250\,\text{cm}^3 of a 0,2 mol⋅dm−30,2\,\text{mol}\cdot\text{dm}^{-3} hydrochloric acid solution according to:     CaCO3(s)+2HCℓ(aq)→CaCℓ2(aq)+H2O(ℓ)+CO2(g)\text{CaCO}_3(s) + 2\text{HC}\ell(aq) \rightarrow \text{CaC}\ell_2(aq) + \text{H}_2\text{O}(\ell) + \text{CO}_2(g)

    • a) Calculate which reactant acts as the limiting agent.

    • b) Calculate the volume of CO2(g)\text{CO}_2(g) produced at STP.

    • Exercise 2: A mass of 5 g5\,\text{g} of sodium carbonate powder is added to 400 cm3400\,\text{cm}^3 of a 0,3 mol⋅dm−30,3\,\text{mol}\cdot\text{dm}^{-3} hydrochloric acid solution, producing effervescence:     Na2CO3(s)+2HCℓ(aq)→2NaCℓ(aq)+H2O(ℓ)+CO2(g)\text{Na}_2\text{CO}_3(s) + 2\text{HC}\ell(aq) \rightarrow 2\text{NaC}\ell(aq) + \text{H}_2\text{O}(\ell) + \text{CO}_2(g)     Determine whether all the sodium carbonate powder will be completely consumed, showing all calculations.

  • Percentage Yield Principles and Computations:

    • Definition: Yield measures the extent of a chemical reaction, evaluated by comparing the actual quantity of product obtained against the maximum theoretical quantity possible.

    • Industrial Significance: Yield analysis is critical in commercial chemical manufacturing to assess whether a production process is economically viable and profitable.

    • Theoretical versus Actual Yield: In approximately 99%99\% of practical reactions, actual yield differs from theoretical yield because reactants are rarely completely converted into target products.

    • Diagnostic Interpretation: A high percentage value reflects high conversion efficiency, indicating that industrial continuation is practical.

    • Percentage Yield Formula:     % yield=actual yieldtheoretical yield×100\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

    • Worked Example: For the chemical decomposition:     4HNO3→4NO2+2H2O+O24\text{HNO}_3 \rightarrow 4\text{NO}_2 + 2\text{H}_2\text{O} + \text{O}_2     If the reaction of 17,8 g17,8\,\text{g} of HNO3\text{HNO}_3 yields 11,6 g11,6\,\text{g} of NO2\text{NO}_2, calculate the percentage yield.

    • Exercise 1: Zinc displaces copper from a solution of copper sulphate:     Zn+CuSO4→ZnSO4+Cu\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}     If 21,3 g21,3\,\text{g} of copper is displaced from 150 cm3150\,\text{cm}^3 of copper sulphate solution with a concentration of 3 mol⋅dm−33\,\text{mol}\cdot\text{dm}^{-3}, calculate the percentage yield.

    • Exercise 2: For the decomposition reaction:     (NH4)2Cr2O7→Cr2O3+N2+4H2O(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \rightarrow \text{Cr}_2\text{O}_3 + \text{N}_2 + 4\text{H}_2\text{O}     If reacting 48,3 g48,3\,\text{g} of (NH4)2Cr2O7(\text{NH}_4)_2\text{Cr}_2\text{O}_7 produces 2,35 g2,35\,\text{g} of N2\text{N}_2, calculate the percentage yield.

  • Percentage Purity Principles and Computations:

    • Evaluates the proportion of pure chemically active substance in an impure raw material sample:     % purity=pure massimpure mass×100\%\,\text{purity} = \frac{\text{pure mass}}{\text{impure mass}} \times 100

    • Worked Example: Excess dilute sulphuric acid reacts with 8 g8\,\text{g} of impure aluminium, producing 6,72 dm36,72\,\text{dm}^3 of dry hydrogen gas at STP:     2Aℓ+3H2SO4→Aℓ2(SO4)3+3H22\text{A}\ell + 3\text{H}_2\text{SO}_4 \rightarrow \text{A}\ell_2(\text{SO}_4)_3 + 3\text{H}_2     Calculate the percentage purity of the aluminium sample.

    • Exercise 1: When 40 g40\,\text{g} of impure sodium nitrate is heated until reaction ceases, 4,48 dm34,48\,\text{dm}^3 of oxygen gas is collected at STP:     2NaNO3→2NaNO2+O22\text{NaNO}_3 \rightarrow 2\text{NaNO}_2 + \text{O}_2     Calculate the percentage purity of the sodium nitrate.

    • Exercise 2: If 10 g10\,\text{g} of impure copper (II) oxide reacts completely with 100 cm3100\,\text{cm}^3 of H2SO4\text{H}_2\text{SO}_4 at a concentration of 0,8 mol⋅dm−30,8\,\text{mol}\cdot\text{dm}^{-3}, determine the percentage purity of the copper (II) oxide:     CuO+H2SO4→CuSO4+H2O\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}

Advanced Stoichiometric Examination Problems

  • Problem 1 — Sodium Combustion:

    • A mass of 5,52 g5,52\,\text{g} of Na\text{Na} reacts completely with oxygen according to the equation:     4Na+O2→2Na2O4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}

    • Calculate:

    • a) The mass of oxygen consumed.

    • b) The mass of Na2O\text{Na}_2\text{O} produced.

  • Problem 2 — Lithium Hydroxide Formation:

    • Calculate the mass of lithium oxide (Li2O\text{Li}_2\text{O}) required to react with excess water to yield 28,8 g28,8\,\text{g} of LiOH\text{LiOH} according to:     Li2O+H2O→2LiOH\text{Li}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{LiOH}

  • Problem 3 — Calcium Reaction with Water:

    • A mass of 1,61 g1,61\,\text{g} of calcium metal is added to excess water, reacting to produce calcium hydroxide and hydrogen gas.

    • 3.1 Write a balanced chemical equation for the reaction.

    • 3.2 Calculate the mass of calcium hydroxide formed.

    • 3.3 Calculate the volume of hydrogen gas produced at STP conditions.

  • Problem 4 — Sulphur Dioxide Production:

    • Sulphur reacts with oxygen according to:     S+O2→SO2\text{S} + \text{O}_2 \rightarrow \text{SO}_2

    • If 12,8 g12,8\,\text{g} of sulphur reacts completely with oxygen, calculate the volume of SO2\text{SO}_2 produced at STP.

  • Problem 5 — Reduction of Copper (II) Oxide:

    • Hydrogen gas reacts with copper (II) oxide according to the balanced equation:     CuO+H2→Cu+H2O\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}

    • Calculate:

    • 5.1 The volume of hydrogen gas at STP required to react completely with 31,8 g31,8\,\text{g} of copper (II) oxide.

    • 5.2 The mass of copper metal produced.

  • Problem 6 — Acid-Carbonate Reaction:

    • A sample of sodium carbonate reacts with hydrochloric acid to produce sodium chloride, carbon dioxide, and water.

    • If 672 cm3672\,\text{cm}^3 of carbon dioxide is generated at STP, calculate the mass of sodium carbonate reacted.

  • Problem 7 — Magnesium Chloride Concentration Adjustment:

    • A mass of 130 g130\,\text{g} of magnesium chloride (MgCℓ2\text{MgC}\ell_2) is dissolved in 300 mℓ300\,\text{m}\ell of water.

    • 7.1 Calculate the concentration of the solution.

    • 7.2 Calculate the additional mass of magnesium chloride that must be added to increase the concentration to 6,7 mol⋅dm−36,7\,\text{mol}\cdot\text{dm}^{-3}.

  • Problem 8 — Industrial Ammonium Sulphate Synthesis:

    • Sulphuric acid reacts with ammonia to produce the fertiliser ammonium sulphate:     H2SO4(aq)+2NH3(g)→(NH4)2SO4(aq)\text{H}_2\text{SO}_4(aq) + 2\text{NH}_3(g) \rightarrow (\text{NH}_4)_2\text{SO}_4(aq)

    • 8.1 Determine the maximum mass of ammonium sulphate that can be obtained from 2,0 kg2,0\,\text{kg} of sulphuric acid and 1,0 kg1,0\,\text{kg} of ammonia.

    • 8.2 A factory operator carries out the reaction with 2,0 kg2,0\,\text{kg} of sulphuric acid and 1,0 kg1,0\,\text{kg} of ammonia and collects 2,5 kg2,5\,\text{kg} of ammonium sulphate. Calculate the percentage yield of the reaction.

  • Problem 9 — Purity Analysis of Limestone:

    • Limestone consists primarily of calcium carbonate (CaCO3\text{CaCO}_3). To analyze the purity of a limestone sample with a mass of 3,5 g3,5\,\text{g}, concentrated hydrochloric acid is added:     CaCO3(s)+2HCℓ(aq)→CO2(g)+CaCℓ2(aq)+H2O(ℓ)\text{CaCO}_3(s) + 2\text{HC}\ell(aq) \rightarrow \text{CO}_2(g) + \text{CaC}\ell_2(aq) + \text{H}_2\text{O}(\ell)

    • If the mass of calcium chloride produced is 3,6 g3,6\,\text{g}, calculate the percentage purity of the limestone sample.

  • Problem 10 — Purity Analysis of Tin Ore:

    • A sample containing tin dioxide (SnO2\text{SnO}_2) with a mass of 6,2 g6,2\,\text{g} is treated with sulphuric acid according to:     SnO2(s)+2H2SO4(aq)→Sn(SO4)2(s)+2H2O(ℓ)\text{SnO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow \text{Sn}(\text{SO}_4)_2(s) + 2\text{H}_2\text{O}(\ell)

    • If the mass of tin sulphate produced is 4,7 g4,7\,\text{g}, determine the percentage purity of the sample.

  • Problem 11 — Agricultural Fertiliser Calculations:

    • Ammonium nitrate (NH4NO3\text{NH}_4\text{NO}_3) and ammonium sulphate ((NH4)2SO4(\text{NH}_4)_2\text{SO}_4) provide essential nitrogen to stimulate plant growth.

    • 11.1 Calculate the number of moles present in a 4 g4\,\text{g} sample of ammonium nitrate.

    • 11.2 The 4 g4\,\text{g} sample of ammonium nitrate is dissolved in water to make a total solution volume of 250 cm3250\,\text{cm}^3. Calculate the concentration of the resulting solution.

    • 11.3 Calculate the percentage of nitrogen present in 1 mole1\,\text{mole} of solid ammonium sulphate.

    • 11.4 Calculate the total number of ammonium ions present when 0,125 mol0,125\,\text{mol} of ammonium sulphate dissolves completely in water, showing all calculation steps.

  • Problem 12 — Artificial Sweetener Analysis and Yield Verification:

    • 12.1 Aspartame (marketed under the brand name NutraSweet by the NutraSweet Company) is a low-calorie sweetener approximately 200 times sweeter than sucrose. Its elemental analysis is 57,14% C57,14\%\,\text{C}, 6,16% H6,16\%\,\text{H}, 9,52% N9,52\%\,\text{N}, and 27,18% O27,18\%\,\text{O}. Demonstrate through calculations that the empirical formula for aspartame is C14H18N2O5\text{C}_{14}\text{H}_{18}\text{N}_2\text{O}_5.

    • 12.2 A reaction is conducted using 9 g9\,\text{g} of aluminium powder and excess chlorine gas, yielding an actual mass of 35,6 g35,6\,\text{g} of aluminium chloride:     2Aℓ+3Cℓ2→2AℓCℓ32\text{A}\ell + 3\text{C}\ell_2 \rightarrow 2\text{A}\ell\text{C}\ell_3

    • 12.2.1 Explain the chemical necessity of supplying chlorine gas in excess.

    • 12.2.2 Calculate the percentage yield of aluminium chloride.

  • Problem 13 — Atmospheric Chemistry and Photochemical Smog:

    • Ozone (O3\text{O}_3) reacts with nitrogen monoxide (NO\text{NO}) to generate nitrogen dioxide (NO2\text{NO}_2). The NO\text{NO} gas originates primarily from internal combustion engine vehicle exhausts and jet aircraft emissions. The resulting NO2\text{NO}_2 gas creates the visible brown smog (smoke and fog) over urban centers and poses severe respiratory hazards to humans:     O3(g)+NO(g)→O2(g)+NO2(g)\text{O}_3(g) + \text{NO}(g) \rightarrow \text{O}_2(g) + \text{NO}_2(g)

    • In a laboratory reaction, 0,74 g0,74\,\text{g} of O3\text{O}_3 is combined with 0,67 g0,67\,\text{g} of NO\text{NO}.

    • 13.1 Calculate the initial number of moles of O3\text{O}_3 and NO\text{NO} present prior to reaction.

    • 13.2 Identify the limiting reagent for the reaction.

    • 13.3 Calculate the volume of NO2\text{NO}_2 gas produced in cm3\text{cm}^3 at STP.