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Chemistry - Mole Concept and Stoichiometry

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Gay-Lussac's Law: When gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volume of the gaseous products, provided temperature and pressure remain constant.

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Avogadro's Law: Equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules.

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Relative Molecular Mass (RMMRMM): The number of times one molecule of a substance is heavier than 112\frac{1}{12} the mass of an atom of Carbon-12 (12C^{12}C).

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Mole Concept: A mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions) as there are atoms in 12 g12\,g of Carbon-12. This number is Avogadro's Number (NA≈6.022×1023N_A \approx 6.022 \times 10^{23}).

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Molar Volume: One mole of any gas at S.T.P. (273 K273\,K and 1 atm1\,atm pressure) occupies a volume of 22.4 dm322.4\,dm^3 or 22.4 litres22.4\,litres.

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Vapour Density (VDVD): The ratio of the mass of a certain volume of a gas to the mass of an equal volume of Hydrogen under similar conditions.

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Empirical Formula: The simplest formula of a compound which shows the lowest whole-number ratio of atoms of the elements present in one molecule.

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Molecular Formula: The chemical formula which represents the actual number of atoms of each element present in one molecule of the compound.

📐Formulae

Number of moles (n)=Given Mass (m)Gram Atomic/Molecular Mass (M)Number\ of\ moles\ (n) = \frac{Given\ Mass\ (m)}{Gram\ Atomic/Molecular\ Mass\ (M)}

Number of moles (n)=Volume of gas at S.T.P.22.4 LNumber\ of\ moles\ (n) = \frac{Volume\ of\ gas\ at\ S.T.P.}{22.4\,L}

Relative Molecular Mass (RMM)=2×Vapour Density (VD)Relative\ Molecular\ Mass\ (RMM) = 2 \times Vapour\ Density\ (VD)

Percentage composition of element=Total mass of the element in the compoundTotal molecular mass of the compound×100Percentage\ composition\ of\ element = \frac{Total\ mass\ of\ the\ element\ in\ the\ compound}{Total\ molecular\ mass\ of\ the\ compound} \times 100

Molecular Formula=n×(Empirical Formula)Molecular\ Formula = n \times (Empirical\ Formula) where n=Molecular MassEmpirical Formula Massn = \frac{Molecular\ Mass}{Empirical\ Formula\ Mass}

Number of molecules=n×NA=n×6.022×1023Number\ of\ molecules = n \times N_A = n \times 6.022 \times 10^{23}

💡Examples

Problem 1:

What volume of Oxygen (O2O_2) is required to burn 200 cm3200\,cm^3 of Methane (CH4CH_4) completely? Also, find the volume of Carbon dioxide (CO2CO_2) formed. Equation: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

Solution:

According to Gay-Lussac's Law: 1 volume of CH4CH_4 reacts with 2 volumes of O2O_2 to produce 1 volume of CO2CO_2. Ratio is 1:2:11 : 2 : 1. Given Volume of CH4=200 cm3CH_4 = 200\,cm^3. Volume of O2O_2 required =2×200=400 cm3= 2 \times 200 = 400\,cm^3. Volume of CO2CO_2 produced =1×200=200 cm3= 1 \times 200 = 200\,cm^3.

Explanation:

Using the stoichiometric coefficients from the balanced equation, we relate the gaseous volumes directly at constant temperature and pressure.

Problem 2:

Calculate the mass of 0.5 moles0.5\,moles of CaCO3CaCO_3. (Atomic masses: Ca=40,C=12,O=16Ca=40, C=12, O=16)

Solution:

First, find the GMMGMM of CaCO3CaCO_3: GMM=40+12+(3×16)=40+12+48=100 g/molGMM = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100\,g/mol. Mass=n×GMM=0.5×100=50 gMass = n \times GMM = 0.5 \times 100 = 50\,g.

Explanation:

The mass of a substance is the product of the number of moles and its molar mass.

Problem 3:

A compound contains 40%40\% Carbon, 6.7%6.7\% Hydrogen and 53.3%53.3\% Oxygen. If the vapour density is 3030, find the molecular formula.

Solution:

  1. Molar ratio: C=4012=3.33C = \frac{40}{12} = 3.33, H=6.71=6.7H = \frac{6.7}{1} = 6.7, O=53.316=3.33O = \frac{53.3}{16} = 3.33.
  2. Simplest ratio: C=3.333.33=1C = \frac{3.33}{3.33} = 1, H=6.73.33≈2H = \frac{6.7}{3.33} \approx 2, O=3.333.33=1O = \frac{3.33}{3.33} = 1.
  3. Empirical Formula =CH2O= CH_2O.
  4. Empirical Formula Mass =12+(2×1)+16=30= 12 + (2 \times 1) + 16 = 30.
  5. Molecular Mass =2×VD=2×30=60= 2 \times VD = 2 \times 30 = 60.
  6. n=6030=2n = \frac{60}{30} = 2.
  7. Molecular Formula =2×(CH2O)=C2H4O2= 2 \times (CH_2O) = C_2H_4O_2.

Explanation:

The empirical formula is determined by finding the simplest atomic ratio. Then, the molecular formula is calculated using the relationship between vapour density and molecular mass.

Mole Concept and Stoichiometry - Revision Notes & Key Formulas | ICSE Class 10 Science