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Stoichiometric Relationships - Introduction to the particulate nature of matter and chemical change

Grade 11IBChemistry

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

🔑Concepts

Matter is defined as anything that has mass and occupies space. It is composed of particles (atoms, molecules, or ions) whose arrangement and movement determine the state of matter: solid (ss), liquid (ll), and gas (gg).

A pure substancepure\ substance has a constant chemical composition and distinct chemical properties. It can be an elementelement (e.g., O2O_2, FeFe) or a compoundcompound (e.g., H2OH_2O, NaClNaCl).

A mixturemixture contains two or more substances that are not chemically combined. Homogeneous mixturesHomogeneous\ mixtures (solutions) have a uniform composition (e.g., NaCl(aq)NaCl(aq)), while heterogeneous mixturesheterogeneous\ mixtures have non-uniform compositions (e.g., a mixture of S(s)S(s) and Fe(s)Fe(s)).

Phase changes are physical processes: meltingmelting (sls \rightarrow l), freezingfreezing (lsl \rightarrow s), vaporizationvaporization (lgl \rightarrow g), condensationcondensation (glg \rightarrow l), sublimationsublimation (sgs \rightarrow g), and depositiondeposition (gsg \rightarrow s).

Chemical changes involve the reorganization of atoms to form new substances. These are represented by chemical equations where the Law of Conservation of MassLaw\ of\ Conservation\ of\ Mass must be obeyed, meaning the number of atoms of each element must be equal on both sides of the equation.

The Kinetic Molecular TheoryKinetic\ Molecular\ Theory postulates that particles in a gas are in constant random motion, have negligible volume compared to the container, and undergo perfectly elastic collisions.

📐Formulae

aA+bBcC+dDaA + bB \rightarrow cC + dD

Temperature(K)=Temperature(C)+273.15Temperature(K) = Temperature(^{\circ}C) + 273.15

H2O(s)H2O(l)H2O(g)H_2O(s) \rightarrow H_2O(l) \rightarrow H_2O(g)

💡Examples

Problem 1:

Balance the following chemical equation for the complete combustion of butane: C4H10(g)+O2(g)CO2(g)+H2O(l)C_4H_{10}(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)

Solution:

2C4H10(g)+13O2(g)8CO2(g)+10H2O(l)2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l)

Explanation:

To balance the equation, we start with Carbon: 4 atoms in C4H10C_4H_{10} require 4CO24CO_2. Then Hydrogen: 10 atoms in C4H10C_4H_{10} require 5H2O5H_2O. This gives 4×2+5×1=134 \times 2 + 5 \times 1 = 13 Oxygen atoms on the right. To avoid fractions (6.5O26.5O_2), we multiply the entire equation by 2 to achieve whole-number coefficients: 22, 1313, 88, and 1010.

Problem 2:

Identify the state symbols and type of mixture formed when excess BaSO4(s)BaSO_4(s) is added to water.

Solution:

BaSO4(s)+H2O(l)BaSO4(s)+H2O(l)BaSO_4(s) + H_2O(l) \rightarrow BaSO_4(s) + H_2O(l) (Heterogeneous Mixture)

Explanation:

Because BaSO4BaSO_4 is insoluble in water, it remains as a solid at the bottom of the container. A mixture with visible boundaries between phases (ss and ll) is classified as a heterogeneous mixtureheterogeneous\ mixture.

Introduction to the particulate nature of matter and chemical change Revision - Grade 11 Chemistry…