Quantitative Chemistry: Chemical Reactions and Stoichiometry (General Chemistry Study Notes)

Quantitative Chemistry: Chemical Reactions and Stoichiometry

Learning Outcomes

• Understand how to balance chemical equations.

• Learn the concept of the mole and how to interconvert between moles, mass, and concentration for atoms, ions, and molecules.

• Apply stoichiometry to calculate the quantities of substances involved in chemical reactions.

Chemical Equations

Representation of Chemical Reactions

Chemical reactions are concisely represented by chemical equations, which show the reactants transforming into products. The numbers in front of chemical formulas are called stoichiometric coefficients and indicate the number of each species involved.

• Molecular formula: Shows the number and type of atoms (e.g., $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$).

• Structural formula: Illustrates the arrangement of atoms.

• Quantitative description: Two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water.

Balancing Chemical Equations

Law of Conservation of Mass

Atoms are neither created nor destroyed in a chemical reaction. Therefore, a chemical equation must have equal numbers of each atom on both sides; it must be balanced.

• Only change the coefficients, never the subscripts in a chemical formula.

• Changing coefficients alters the amount, not the identity, of a substance.

• Changing subscripts changes the chemical identity (e.g., $H_2O$ is water, $H_2O_2$ is hydrogen peroxide).

Table: Meaning of Chemical Symbols

Chemical Symbol	Meaning	Composition
H2O	One molecule of water	Two H atoms, one O atom
2H2O	Two molecules of water	Four H atoms, two O atoms
H2O2	One molecule of hydrogen peroxide	Two H atoms, two O atoms

Hints for Balancing Equations

1. Ensure correct chemical formulas for all reactants and products.

2. Balance elements that appear only once on each side first.

3. Leave H and O atoms until later.

4. Balance other elements in order of those appearing in the fewest formulas, ending with O and H.

5. If a polyatomic ion appears unchanged on both sides, balance it as a group.

Example: Combustion of Propane

• Unbalanced: $C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)$

• Balanced: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$

• Balance C atoms first, then H, then O.

Balancing Equations with Polyatomic Ions

• Example: $Cu(NO_3)_2 + Na_3PO_4 \rightarrow Cu_3(PO_4)_2 + NaNO_3$

• Balance polyatomic ions as units if they appear unchanged.

• Balanced: $3Cu(NO_3)_2 + 2Na_3PO_4 \rightarrow Cu_3(PO_4)_2 + 6NaNO_3$

Atomic Mass, Molecular Mass, Formula Mass

Definitions and Applications

Each element has a constant, characteristic mass known as its atomic mass. This is the basis for quantitative chemistry.

• Relative atomic mass: The mass of an atom compared to a standard (currently $^{12}C$).

• Atomic mass unit (amu): $^{12}C$ is defined as 12 amu.

• Atomic masses on the periodic table are in amu.

Calculating Molecular and Formula Mass

• Molecular mass: Sum of atomic masses in a molecule (e.g., $H_2O$: $2 \times 1.00794$ amu $+ 1 \times 15.9994$ amu $= 18.015$ amu).

• Formula mass: Sum of atomic masses in an ionic compound's formula unit (e.g., $Ca_3(PO_4)_2$: $3 \times 40.078$ amu $+ 2 \times 30.9738$ amu $+ 8 \times 15.9994$ amu $= 310.177$ amu).

Moles and Molar Mass

The Mole Concept

The mole is the SI unit for the amount of substance, defined as the number of atoms in exactly 12 grams of $^{12}C$.

• 1 mole $= 6.022 \times 10^{23}$ objects (Avogadro's number).

• Used to count atoms, molecules, ions, etc.

Interconverting Moles, Mass, and Number of Particles

• Number of moles: $\text{moles} = \dfrac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Mass: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

• Molar mass: $\text{molar mass (g/mol)} = \dfrac{\text{mass (g)}}{\text{moles}}$

• Number of particles: $\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

Molar Mass Triangle

Mass (g)	Moles (mol)	Molar Mass (g/mol)
Top	Bottom left	Bottom right

Examples

• Calculate moles of Na in 0.234 mol: $0.234 \text{ mol} \times 6.022 \times 10^{23} = 1.41 \times 10^{23}$ atoms

• Calculate moles from number of atoms: $\dfrac{8.93 \times 10^{24}}{6.022 \times 10^{23}} = 14.8$ mol

• Calculate moles of H and O in $4.56 \times 10^{22}$ molecules of $H_2O$:

  • Moles of $H_2O$: $\dfrac{4.56 \times 10^{22}}{6.022 \times 10^{23}} = 0.0757$ mol

  • Moles of H: $2 \times 0.0757 = 0.151$ mol

  • Moles of O: $1 \times 0.0757 = 0.0757$ mol

Mole Calculations Using Chemical Formulas

Percent Composition by Mass

The percent composition by mass of an element in a compound is:

• $\text{Mass\%A=Mass A/Total Mass of Compound}\cdot100\%$

• Example for glucose ($C_6H_{12}O_6$):

  • Molar mass $= 180.16$ g/mol

  • Mass % C: $\dfrac{6 \times 12.01}{180.16} \times 100\% = 40.00\%$

  • Mass % H: $\dfrac{12 \times 1.008}{180.16} \times 100\% = 6.71\%$

  • Mass % O: $\dfrac{6 \times 16.00}{180.16} \times 100\% = 53.29\%$

Empirical Formula from Mass Percent Data

• Assume 100 g sample for calculation.

• Convert mass percent to grams, then to moles using molar mass.

• Divide by smallest number of moles to get subscripts.

• Example: 31.3% Ca, 18.8% C, 50% O

  • Moles Ca: $\dfrac{31.3}{40.08} = 0.7809$ mol

  • Moles C: $\dfrac{18.8}{12.01} = 1.565$ mol

  • Moles O: $\dfrac{50}{16.00} = 3.125$ mol

  • Empirical formula: $CaC_2O_4$ (calcium oxalate)

Stoichiometry

Quantitative Relationships in Chemical Reactions

Stoichiometry is the calculation of reactants and products in chemical reactions using balanced equations.

• Subscripts in formulas and coefficients in equations indicate precise quantities.

• Example: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ means 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.

Stoichiometric Calculations

• Use the ratio of coefficients to relate moles of reactants and products.

• General formula: $aA + bB \rightarrow$ products

  • Moles of B $= \dfrac{b}{a} \times$ moles of A

• Example: $4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O$

  • Moles of $N_2$ from 2.5 mol $NH_3$: $2/4 \times 2.5 = 1.25$ mol

  • Moles of $O_2$ needed for 3.61 mol $NH_3$: $3/4 \times 3.61 = 2.71$ mol

Mass Calculations in Stoichiometry

• Convert mass to moles using molar mass.

• Use stoichiometric ratios to find moles of products.

• Convert moles of products to mass.

• Example: Burning 0.0679 g propane ($C_3H_8$) produces:

  • Moles $C_3H_8$: $0.0679/44.09 = 1.54 \times 10^{-3}$ mol

  • Moles $CO_2$: $3 \times 1.54 \times 10^{-3} = 4.62 \times 10^{-3}$ mol

  • Mass $CO_2$: $4.62 \times 10^{-3} \times 44.01 = 0.203$ g

  • Mass $H_2O$: $6.16 \times 10^{-3} \times 18.02 = 0.111$ g

Limiting Reactants and Yield

Limiting Reactant Concept

The limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed.

• Identify the limiting reactant by comparing available moles divided by their stoichiometric coefficients.

• Product yield is determined by the limiting reactant.

Theoretical and Percent Yield

• Theoretical yield: Maximum possible amount of product.

• Actual yield: Amount actually obtained.

• Percent yield: $\text{Percent yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$

• Example: If theoretical yield is 1.29 g and actual yield is 1.05 g, percent yield $= \dfrac{1.05}{1.29} \times 100\% = 81.4\%$

Summary Table: Key Equations

Calculation	Equation (LaTeX)
Moles from mass	$\text{moles} = \dfrac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
Mass from moles	$\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$
Number of particles	$\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$
Percent composition	$\text{Mass \%A=Mass A/Total mass of compound}\cdot100\%$
Percent yield	$\text{Percent yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$

Additional info: These notes cover the essential quantitative aspects of chemical reactions, including balancing equations, mole concept, mass-mole-particle conversions, percent composition, empirical formula determination, stoichiometry, limiting reactants, and yield calculations. All examples and equations are foundational for General Chemistry students preparing for exams or laboratory work.