Quantities in Chemical Reactions and Chemical Composition: Study Notes

Quantities in Chemical Reactions

Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. It allows chemists to count entities at the atomic and molecular scale by relating mass to number of particles.

• Definition: One mole contains $6.022 \times 10^{23}$ entities (Avogadro's number).

• Application: Used to convert between mass, number of particles, and volume (for gases).

• Example: To find the number of moles in 28 g of $H_2O$, use the molar mass of water ($18.02$ g/mol): $\text{Moles of } H_2O = \frac{28\ \text{g}}{18.02\ \text{g/mol}} = 1.55\ \text{mol}$

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule.

• Formula: $\text{Molar Mass} = \sum \text{Atomic Masses of Elements in Compound}$

• Example: For $CO_2$, molar mass = $12.01$ (C) + $2 \times 16.00$ (O) = $44.01$ g/mol

Stoichiometry

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

• Steps:

  1. Write the balanced chemical equation.

  2. Convert given quantities to moles.

  3. Use mole ratios from the equation to find moles of desired substance.

  4. Convert moles back to grams or other units as needed.

• Example: If 23 mol of $NaCl$ are produced, how many moles of $Na$ were used? (Assuming $2Na + Cl_2 \rightarrow 2NaCl$) From the equation, 2 mol $Na$ produce 2 mol $NaCl$, so 23 mol $NaCl$ requires 23 mol $Na$.

Chemical Composition

Percent Composition

Percent composition refers to the percentage by mass of each element in a compound.

• Formula: $\%\ \text{Element} = \frac{\text{Mass of Element in 1 mol of Compound}}{\text{Molar Mass of Compound}} \times 100$

• Example: For $H_2O$, mass of H = $2 \times 1.008$ g, mass of O = $16.00$ g, molar mass = $18.02$ g/mol. $\%\ H = \frac{2.016}{18.02} \times 100 = 11.19\%$ $\%\ O = \frac{16.00}{18.02} \times 100 = 88.81\%$

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule.

• Empirical Formula: Simplest ratio (e.g., $CH_2O$ for glucose)

• Molecular Formula: Actual formula (e.g., $C_6H_{12}O_6$ for glucose)

• Relationship: $\text{Molecular Formula} = n \times \text{Empirical Formula}$, where $n$ is an integer

Tables

Sample Table: Molar Masses of Common Compounds

Compound	Formula	Molar Mass (g/mol)
Water	H2O	18.02
Carbon Dioxide	CO2	44.01
Sodium Chloride	NaCl	58.44
Glucose	C6H12O6	180.16
Ammonia	NH3	17.03

Additional info: Table entries inferred for clarity and completeness.