Chemical Formulas, Molar Mass, and Percent Composition

Chemical Formula Types and Representations

• Molecular Formula:

  • Specifies the exact identity and number of atoms of each element present in a single individual molecule of a substance.

  • Example: The molecular formula of ribose is $C_5H_{10}O_5$, indicating that each ribose molecule consists of 5 carbon atoms, 10 hydrogen atoms, and 5 oxygen atoms.

• Empirical Formula:

  • Indicates the relative ratio of each kind of atom in a compound expressed using the smallest integer (whole-number) ratio.

  • Determined by dividing the subscript numbers in a molecular formula by their greatest common factor.

  • Example: Dividing the subscripts of ribose ($C_5H_{10}O_5$) by 5 yields its empirical formula, $CH_2O$.

  • The empirical formula $CH_2O$ reveals that the substance is composed of carbon and water ($C + H_2O$), which gives rise to the term "carbohydrate".

  • It provides the relative atomic proportions but does not convey the actual number of atoms in a discrete molecule.

• Structural Formula:

  • Illustrates the specific arrangement and chemical bonding connections between the constituent atoms within a molecule.

• Three-Dimensional (3D) Model:

  • Provides a spatial representation showing the actual 3D geometry, bond angles, and relative positions of atoms.

Molecular Weight, Molar Mass, and Stoichiometric Interconversions

• Molecular Weight and Formula Weight:

  • Molecular Weight: The total sum of the atomic weights of all atoms in a discrete molecule.

  • Formula Weight: Computed for non-molecular substances (such as ionic lattices) or empirical formulas by summing the atomic weights of all atoms in the stated formula unit.

  • Expressed in units of atomic mass units ($\text{amu}$).

• Molar Mass:

  • Defined as the mass in grams of $1\,\text{mole}$ ($6.022 \times 10^{23}$ units) of a substance.

  • Units are expressed in grams per mole ($\text{g/mol}$ or $\text{g\,mol}^{-1}$).

  • Numerically identical to the molecular weight or formula weight expressed in $\text{amu}$.

  • In practical chemical applications, molecular weight and molar mass are used interchangeably.

• Comprehensive Calculation of Ribose Molar Mass:

  • Single molecule mass of ribose ($C_5H_{10}O_5$) = $150.13\,\text{amu}$.

  • Molar mass of ribose = $150.13\,\text{g/mol}$.

  • Detailed step-by-step summation of constituent atomic masses:

  • Carbon: $(5\,\text{mol C atoms}) \times (12.011\,\text{g/mol C}) = 60.055\,\text{g}$

  • Hydrogen: $(10\,\text{mol H atoms}) \times (1.008\,\text{g/mol H}) = 10.080\,\text{g}$

  • Oxygen: $(5\,\text{mol O atoms}) \times (15.999\,\text{g/mol O}) = 79.995\,\text{g}$

  • Total Molar Mass: $60.055\,\text{g} + 10.080\,\text{g} + 79.995\,\text{g} = 150.13\,\text{g/mol}$

• Conversion Procedures between Mass, Moles, and Particles:

  • Mass to Moles: Divide mass in grams by molar mass ($\text{g/mol}$):     $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

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

  • Moles to Particles: Multiply moles by Avogadro's number ($N_A$):     $\text{particles} = \text{moles} \times N_A$

  • Particles to Moles: Divide particle count by Avogadro's number ($N_A$):     $\text{moles} = \frac{\text{particles}}{N_A}$

Percent Composition and Mass Percentages

• Definition of Percent Composition:

  • Represents the percentage by mass contributed by each individual element present within a compound.

• Mass Percent Calculation from Sample Data:

  • Defined as the fraction of the sample containing a specific element multiplied by $100\%$:     $\text{mass \% element} = \frac{\text{mass of element in the sample}}{\text{total mass of the sample}} \times 100\%$

  • Sample Problem: A $100.0\,\text{g}$ sample of ribose contains $40.0\,\text{g}$ of carbon.     $\text{\% Carbon} = \frac{40.0\,\text{g carbon}}{100.0\,\text{g ribose}} \times 100\% = 40.0\%$

• Mass Percent Calculation from Chemical Formula:

  • Determined using the atomic weights and chemical formula of the compound:     $\text{mass \% element} = \frac{(\text{\# atoms of an element}) \times (\text{atomic weight of element})}{\text{molecular weight of compound}} \times 100\%$

  • Sample Problem (Oxygen in Ribose):

  • Molecular weight of ribose = $150.13\,\text{g/mol}$.

  • Ribose contains 5 oxygen atoms, each with an atomic weight of $15.999\,\text{g/mol}$.     $\text{\% Oxygen} = \frac{5 \times 15.999\,\text{g/mol}}{150.13\,\text{g/mol}} \times 100\% = 53.284\%$

• Total Mass Percentage Rule:

  • The sum of all mass percentages of all elements in any compound must equal $100\%$     $\sum (\text{mass \% of all elements}) = 100\%$

  • Sample Problem (Hydrogen in Ribose via Subtraction):

  • Given that carbon is $40.0\%$ and oxygen is $53.284\%$     $\text{\% Hydrogen} = 100\% - 40.0\% - 53.284\% = 6.7\%$

  • Calculating the mass percent of hydrogen using the formula-based method yields the identical result.