8.2 The Mole Concept, Molar Mass & Empirical/Molecular Formulas

8.2 The Mole Concept, Molar Mass & Empirical/Molecular Formulas

Quick Summary: The mole provides the quantitative bridge between atomic-scale entities and macroscopic laboratory samples. One mole contains Avogadro's number ($6.02214076 \times 10^{23}$) of particles. The molar mass in grams per mole equals the substance's formula mass in atomic mass units. Percent mass composition enables empirical formula calculation via a four-step algorithm, while experimental molar mass establishes the molecular formula. Combustion analysis and gravimetric hydrate analysis provide standard laboratory avenues for formula determination.

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1. The Mole Concept and Avogadro's Constant

Because individual atoms have masses on the order of $10^{-24}\text{ to } 10^{-22}\text{ grams}$, counting atoms directly requires a macroscopic scaling unit. The International System of Units (SI) defines the mole (symbol: $\text{mol}$) by fixing Avogadro's constant ($N_A$) to an exact value: $N_A = 6.02214076 \times 10^{23}\text{ entities/mol}$ Elementary entities represent atoms, molecules, ions, or formula units, depending on the chemical structure.

Bridging Atomic Mass Units to Grams

The unified atomic mass unit ($1\text{ u}$ or $1\text{ amu}$) equals $1/12$ the mass of an unbound $^{12}\text{C}$ atom ($1.66054 \times 10^{-24}\text{ g}$). Because $N_A \times (1.66054 \times 10^{-24}\text{ g}) = 1.000\text{ g}$, an atom's average mass in $\text{amu}$ corresponds numerically to the mass of one mole of those atoms in $\text{grams}$:

• One $^{12}\text{C}$ atom has a mass of $12\text{ amu}$; one mole of $^{12}\text{C}$ atoms has a mass of $12.000\text{ g}$.

Mass Terminology

• Molecular Mass: Sum of atomic masses in a discrete covalent molecule (e.g., $\text{H}_2\text{O} = 18.02\text{ amu}$).
• Formula Mass: Sum of atomic masses in an empirical formula unit of an ionic crystal (e.g., $\text{NaCl} = 58.44\text{ amu}$).
• Molar Mass ($M$): Mass of one mole of substance in grams per mole ($\text{g/mol}$).

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2. Quantitative Conversion Pathways

Chemical calculations convert between mass, moles, and particle count: $\text{Mass (g)} \xleftrightarrow{\div M}{\times M} \text{Moles (mol)} \xleftrightarrow{\times N_A}{\div N_A} \text{Number of Particles}$

• $\text{Moles} = m / M$
• $\text{Particles} = n \times (6.022 \times 10^{23}\text{ particles/mol})$

Sub-Formula Stoichiometry

A formula also dictates internal molar ratios. In one mole of aluminum sulfate, $\text{Al}_2(\text{SO}_4)_3$ ($M = 342.15\text{ g/mol}$):

• $2\text{ mol } \text{Al}^{3+}$ ions ($1.204 \times 10^{24}\text{ ions}$)
• $3\text{ mol } \text{SO}_4^{2-}$ ions ($1.807 \times 10^{24}\text{ ions}$)
• $12\text{ mol } \text{O}$ atoms ($192.00\text{ g}$, or $7.226 \times 10^{24}\text{ atoms}$)

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3. Percent Composition and Formula Determination

Percent Composition by Mass defines the percentage of total mass contributed by each element in a pure compound: $\text{Mass } \% = \left( \frac{n \times M_{\text{element}}}{M_{\text{compound}}} \right) \times 100\%$ where $n$ is the moles of element per mole of compound.

Empirical versus Molecular Formulas

• Empirical Formula: The simplest, lowest whole-number ratio of atoms in a substance (e.g., $\text{CH}_2\text{O}$).
• Molecular Formula: The actual atom count in a discrete molecule (e.g., $\text{C}_6\text{H}_{12}\text{O}_6$).

Four-Step Empirical Formula Algorithm

1. Assume a 100.0 g Sample: Convert percentages directly to grams ($40.0\% \to 40.0\text{ g}$), or use given elemental masses directly.
2. Convert Grams to Moles: Divide each element's mass by its periodic table molar mass.
3. Divide by Smallest Mole Value: Divide all molar values by the smallest value to produce normalized ratios.
4. Scale to Integers: If normalized values contain fractions, multiply all ratios by a common integer ($0.50 \times 2$, $0.33 \times 3$, $0.25 \times 4$).

Determining Molecular Formulas

The molecular formula is an integer multiple ($n$) of the empirical formula: $n = \frac{\text{Experimental Molar Mass}}{\text{Empirical Formula Mass}} \implies \text{Molecular Formula} = (\text{Empirical Formula})_n$

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4. Combustion Analysis

Combustion analysis determines the empirical formula of organic compounds containing carbon, hydrogen, and oxygen. A weighed sample is combusted in a pure oxygen stream. Water vapor is trapped in magnesium perchlorate, $\text{Mg(ClO}_4)_2$, and carbon dioxide is absorbed in sodium hydroxide, $\text{NaOH}$:

• Carbon mass from $\text{CO}_2$: $m_{\text{C}} = m_{\text{CO}_2} \times (12.011 / 44.01)$
• Hydrogen mass from $\text{H}_2\text{O}$: $m_{\text{H}} = m_{\text{H}_2\text{O}} \times (2.016 / 18.015)$
• Oxygen mass by difference: $m_{\text{O}} = m_{\text{sample}} - (m_{\text{C}} + m_{\text{H}})$
• For nitrogen-bearing compounds, combustion gases pass over hot copper to reduce nitrogen oxides back to elemental $\text{N}_2$ for quantification.

Worked Example: Organic Combustion

A $3.870\text{ g}$ sample of a liquid containing carbon, hydrogen, and oxygen produces $6.715\text{ g } \text{CO}_2$ and $3.665\text{ g } \text{H}_2\text{O}$. Experimental molar mass is $76.10\text{ g/mol}$:

1. $m_{\text{C}} = 6.715 \times (12.011 / 44.01) = 1.8326\text{ g C} \implies 0.15258\text{ mol C}$
2. $m_{\text{H}} = 3.665 \times (2.016 / 18.015) = 0.4101\text{ g H} \implies 0.4069\text{ mol H}$
3. $m_{\text{O}} = 3.870 - (1.8326 + 0.4101) = 1.6273\text{ g O} \implies 0.1017\text{ mol O}$
4. Normalized ratios: $\text{C} = 1.50$, $\text{H} = 4.00$, $\text{O} = 1.00$.
5. Multiplying by $2$ yields empirical formula $\text{C}_3\text{H}_8\text{O}_2$ (formula mass $= 76.10\text{ g/mol}$). The molecular formula is also $\text{C}_3\text{H}_8\text{O}_2$.

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5. Hydrate Formula Determination

Hydrates are crystalline salts containing stoichiometric water in their lattice ($\text{Salt} \cdot n\,\text{H}_2\text{O}$). Thermal dehydration drives off water as vapor, leaving anhydrous salt behind: $n = \frac{\text{Moles of } \text{H}_2\text{O Lost}}{\text{Moles of Anhydrous Salt Remaining}}$ For example, heating $4.108\text{ g}$ of hydrated magnesium sulfate leaves $2.007\text{ g}$ of anhydrous $\text{MgSO}_4$ ($M = 120.37\text{ g/mol}$, $0.01667\text{ mol}$) and drives off $2.101\text{ g}$ of water ($M = 18.015\text{ g/mol}$, $0.1166\text{ mol}$). The ratio $n = 0.1166 / 0.01667 = 7.00$, confirming $\text{MgSO}_4 \cdot 7\,\text{H}_2\text{O}$.