Chapter 6: Chemical Composition – The Mole Concept and Chemical Formulas

Chapter 6: Chemical Composition

The Mole Concept and Chemical Formulas

This chapter introduces the concept of the mole, a fundamental unit in chemistry for counting particles, and explains how chemical formulas are used to relate masses, moles, and numbers of atoms or molecules. The following study notes cover key conversion techniques and the use of chemical formulas in quantitative chemical analysis.

Converting Between Moles and Number of Atoms

Understanding the relationship between moles and the number of atoms is essential for quantifying substances in chemistry.

• Mole: The mole is a counting unit in chemistry, defined as the amount of substance containing as many entities (atoms, molecules, ions) as there are atoms in 12 grams of carbon-12.

• Avogadro's Number: $6.022 \times 10^{23}$ is the number of particles in one mole of a substance.

• Conversion Example: To convert atoms to moles, divide the number of atoms by Avogadro's number.

Example: A silver ring contains $1.1 \times 10^{22}$ silver atoms. To find the number of moles of silver:

• Given: $1.1 \times 10^{22}$ Ag atoms

• Required: mol Ag

• Conversion: $\frac{1.1 \times 10^{22} \text{ Ag atoms}}{6.022 \times 10^{23} \text{ Ag atoms/mol}} = 1.8 \times 10^{-2}$ mol Ag$

Key Point: The number of moles is much smaller than the number of atoms because a mole contains a very large number of atoms.

The Mole Concept—Converting Between Grams and Moles

Chemists often need to convert between the mass of a substance and the number of moles. This requires knowledge of the molar mass.

• Molar Mass: The mass of one mole of a substance, usually expressed in grams per mole (g/mol).

• Conversion Formula: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Example: Calculate the number of moles of sulfur in 57.8 g of sulfur (molar mass of S = 32.06 g/mol): $\frac{57.8 \text{ g S}}{32.06 \text{ g/mol}} = 1.80 \text{ mol S}$

Key Point: The number of moles is proportional to the mass and inversely proportional to the molar mass.

Converting Between Grams and Number of Atoms

To determine the number of atoms in a given mass, a two-step conversion is required: grams to moles, then moles to atoms.

• Step 1: Convert grams to moles using molar mass.

• Step 2: Convert moles to atoms using Avogadro's number.

• Formula: $\text{Number of atoms} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \times 6.022 \times 10^{23}$

• Example: For 16.2 g of aluminum (molar mass = 26.98 g/mol): $\frac{16.2}{26.98} \times 6.022 \times 10^{23} = 3.62 \times 10^{23}$ Al atoms$

Key Point: The number of atoms in a macroscopic sample is extremely large.

Converting Between Grams and Moles for Compounds

For compounds, the molar mass is calculated by summing the atomic masses of all atoms in the chemical formula.

• Example: Water (H2O): Molar mass = $2 \times 1.01 + 16.00 = 18.02$ g/mol

• Conversion: To find the mass of 1.75 mol of water: $1.75 \text{ mol} \times 18.02 \text{ g/mol} = 31.5 \text{ g}$

Key Point: Always use the correct chemical formula to determine molar mass.

Converting Between Mass of a Compound and Number of Molecules

To find the mass from the number of molecules, convert molecules to moles, then moles to grams.

• Step 1: Number of molecules to moles: $\text{moles} = \frac{\text{number of molecules}}{6.022 \times 10^{23}}$

• Step 2: Moles to grams: $\text{mass} = \text{moles} \times \text{molar mass}$

• Example: For $4.78 \times 10^{23}$ NO2 molecules (molar mass = 46.01 g/mol): $\frac{4.78 \times 10^{23}}{6.022 \times 10^{23}} \times 46.01 = 36.5$ g NO2$

Chemical Formulas as Conversion Factors

Chemical formulas indicate the ratio of elements in a compound, which can be used to convert between moles of compound and moles of constituent elements.

• Example: CaCO3 contains 3 moles of O per mole of CaCO3.

• Conversion: $1.7 \text{ mol CaCO}_3 \times \frac{3 \text{ mol O}}{1 \text{ mol CaCO}_3} = 5.1 \text{ mol O}$

Key Point: Use subscripts in the chemical formula as exact conversion factors.

Converting Between Grams of a Compound and Grams of a Constituent Element

To find the mass of an element in a compound, convert grams of compound to moles of compound, then to moles of element, and finally to grams of element.

• Example: Carvone (C10H14O): Find mass of C in 55.4 g carvone. Molar mass carvone = 150.2 g/mol; 10 mol C per mol carvone; molar mass C = 12.01 g/mol. $\frac{55.4}{150.2} \times 10 \times 12.01 = 44.3$ g C$

Key Point: The mass of a constituent element is always less than the mass of the compound.

Using Mass Percent Composition as a Conversion Factor

Mass percent composition allows conversion between the mass of an element and the mass of the compound.

• Formula: $\text{mass percent} = \frac{\text{mass of element in 1 mol compound}}{\text{mass of 1 mol compound}} \times 100\%$

• Example: Sodium chloride is 39% sodium by mass. To find the mass of NaCl containing 2.3 g Na: $\frac{2.3 \text{ g Na}}{39 \text{ g Na}} \times 100 \text{ g NaCl} = 5.9 \text{ g NaCl}$

Determining Mass Percent Composition from a Chemical Formula

Calculate the mass percent of an element in a compound using the chemical formula and atomic masses.

• Example: For K2O: Molar mass K2O = $2 \times 39.10 + 16.00 = 94.20$ g/mol Mass % K = $\frac{2 \times 39.10}{94.20} \times 100\% = 83.1\%$

Empirical and Molecular Formulas from Experimental Data

Empirical formulas show the simplest whole-number ratio of elements in a compound. Molecular formulas show the actual number of atoms in a molecule.

• Steps to Determine Empirical Formula:

  1. Obtain masses (or mass percent) of each element.

  2. Convert masses to moles using atomic masses.

  3. Write a pseudoformula using mole values as subscripts.

  4. Divide all subscripts by the smallest value to get whole numbers.

  5. If necessary, multiply by a small integer to obtain whole-number subscripts.

• Example: A compound contains 24.5 g N and 70.0 g O. Moles N: $\frac{24.5}{14.01} = 1.75$ Moles O: $\frac{70.0}{16.00} = 4.38$ Ratio: $\frac{1.75}{1.75} : \frac{4.38}{1.75} = 1 : 2.5$ Multiply by 2: $2 : 5$ Empirical formula: N2O5

• Determining Molecular Formula:

  1. Calculate empirical formula molar mass.

  2. Divide the compound's molar mass by the empirical formula mass to get n.

  3. Multiply subscripts in empirical formula by n.

• Example: Naphthalene has empirical formula C5H4, molar mass 128.16 g/mol. Empirical formula mass: $5 \times 12.01 + 4 \times 1.01 = 64.09$ g/mol $n = \frac{128.16}{64.09} = 2$ Molecular formula: C10H8

Summary Table: Key Conversion Relationships

Conversion	Factor/Formula	Example
Atoms → Moles	$\text{moles} = \frac{\text{atoms}}{6.022 \times 10^{23}}$	$1.1 \times 10^{22}$ Ag atoms → $1.8 \times 10^{-2}$ mol Ag
Grams → Moles	$\text{moles} = \frac{\text{grams}}{\text{molar mass}}$	57.8 g S → 1.80 mol S
Moles → Atoms	$\text{atoms} = \text{moles} \times 6.022 \times 10^{23}$	4.8 mol Cu → $2.9 \times 10^{24}$ Cu atoms
Compound → Element (moles)	Use subscripts from formula	1.7 mol CaCO3 → 5.1 mol O
Mass Percent	$\text{mass percent} = \frac{\text{mass of element}}{\text{mass of compound}} \times 100\%$	K2O: 83.1% K

Additional info: These notes are based on textbook examples and practice problems from "Introductory Chemistry" by Nivaldo J. Tro, 7th Edition, Chapter 6. All conversion factors and relationships are standard in introductory chemistry courses.