The Mole, Molar Mass, and Stoichiometry in Chemical Reactions

7.4 The Mole

Counting Units in Chemistry

In chemistry, counting terms such as dozen, gross, and ream are used to specify exact quantities of items. The mole is the standard counting unit for atoms, molecules, and ions, analogous to these everyday terms.

• Dozen: 12 items

• Gross: 144 items

• Ream: 500 sheets (commonly used for paper)

• Mole: 6.02 × 1023 items (Avogadro's number)

Avogadro’s Number

The mole is defined as the amount of substance that contains exactly $6.02 \times 10^{23}$ elementary entities (atoms, molecules, or ions). This value is known as Avogadro’s number, named after Amedeo Avogadro.

• 1 mole = $6.02 \times 10^{23}$ particles

• Used to count extremely small particles such as atoms and molecules

Mole of Atoms

One mole of any element contains $6.02 \times 10^{23}$ atoms of that element. For example:

• 1 mole of carbon = $6.02 \times 10^{23}$ carbon atoms

• 1 mole of sodium = $6.02 \times 10^{23}$ sodium atoms

• 1 mole of sulfur = $6.02 \times 10^{23}$ sulfur atoms

Conversion Factor: Avogadro’s Number

Avogadro’s number can be used as a conversion factor between moles and particles:

• Equality: $1\ \text{mole} = 6.02 \times 10^{23}\ \text{particles}$

• Conversion factors:

  • $\frac{1\ \text{mole}}{6.02 \times 10^{23}\ \text{particles}}$

  • $\frac{6.02 \times 10^{23}\ \text{particles}}{1\ \text{mole}}$

Converting Moles to Molecules

To find the number of molecules in a given amount of moles, multiply by Avogadro’s number. For example, to find the number of CO2 molecules in 0.500 mole:

• $0.500\ \text{mole} \times 6.02 \times 10^{23}\ \text{molecules/mole} = 3.01 \times 10^{23}\ \text{molecules}$

Moles of Elements in a Chemical Compound

Subscripts and Mole Ratios

Subscripts in a chemical formula indicate the number of moles of each element in one mole of the compound. For example, in aspirin (C9H8O4):

• 1 mole of C9H8O4 contains:

  • 9 moles of C

  • 8 moles of H

  • 4 moles of O

7.5 Molar Mass

Definition of Molar Mass

The molar mass of an element or compound is the mass in grams of one mole of that substance. For elements, the molar mass (in g/mol) is numerically equal to the atomic mass (in amu) found on the periodic table.

• Example: 1 mole of sodium (Na) has a molar mass of 22.99 g.

Determining Molar Mass from the Periodic Table

The molar mass of a compound is calculated by summing the molar masses of its constituent elements, each multiplied by its subscript in the formula.

• Example: For Li2CO3 (lithium carbonate):

  • 2 moles Li × 6.941 g/mol = 13.88 g

  • 1 mole C × 12.01 g/mol = 12.01 g

  • 3 moles O × 16.00 g/mol = 48.00 g

  • Total molar mass = 13.88 + 12.01 + 48.00 = 73.89 g/mol

Molar Mass Conversion Factors

Molar mass is used to convert between grams and moles:

• Equality: 1 mole CH4 = 16.05 g CH4

• Conversion factors:

  • $\frac{1\ \text{mole CH}_4}{16.05\ \text{g CH}_4}$

  • $\frac{16.05\ \text{g CH}_4}{1\ \text{mole CH}_4}$

7.6 Calculations Using Molar Mass

Converting Mass to Moles

To determine the number of moles in a given mass of a substance, divide the mass by the molar mass:

• Example: A box of NaCl (table salt) contains 737 g. Moles of NaCl = $\frac{737\ \text{g}}{58.44\ \text{g/mol}} = 12.6\ \text{moles}$

Mass–Moles–Particles Map

This flowchart summarizes the relationships between mass, moles, and particles for both elements and compounds:

• Grams ↔ Moles (via molar mass)

• Moles ↔ Particles (via Avogadro’s number)

• Formula subscripts relate moles of elements to moles of compounds

7.7 Mole Relationships in Chemical Equations

Law of Conservation of Mass

In a chemical reaction, the total mass of reactants equals the total mass of products. Matter is neither created nor destroyed.

• Example: $2\text{Fe}(s) + 3\text{S}(s) \rightarrow \text{Fe}_2\text{S}_3(s)$

Mole–Mole Factors from an Equation

Balanced chemical equations provide the ratios (mole–mole factors) for converting between moles of reactants and products.

• For $2\text{Fe} + 3\text{S} \rightarrow \text{Fe}_2\text{S}_3$:

  • 2 moles Fe : 3 moles S : 1 mole Fe2S3

  • Mole–mole factors can be written as fractions for calculations.

Calculations with Mole Factors

To determine the amount of product formed or reactant needed, use the mole–mole factors from the balanced equation. For example:

• Given 6.0 moles O2 in $4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3$:

• $6.0\ \text{moles O}_2 \times \frac{2\ \text{moles Fe}_2\text{O}_3}{3\ \text{moles O}_2} = 4.0\ \text{moles Fe}_2\text{O}_3$

7.8 Mass Calculations for Chemical Reactions

Mass Relationships in Reactions

To calculate the mass of a reactant or product in a chemical reaction:

1. Convert mass of given substance to moles (using molar mass).

2. Use mole–mole factor from the balanced equation to find moles of desired substance.

3. Convert moles of desired substance to mass (using molar mass).

7.9 Limiting Reactants and Percent Yield

Limiting and Excess Reactants

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, thus limiting the amount of product formed. The excess reactant is the substance that remains after the reaction is complete.

• To identify the limiting reactant, calculate the amount of product formed from each reactant; the one producing the least product is limiting.

Calculating Product from Limiting Reactant

When quantities of reactants are given in grams:

1. Convert grams of each reactant to moles.

2. Use mole–mole factors to determine moles of product from each reactant.

3. The reactant that produces the least amount of product is limiting.

4. Convert moles of product to grams if required.

Theoretical, Actual, and Percent Yield

Not all reactions go to completion. The theoretical yield is the maximum amount of product predicted by stoichiometry. The actual yield is the amount actually obtained. Percent yield is calculated as:

• $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Example: If 50.0 g of LiOH produces 72.8 g of LiHCO3, calculate the percent yield using the above formula.