Calculate Amount of Substance, Mass, Molar Mass

Introduction

Key Concepts

Amount of Substance

The **amount of substance** is a measure of the number of specified elementary entities (such as atoms, molecules, ions, etc.) present in a given sample. In the International System of Units (SI), the unit of amount of substance is the mole (symbol: mol). One mole contains exactly $6.022 \times 10^{23}$ elementary entities, a number known as Avogadro's number.

The formula to calculate the amount of substance is:

However, in practical chemistry, we often deal with moles rather than individual entities. For example, 1 mole of carbon atoms contains $6.022 \times 10^{23}$ carbon atoms.

Mass

**Mass** is a measure of the amount of matter in an object or substance and is typically measured in grams (g) in chemistry. The mass of a substance can be calculated if the amount of substance and the molar mass are known.

The relationship between mass, amount of substance, and molar mass is given by the equation:

This equation allows chemists to determine how much of a substance is needed or produced in a chemical reaction.

Molar Mass

**Molar mass** is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). It is numerically equivalent to the average atomic mass of the element in grams. For compounds, the molar mass is calculated by summing the molar masses of all the atoms in the molecular formula.

For example, the molar mass of water ($\mathrm{H}_2\mathrm{O}$) is calculated as follows:

Calculating Amount of Substance

To calculate the amount of substance when mass and molar mass are known, rearrange the mass formula:

**Example:** Calculate the amount of substance in 36.04 grams of water.

Given:

• Mass (m) = 36.04 g
• Molar mass (M) of $\mathrm{H}_2\mathrm{O}$ = 18.02 g/mol

Calculation:

Therefore, 36.04 grams of water is equivalent to 2 moles of water.

Calculating Mass

To find the mass when the amount of substance and molar mass are known, use the mass formula:

**Example:** Calculate the mass of 3 moles of carbon dioxide ($\mathrm{CO}_2$).

Given:

• Amount of substance (n) = 3 mol
• Molar mass (M) of $\mathrm{CO}_2$ = 44.01 g/mol

Calculation:

Thus, 3 moles of carbon dioxide have a mass of 132.03 grams.

Calculating Molar Mass

The molar mass of a compound is calculated by adding the molar masses of all the atoms in its molecular formula.

**Example:** Calculate the molar mass of sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$).

Given:

• Hydrogen (H) = 1.01 g/mol
• Sulfur (S) = 32.07 g/mol
• Oxygen (O) = 16.00 g/mol

Calculation:

Therefore, the molar mass of sulfuric acid is 98.09 grams per mole.

Significant Figures and Precision

When performing calculations involving the amount of substance, mass, and molar mass, it's crucial to maintain the correct number of significant figures to ensure precision. The final answer should reflect the least number of significant figures in the given data.

**Example:** If 5.00 grams of a substance are used, and the molar mass is 24.305 g/mol, the amount of substance should be reported with three significant figures:

Mole Concept in Chemical Reactions

The mole concept is integral to balancing chemical equations and performing stoichiometric calculations. It allows chemists to quantify the relationships between reactants and products in a reaction.

For a balanced chemical equation:

- $a$, $b$, $c$, and $d$ are the stoichiometric coefficients. - The mole ratio between reactants and products is derived from these coefficients.

**Example:** In the reaction $\mathrm{2H}_2 + \mathrm{O}_2 \rightarrow \mathrm{2H}_2\mathrm{O}$, the mole ratio of hydrogen to oxygen is 2:1.

This ratio is used to calculate how much of each reactant is needed or how much product will be formed.

Applications of the Mole Concept

The mole concept is applied in various aspects of chemistry, including:

• Stoichiometry: Calculating the quantities of reactants and products in chemical reactions.
• Gas Laws: Relating the amount of gas to volume, pressure, and temperature.
• Solution Concentration: Determining molarity, which is moles of solute per liter of solution.
• Empirical and Molecular Formulas: Deriving the simplest and actual ratios of elements in compounds.

Examples and Practice Problems

To solidify understanding, let's explore several examples and practice problems.

1. Example 1: Calculate the amount of substance in 50 grams of sodium chloride (NaCl).
   Given:
   Molar mass of NaCl = 58.44 g/mol

   Solution:

   $$ n = \frac{50 \, \text{g}}{58.44 \, \text{g/mol}} \approx 0.855 \, \text{mol} $$

2. Example 2: Determine the mass of 0.5 moles of glucose ($\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$).
   Given:
   Molar mass of glucose = 180.16 g/mol

   Solution:

   $$ m = 0.5 \, \text{mol} \times 180.16 \, \text{g/mol} = 90.08 \, \text{g} $$

3. Example 3: Find the molar mass of calcium carbonate ($\mathrm{CaCO}_3$).
   Given:
   Calcium (Ca) = 40.08 g/mol, Carbon (C) = 12.01 g/mol, Oxygen (O) = 16.00 g/mol

   Solution:

   $$ \text{Molar mass of } \mathrm{CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol} $$

Advanced Concepts

In-depth Theoretical Explanations

Delving deeper into the mole concept involves understanding its historical development and its connection to other fundamental principles in chemistry. The mole was introduced to bridge the gap between the atomic scale and the macroscopic quantities we observe.

**Avogadro's Law** states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. This law is foundational in defining the mole and establishing the relationship between atomic mass units and grams.

Mathematically, Avogadro's number ($N_A$) connects the macroscopic scale to the atomic scale:

Furthermore, the mole bridges the gap between mass and energy through the concept of molar energy. This is particularly important in thermodynamics and kinetic studies.

Complex Problem-Solving

Advanced stoichiometric problems often involve limiting reagents, percent yield, and multi-step reactions. These problems require a comprehensive understanding of the mole concept and its application in various scenarios.

Problem 1: When 10 grams of hydrogen react with 80 grams of oxygen to form water, determine the limiting reagent and the theoretical yield of water.

Solution:

• Write the balanced equation: $\mathrm{2H}_2 + \mathrm{O}_2 \rightarrow \mathrm{2H}_2\mathrm{O}$
• Calculate the moles of each reactant:
  • Molar mass of $\mathrm{H}_2$ = 2.02 g/mol
  • Molar mass of $\mathrm{O}_2$ = 32.00 g/mol
  • Moles of $\mathrm{H}_2$ = $10 \, \text{g} / 2.02 \, \text{g/mol} \approx 4.95 \, \text{mol}$
  • Moles of $\mathrm{O}_2$ = $80 \, \text{g} / 32.00 \, \text{g/mol} = 2.5 \, \text{mol}$
• Determine the limiting reagent using the mole ratio:
  • From the equation, 2 mol $\mathrm{H}_2$ reacts with 1 mol $\mathrm{O}_2$
  • Moles of $\mathrm{O}_2$ required for 4.95 mol $\mathrm{H}_2$ = $4.95 / 2 = 2.475 \, \text{mol}$
  • Available $\mathrm{O}_2$ = 2.5 mol
  • Since 2.475 mol $\mathrm{O}_2$ are needed and 2.5 mol are available, $\mathrm{H}_2$ is the limiting reagent.
• Calculate the theoretical yield of $\mathrm{H}_2\mathrm{O}$:
  • Mole ratio of $\mathrm{H}_2$ to $\mathrm{H}_2\mathrm{O}$ is 2:2 or 1:1
  • Moles of $\mathrm{H}_2\mathrm{O}$ = 4.95 mol
  • Molar mass of $\mathrm{H}_2\mathrm{O}$ = 18.02 g/mol
  • Mass of $\mathrm{H}_2\mathrm{O}$ = $4.95 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 89.1 \, \text{g}$

Problem 2: A 25.0 mL solution of hydrochloric acid (HCl) has a concentration of 0.100 M. How many grams of HCl are present in the solution?

Solution:

• First, calculate moles of HCl: $$ n = M \times V = 0.100 \, \text{mol/L} \times 0.0250 \, \text{L} = 0.00250 \, \text{mol} $$
• Calculate mass using molar mass of HCl (36.46 g/mol): $$ m = 0.00250 \, \text{mol} \times 36.46 \, \text{g/mol} = 0.09115 \, \text{g} $$
• Therefore, approximately 0.091 grams of HCl are present.

Mathematical Derivations and Proofs

Understanding the derivation of the mole concept from Avogadro's hypothesis provides deeper insight into its significance. Avogadro's hypothesis asserts that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. This principle forms the basis for the mole and allows the mole to serve as a bridge between the atomic scale and macroscopic measurements.

The derivation starts by considering one mole of an ideal gas at standard temperature and pressure (STP: 0°C and 1 atm), which occupies 22.414 liters. Using the ideal gas law:

Where:

• P = pressure (1 atm)
• V = volume (22.414 L)
• n = amount of substance (1 mol)
• R = ideal gas constant (0.0821 L.atm/mol.K)
• T = temperature (273.15 K)

Substituting the values:

Solving confirms that 1 mole corresponds to $6.022 \times 10^{23}$ entities, thus validating Avogadro's number.

Interdisciplinary Connections

The mole concept is not confined to chemistry alone; it plays a critical role in various scientific and engineering disciplines.

• Biology: Quantifying molecules such as DNA, proteins, and enzymes in biological systems.
• Pharmacology: Determining dosages based on molecular interactions within the body.
• Environmental Science: Measuring pollutant concentrations in ecosystems.
• Material Science: Calculating the composition of alloys and composite materials.
• Physics: Linking macroscopic properties of materials to their molecular structures.

Moreover, in industrial chemistry, the mole concept is vital for scaling reactions from the laboratory to production levels, ensuring efficiency and cost-effectiveness.

Advanced Applications

Advanced applications of the mole concept include:

• Thermodynamics: Calculating energy changes in reactions based on mole quantities.
• Quantum Chemistry: Relating mole quantities to particle behavior at the quantum level.
• Stoichiometric Calculations in Reactions: Optimizing reactant use to maximize product yield.
• Electrochemistry: Determining the amount of substance involved in redox reactions.

Limitations of the Mole Concept

While the mole concept is a powerful tool, it has limitations:

• Assumption of Ideal Behavior: The concept assumes ideal conditions, which may not hold in real-world scenarios.
• Complex Stoichiometry: Reactions with multiple steps or catalysts can complicate mole-based calculations.
• Intermolecular Forces: These can affect the behavior of substances, making precise calculations challenging.

Case Studies

**Case Study 1:** Determining Fuel Requirements for Combustion
In automotive engineering, calculating the amount of fuel required for complete combustion involves stoichiometric calculations using the mole concept. This ensures engines run efficiently without excess fuel or oxygen.

**Case Study 2:** Pharmaceutical Formulation
Pharmaceutical companies use the mole concept to determine the correct proportions of active ingredients and excipients, ensuring the efficacy and safety of medications.

Comparison Table

Summary and Key Takeaways