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Stoichiometry And Stoichiometric Calculations

Percentage Composition

Percentage composition is a quantitative measure of the composition of a chemical compound. It expresses the proportion of each element in the compound as a percentage by mass.

Percentage Composition of a Compound

Definition: The percentage composition of a compound lists the percentage by mass of each element present in the compound. It is calculated by dividing the mass of the element in one mole of the compound by the molar mass of the compound and multiplying by 100.

Formula:

$$\text{Percentage of an element} = \frac{\text{Mass of the element in one mole of the compound}}{\text{Molar mass of the compound}} \times 100\%$$

Calculation Steps:

  1. Determine the molar mass of the compound.
  2. For each element in the compound, determine the total mass contributed by that element (atomic mass $\times$ number of atoms of that element in the formula).
  3. Divide the total mass of each element by the molar mass of the compound and multiply by 100 to get the percentage composition of that element.

Example: Calculate the percentage composition of water ($H_2O$).

Example 1. Calculate the percentage composition of water ($H_2O$).

Answer:

Step 1: Calculate the molar mass of $H_2O$.

Atomic mass of H $\approx$ 1.0 g/mol

Atomic mass of O $\approx$ 16.0 g/mol

Molar mass of $H_2O$ = (2 $\times$ 1.0) + 16.0 = 18.0 g/mol

Step 2: Calculate the percentage of Hydrogen (H).

Mass of Hydrogen in one mole of $H_2O$ = 2 $\times$ 1.0 g = 2.0 g

Percentage of H = $\frac{2.0 \text{ g}}{18.0 \text{ g/mol}} \times 100\% = 11.11\%$

Step 3: Calculate the percentage of Oxygen (O).

Mass of Oxygen in one mole of $H_2O$ = 1 $\times$ 16.0 g = 16.0 g

Percentage of O = $\frac{16.0 \text{ g}}{18.0 \text{ g/mol}} \times 100\% = 88.89\%$

Check: 11.11% + 88.89% = 100%

Thus, water is approximately 11.11% hydrogen and 88.89% oxygen by mass.

Empirical Formula vs. Molecular Formula

Molecular Formula: The molecular formula of a compound indicates the actual number of atoms of each element in one molecule of the compound. For example, the molecular formula of glucose is $C_6H_{12}O_6$.

Empirical Formula: The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present in the compound. For example, the empirical formula of glucose ($C_6H_{12}O_6$) is $CH_2O$, because the ratio 6:12:6 simplifies to 1:2:1.

Relationship Between Empirical and Molecular Formulas:

The molecular formula is always a whole-number multiple of the empirical formula.

Molecular Formula = (Empirical Formula)$_n$

Where $n$ is a positive integer. This integer can be found by:

$$n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}$$

Determining Empirical Formula from Percentage Composition:

This is a crucial skill in chemistry, often used when analyzing an unknown compound.

Steps:

  1. Assume a 100g sample: If given percentage composition, assume a 100g sample. This converts percentages directly into grams (e.g., 11.11% H becomes 11.11g H).
  2. Convert grams to moles: Use the molar mass of each element to convert the mass of each element into moles.
  3. Find the simplest mole ratio: Divide the number of moles of each element by the smallest number of moles calculated. This gives the simplest whole-number ratio of atoms.
  4. Write the empirical formula: Use these simplest whole-number ratios as subscripts in the chemical formula. If the ratios are not whole numbers, multiply them by a small integer (usually 2, 3, or 5) to obtain whole numbers.

Example: A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass. Determine its empirical formula.

Example 2. A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass. Determine its empirical formula.

Answer:

Step 1: Assume a 100g sample. This gives us:

  • Mass of C = 40.0 g
  • Mass of H = 6.7 g
  • Mass of O = 53.3 g

Step 2: Convert grams to moles (using approximate atomic masses: C=12.0 g/mol, H=1.0 g/mol, O=16.0 g/mol).

  • Moles of C = $\frac{40.0 \text{ g}}{12.0 \text{ g/mol}} = 3.33$ mol
  • Moles of H = $\frac{6.7 \text{ g}}{1.0 \text{ g/mol}} = 6.7$ mol
  • Moles of O = $\frac{53.3 \text{ g}}{16.0 \text{ g/mol}} = 3.33$ mol

Step 3: Find the simplest mole ratio.

The smallest number of moles is 3.33 mol (for both C and O).

  • Ratio of C = $\frac{3.33 \text{ mol}}{3.33 \text{ mol}} = 1$
  • Ratio of H = $\frac{6.7 \text{ mol}}{3.33 \text{ mol}} \approx 2$
  • Ratio of O = $\frac{3.33 \text{ mol}}{3.33 \text{ mol}} = 1$

Step 4: Write the empirical formula.

The simplest whole-number ratio of C:H:O is 1:2:1. Therefore, the empirical formula is $CH_2O$.

Determining Molecular Formula from Empirical Formula and Molecular Mass:

Steps:

  1. Determine the empirical formula using the percentage composition or other analytical data.
  2. Calculate the empirical formula mass.
  3. Divide the given molecular mass by the empirical formula mass to find the integer 'n'.
  4. Multiply the subscripts in the empirical formula by 'n' to get the molecular formula.

Example: The empirical formula of a compound is $CH_2O$, and its molecular mass is 180 g/mol. Determine its molecular formula.

Example 3. The empirical formula of a compound is $CH_2O$, and its molecular mass is 180 g/mol. Determine its molecular formula.

Answer:

Step 1: Empirical formula is $CH_2O$.

Step 2: Calculate the empirical formula mass.

Empirical formula mass of $CH_2O$ = (1 $\times$ 12.0) + (2 $\times$ 1.0) + (1 $\times$ 16.0) = 12.0 + 2.0 + 16.0 = 30.0 g/mol

Step 3: Find the integer 'n'.

Molecular mass = 180 g/mol

$n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{180 \text{ g/mol}}{30.0 \text{ g/mol}} = 6$

Step 4: Determine the molecular formula.

Molecular Formula = (Empirical Formula)$_n$ = ($CH_2O$)$_6$ = $C_6H_{12}O_6$

Thus, the molecular formula of the compound is glucose ($C_6H_{12}O_6$).


Stoichiometry And Stoichiometric Calculations

Stoichiometry: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It is based on the law of conservation of mass and the law of definite proportions.

Stoichiometric Calculations: These calculations allow us to predict the amounts of reactants needed or products formed in a chemical reaction, given the amount of one reactant or product. These calculations typically involve:

  1. Balancing the chemical equation.
  2. Converting the given amount of substance (usually mass) to moles.
  3. Using the mole ratios from the balanced equation to find the moles of the desired substance.
  4. Converting moles of the desired substance back to mass (or other units).

Limiting Reagent

Definition: In a chemical reaction, reactants are often mixed in proportions that are not exactly stoichiometric. The limiting reagent (or limiting reactant) is the reactant that is completely consumed first in a chemical reaction. It determines the maximum amount of product that can be formed.

Excess Reagent: The reactant(s) that are not completely consumed in a chemical reaction are called excess reagents. Some amount of these reactants will remain unreacted after the reaction is complete.

Determining the Limiting Reagent:

To identify the limiting reagent, follow these steps:

  1. Ensure the chemical equation is balanced.
  2. Convert the given masses of each reactant into moles.
  3. For each reactant, calculate the amount of product that would be formed if that reactant were completely consumed. This is done by using the mole ratio between the reactant and the product from the balanced equation.
  4. The reactant that produces the smallest amount of product is the limiting reagent.

Example: Consider the reaction between nitrogen ($N_2$) and hydrogen ($H_2$) to form ammonia ($NH_3$):

$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$

If 28 g of $N_2$ and 9 g of $H_2$ are mixed, identify the limiting reagent and calculate the maximum amount of $NH_3$ that can be formed.

Example 1. Consider the reaction: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. If 28 g of $N_2$ and 9 g of $H_2$ are mixed, identify the limiting reagent and calculate the maximum amount of $NH_3$ that can be formed.

Answer:

Step 1: Balanced equation is already provided: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

Step 2: Convert masses to moles.

  • Molar mass of $N_2$ = 2 $\times$ 14.0 g/mol = 28.0 g/mol
  • Molar mass of $H_2$ = 2 $\times$ 1.0 g/mol = 2.0 g/mol
  • Moles of $N_2 = \frac{28 \text{ g}}{28.0 \text{ g/mol}} = 1.0$ mol
  • Moles of $H_2 = \frac{9 \text{ g}}{2.0 \text{ g/mol}} = 4.5$ mol

Step 3: Calculate the amount of $NH_3$ produced from each reactant.

  • From $N_2$: From the balanced equation, 1 mole of $N_2$ produces 2 moles of $NH_3$.

    • Moles of $NH_3$ from $N_2$ = 1.0 mol $N_2 \times \frac{2 \text{ mol } NH_3}{1 \text{ mol } N_2} = 2.0$ mol $NH_3$

  • From $H_2$: From the balanced equation, 3 moles of $H_2$ produce 2 moles of $NH_3$.

    • Moles of $NH_3$ from $H_2$ = 4.5 mol $H_2 \times \frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2} = 3.0$ mol $NH_3$

Step 4: Identify the limiting reagent and maximum product.

Since $N_2$ produces fewer moles of $NH_3$ (2.0 mol vs 3.0 mol), Nitrogen ($N_2$) is the limiting reagent.

The maximum amount of $NH_3$ that can be formed is 2.0 moles.

Step 5: Convert moles of $NH_3$ to mass.

Molar mass of $NH_3$ = 14.0 + (3 $\times$ 1.0) = 17.0 g/mol

Mass of $NH_3$ = 2.0 mol $\times$ 17.0 g/mol = 34.0 g

Therefore, the maximum amount of ammonia that can be formed is 34.0 grams.

Reactions in Solutions

Many chemical reactions occur in aqueous solutions. To perform stoichiometric calculations for these reactions, we need to understand concentrations and how to relate them to the amount of solute in moles.

Concentration: Concentration describes the amount of solute dissolved in a given amount of solvent or solution.

Molarity (M): Molarity is the most common unit of concentration in chemistry. It is defined as the number of moles of solute per liter of solution.

$$ \text{Molarity (M)} = \frac{\text{Moles of solute (mol)}}{\text{Volume of solution (L)}} $$

Calculations involving Molarity:

  1. Finding moles from Molarity and Volume:
    2. $\text{Moles of solute} = \text{Molarity} \times \text{Volume of solution (L)}$
  2. Using stoichiometry with solutions:

    3. a. Convert the given volume and molarity of a reactant/product to moles.

    b. Use the mole ratio from the balanced chemical equation to find moles of other reactants/products.

    c. Convert moles back to desired units (e.g., volume, mass, or molarity).

Example: How many grams of calcium carbonate ($CaCO_3$) will react with 500 mL of a 0.5 M hydrochloric acid (HCl) solution?

The reaction is: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$

Example 2. How many grams of calcium carbonate ($CaCO_3$) will react with 500 mL of a 0.5 M hydrochloric acid (HCl) solution? The reaction is: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$

Answer:

Step 1: Balanced chemical equation is given.

Step 2: Calculate moles of HCl from the given solution.

  • Volume of HCl solution = 500 mL = 0.500 L
  • Molarity of HCl solution = 0.5 M = 0.5 mol/L
  • Moles of HCl = Molarity $\times$ Volume
  • Moles of HCl = 0.5 mol/L $\times$ 0.500 L = 0.25 mol

Step 3: Use the mole ratio to find moles of $CaCO_3$.

From the balanced equation, 2 moles of HCl react with 1 mole of $CaCO_3$.

Moles of $CaCO_3$ = 0.25 mol HCl $\times \frac{1 \text{ mol } CaCO_3}{2 \text{ mol } HCl} = 0.125$ mol $CaCO_3$

Step 4: Convert moles of $CaCO_3$ to grams.

Molar mass of $CaCO_3$ = 40.1 (Ca) + 12.0 (C) + (3 $\times$ 16.0) (O) = 40.1 + 12.0 + 48.0 = 100.1 g/mol

Mass of $CaCO_3$ = Moles $\times$ Molar mass

Mass of $CaCO_3$ = 0.125 mol $\times$ 100.1 g/mol = 12.5125 g

Therefore, approximately 12.5 grams of calcium carbonate will react with 500 mL of 0.5 M HCl solution.