5 Chemical Reactions and Stoichiometry

Reading chemical formulas

Chemical formulas show which elements make up a substance and how many atoms or ions of each are represented. For example, $\mathrm{H_2O}$ contains two hydrogen atoms and one oxygen atom per molecule. A missing subscript means one. Parentheses group atoms or ions, so $\mathrm{Ca(OH)_2}$ contains one calcium atom, two oxygen atoms, and two hydrogen atoms. For an ionic compound such as $\mathrm{NaCl}$, the formula represents the simplest whole-number ratio of ions, called a formula unit.

A Subscript is part of a substance’s identity: $\mathrm{H_2O}$ and $\mathrm{H_2O_2}$ are different substances. A Coefficient placed before a formula counts how many of that substance are involved. For instance, $\mathrm{2H_2O}$ represents two water molecules or, in a mole-based calculation, two moles of water.

Keep the distinction clear: subscripts describe composition, while coefficients describe relative amounts.

Moles and molar mass

The Mole is the SI unit for amount of substance. One mole contains exactly $6.02214076 \times 10^{23}$ specified entities. Always identify what is being counted: one mole of $\mathrm{O_2}$ molecules is not the same particle count as one mole of oxygen atoms.

Molar mass is the mass of one mole, commonly expressed in $\mathrm{g\,mol^{-1}}$. To find it, add the atomic molar masses of every atom in a formula, including the atoms indicated by subscripts and parentheses. For example:

• $\mathrm{H_2O}$: $2(1.008) + 16.00 = 18.016\,\mathrm{g\,mol^{-1}}$

• $\mathrm{NaCl}$: $22.99 + 35.45 = 58.44\,\mathrm{g\,mol^{-1}}$

For a pure sample, convert between mass and amount using:

Here, $n$ is the amount in moles, $m$ is the mass in grams, and $M$ is the molar mass. The rearranged relationship is $m = nM$. To convert an amount in moles to a particle count, use $N = nN_{\mathrm{A}}$, where $N_{\mathrm{A}} = 6.02214076 \times 10^{23}\,\mathrm{mol^{-1}}$.

Example: A $36.0\,\mathrm{g}$ water sample contains:

Molar mass links measurable mass to the amount used in reaction calculations.

Writing and balancing equations

A chemical equation places reactants to the left of the arrow and products to the right. A balanced chemical equation has the same number of atoms of each element on both sides. For ionic equations, total charge must also be conserved.

To balance an equation:

1. Write the correct formulas for all reactants and products.

2. Count each kind of atom on each side.

3. Adjust coefficients only until the counts match.

4. Reduce the coefficients to the smallest whole-number ratio and check the counts again.

For propane combustion, start with:

Balance carbon and hydrogen, then oxygen:

Each side now has three carbon atoms, eight hydrogen atoms, and ten oxygen atoms. The coefficients express the relative ratio $1:5:3:4$. They do not give mass ratios. Never change subscripts to balance an equation, because that would change the substances themselves.

The key check is conservation: each element—and, for ionic equations, charge—must balance across the reaction arrow.

Recognizing reaction patterns

Reaction categories describe common patterns. They are useful for recognizing reactions, but one reaction may fit more than one category.

• Synthesis (combination): simpler substances form a more complex product. Example: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$.

• Decomposition: one substance breaks into simpler substances. Example: $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$.

• Single replacement: one element replaces another in a compound. Example: $\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2}$.

• Double replacement: ions in two compounds exchange partners. Example: $\mathrm{AgNO_3 + NaCl \rightarrow AgCl + NaNO_3}$.

• Combustion: a fuel reacts with oxygen; a hydrocarbon often produces carbon dioxide and water. Example: $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$.

In aqueous solutions, a double-replacement reaction may form a precipitate, water, or a gas. A redox reaction involves electron transfer and is identified by changes in oxidation states; combustion is a common example.

Recognizing a reaction pattern helps describe what changes, while balancing still ensures atoms and charge are conserved.

Stoichiometry and limiting reactants

Coefficients in a balanced equation give the mole ratios used to relate reactants and products. For example:

This equation relates $2\,\mathrm{mol}$ of hydrogen to $1\,\mathrm{mol}$ of oxygen and $2\,\mathrm{mol}$ of water. A Mole ratio comes directly from these coefficients.

Use this sequence to find a quantity of one substance from another:

1. Convert the given quantity to moles, if needed.

2. Use the balanced equation’s mole ratio to find moles of the target substance.

3. Convert those moles to the requested quantity, if needed.

If converting between mass and moles, use molar mass. Track units so that units you do not need cancel. Use only coefficients from the balanced equation.

Example: Find the mass of water that can form from $4.00\,\mathrm{g}$ of $\mathrm{H_2}$, assuming oxygen is in excess.

1. Convert hydrogen to moles:

   $$\frac{4.00\,\mathrm{g\ H_2}}{2.016\,\mathrm{g\,mol^{-1}}} = 1.984\,\mathrm{mol\ H_2}$$

2. Apply the equation’s $2:2$ hydrogen-to-water mole ratio:

   $$1.984\,\mathrm{mol\ H_2} \times \frac{2\,\mathrm{mol\ H_2O}}{2\,\mathrm{mol\ H_2}} = 1.984\,\mathrm{mol\ H_2O}$$

3. Convert water to mass:

   $$1.984\,\mathrm{mol\ H_2O} \times 18.016\,\mathrm{g\,mol^{-1}} = 35.7\,\mathrm{g\ H_2O}$$

When amounts of multiple reactants are given, calculate the product amount each reactant could form. The Limiting reactant is the one that produces the least product; it is consumed first and sets the maximum product amount. The other reactants are in excess.

The central strategy is to convert through moles, using the balanced equation to connect substances and molar mass to connect mass and amount.