Chemistry: Week 3 Pt 1 - Moles, Balancing Equations and Stoichiometry

Balancing Chemical Equations

• Balancing chemical equations is a skill that often involves trial and error.
• Coefficients in balanced equations represent the recipe for chemical reactions, indicating the required amounts of reactants for a desired amount of product.
• Coefficients can be used to calculate the amounts of reactants needed or products formed, applicable in healthcare for dosage calculations.

Terminology

• Reactants: Substances that react together, located on the left side of the arrow in a chemical equation.
• Products: Substances produced as a result of the chemical reaction, located on the right side of the arrow.
• Chemical Formula:
  • Subscripts indicate the number of atoms of an element in the formula (e.g., $N_2$ has two nitrogen atoms).
  • Coefficients indicate the number of molecules (e.g., $2NH_3$ means two molecules of ammonia).
• State of Matter:
  • (g) - gas
  • (l) - liquid
  • (s) - solid
  • (aq) - aqueous (dissolved in liquid)

Reading Chemical Equations

• Equations can be read in terms of moles: "One mole of $N<em>2$ reacts with three moles of $H</em>2$ to yield two moles of $NH_3$".

Understanding Coefficients Visually

• $2NH<em>3$ means there are two molecules of $NH</em>3$.
• To find the number of atoms of each element, multiply the coefficient by the subscript in the formula:
  • For $2NH_3$, there are $2 \times 1 = 2$ nitrogen atoms and $2 \times 3 = 6$ hydrogen atoms.

Practice

• $4H<em>2$ represents four molecules of $H</em>2$.
• It contains $4 \times 2 = 8$ atoms of hydrogen.

Law of Conservation of Mass

• Matter is neither created nor destroyed in a chemical reaction; atoms rearrange and rebond.
• Balancing equations involves using coefficients to ensure the number of atoms of each element is the same on both sides of the equation.

Example

• In the balanced equation, every atom is accounted for on both sides, demonstrating the conservation of mass.

More Practice

• Coefficients and subscripts distribute to all atoms within a molecule or polyatomic ion; for example, $2(OH)$ means each element is multiplied by two.

Balancing Unbalanced Chemical Equations

1. Count the atoms of each element on both sides of the equation.
2. Treat polyatomic ions as a single unit if they appear on both sides of the equation.
3. Add coefficients to balance the number of atoms, multiplying through the entire molecule.
4. Simplify coefficients to the lowest terms and double-check the answer.

Example

• Balancing $Al<em>2(SO</em>4)<em>3 + CaO \rightarrow Al</em>2O<em>3 + CaSO</em>4$
  • Start with the most complex part, such as sulfates.
  • Add coefficients to balance sulfates, then balance other elements accordingly.

Example 2

• Balancing $FeCl<em>3 + MgO \rightarrow Fe</em>2O<em>3 + MgCl</em>2$
  • Balance chlorine first by using coefficients that are multiples of each other.
  • Adjust other coefficients to balance all elements in the equation.

Key Points for Balancing Equations

• Follow the law of conservation of mass.
• States of matter (gas, liquid, solid, aqueous) do not affect the balancing process.
• Reactants are on the left side of the arrow, and products are on the right.
• Coefficients represent the number of molecules and should be adjusted to balance the equation without changing the chemical formulas.

Real-World Relevance

• Balanced equations are seen in central respiration (glucose + oxygen -> carbon dioxide + water), essential for energy production in our bodies.

Stoichiometry

• Stoichiometry uses coefficients from balanced chemical equations to create mole-to-mole ratios for further calculations.
• Chemical equations are like chemical recipes; coefficients dictate the proportions of reactants and products.

Avogadro's Number and the Mole

• A mole is a chemist's way of counting a large number of tiny things (atoms, molecules, compounds).
• Avogadro's number is $6.022 \times 10^{23}$, representing the number of items in one mole.
• One mole of any substance contains $6.022 \times 10^{23}$ units of that substance.

Molar Mass

• The atomic mass from the periodic table (grams per mole) is used as the atomic weight.
• The molar mass (also known as molecular weight, formula weight, molar weight, formula mass) is calculated by adding up the atomic masses of each atom in the chemical formula.

Example

• Calculating the molar mass of $FeCl_2$
  • Iron (Fe): 1 atom $\times$ 55.845 g/mol = 55.845 g/mol
  • Chlorine (Cl): 2 atoms $\times$ 35.453 g/mol = 70.906 g/mol
  • Molar mass of $FeCl_2$ = 55.845 + 70.906 = 126.751 g/mol

Converting Between Grams and Moles

• Use the molecular mass as a conversion factor, as its units are grams per mole.

Practice

• How many moles of $O<em>2$ are present in a 92-gram sample of $O</em>2$?
  • Molecular mass of $O_2$ = 32 g/mol
  • $92 \text{ grams } O<em>2 \times \frac{1 \text{ mole } O</em>2}{32 \text{ grams } O<em>2} = 2.9 \text{ moles } O</em>2$

Triangle Method

• A method to rearrange equations involving molar weight, grams, and moles:
  • Molecular weight (MW) = Grams / Moles
  • Grams = MW x Moles
  • Moles = Grams / MW

Example

• Given a 2.5 mole sample of NaCl with a molecular weight of 58.44 g/mol, find the mass in grams.
  • Mass = 58.44 g/mol $\times$ 2.5 moles = 146.1 grams

Practice

• How many moles are present in a 75-gram sample of $CaCO_3$ (molecular weight = 100.087 g/mol)?
  • Moles = 75 grams / 100.087 g/mol = 0.749 moles

Using Chemical Equations as Recipes

• Coefficients in chemical equations are used to determine the ratios of reactants and products needed.

Stoichiometry Problem

• Balanced chemical equation is key
• The numbers, or coefficients, represent moles.
• Mole to mole ratio; use the coefficients to change one compound to another

Example Calculation:

• How many moles of $Na<em>3N$ will be produced if the reaction starts with 0.7 moles of $NaBr$ given the balanced equation: $Mg</em>3N<em>2 + 6NaBr \rightarrow 3MgBr</em>2 + 2Na_3N$
  1. List what's given and what formula to follow.
  2. Given $NaBr$ = 6.
  3. Want $Na_3N$ = 2.
  4. Set up the dimensional analysis equation. Start with what you're given!

$\frac{0.7 \text{ moles }}{1} \times \frac{2 \text{ moles }Na<em>3N}{6 \text{ moles }NaBr} = .233\text{ moles }Na</em>3N$
Remember to cancel same units diagonally.

Key Points

• A mole is a unit for counting atoms, molecules, and compounds.
• Molar mass (grams/mole) is used to convert between grams and moles.
• Coefficients from balanced equations give mole ratios for stoichiometry.
• Dimensional analysis is essential for stoichiometry to cancel units in mol-to-mol ratios.