Exam Prep: Balancing Equations, Dilutions, and Reaction Types

Molar Mass and Empirical Mass

• The molar mass of a compound cannot be determined without experimental data, typically obtained through mass spectrometry ($\text{mass spec}$).

• If mass spectrometry data is unavailable, one must rely on empirical mass, which is derived from the simplest whole-number ratio of atoms in a compound.

Dilution Calculations

• Context: Laboratories often dilute highly concentrated "stock solutions" of acids and bases (e.g., $12.1$ Molar HCl) to more usable concentrations (e.g., $0.1$ to $0.2$ Molar) for experiments.

• Dilution Formula: The principle of dilution is based on the idea that the total number of moles of solute remains constant before and after dilution. Therefore, the product of molarity and volume for the initial (concentrated) solution equals that for the final (diluted) solution. ${M<em>{\text{concentrated}}V</em>{\text{concentrated}}} = {M<em>{\text{diluted}}V</em>{\text{diluted}}}$ or ${M<em>1V</em>1} = {M<em>2V</em>2}$

  • $M_1$: Molarity of the stock (concentrated) solution.

  • $V_1$: Volume of the stock solution needed.

  • $M_2$: Molarity of the desired (diluted) solution.

  • $V_2$: Volume of the desired (diluted) solution.

• Units: It is not necessary to convert units like milliliters (mL) to liters (L) within the formula, as long as the volume units are consistent on both sides of the equation. If $V<em>1$ is in mL, then $V</em>2$ will be in mL, and vice-versa.

• Significance: The product of molarity and volume ($M \times V$) directly yields the number of moles of a substance, provided the volume is in liters ($L$). This formula is incredibly important and will be used repeatedly in acid-base titrations and other quantitative chemistry applications.

Example 1: Diluting HCl

• Question: How much $12.1$ Molar HCl is needed to make $150$ mL of $0.2$ Molar HCl?

• Given:

  • $M_1 = 12.1 \; \text{M}$ (what we have)

  • $V_1 = \text{unknown}$ (what we need from the concentrated stock)

  • $M_2 = 0.2 \; \text{M}$ (what we want)

  • $V_2 = 150 \; \text{mL}$ (what we want)

• Calculation:
  ${(12.1 \; \text{M})} \cdot {V<em>1} = {(0.2 \; \text{M})} \cdot {(150 \; \text{mL})}$ ${12.1 V</em>1} = {30 \; \text{mL}}$
  ${V<em>1} = \frac{30 \; \text{mL}}{12.1}$ ${V</em>1} \approx 2.48 \; \text{mL}$

Example 2: Diluting $NaNO_3$

• Question: What volume of a $6$ Molar solution of $NaNO<em>3$ is needed to make $0.525$ Liters of $1.20$ Molar $NaNO</em>3$?

• Given:

  • $M_1 = 6 \; \text{M}$

  • $V_1 = \text{unknown}$

  • $M_2 = 1.20 \; \text{M}$

  • $V_2 = 0.525 \; \text{L}$

• Calculation:
  ${(6 \; \text{M})} \cdot {V<em>1} = {(1.20 \; \text{M})} \cdot {(0.525 \; \text{L})}$ ${6 V</em>1} = {0.63 \; \text{L}}$
  ${V<em>1} = \frac{0.63 \; \text{L}}{6}$ {V1} = 0.105 \; \text{L}}

• Note on Units: Since the desired volume was in liters, the answer is in liters. This can be converted to milliliters ($105 \; \text{mL}$) if required by the question, but the crucial point is consistency within the calculation.

Balancing Chemical Reactions

• Law of Conservation of Mass: Matter cannot be created or destroyed. This means that in any chemical reaction, the total number of atoms of each element must be the same on both the reactant side and the product side.

  • Reactant Side: The starting materials of a chemical reaction.

  • Product Side: The substances formed during a chemical reaction.

• Stoichiometric Coefficients: These are the numbers placed in front of chemical formulas in a balanced equation. They represent the relative number of moles (or molecules) of each substance involved in the reaction. These coefficients act like a recipe, indicating the proportions of reactants needed and products formed.

• Process (Trial and Error with Hints):

  1. Count the number of atoms of each element on both sides of the unbalanced equation.

  2. Add coefficients to balance the number of atoms. You cannot change the subscripts within a chemical formula (e.g., changing $H<em>2O$ to $H</em>2O_2$ changes the substance from water to hydrogen peroxide).

  3. Helpful Hint: If an element appears by itself (e.g., $O<em>2$, $N</em>2$, $H_2$), save it for the last step in balancing.

Example 1: Formation of Water

• Unbalanced: ${H<em>2} + {O</em>2} \rightarrow {H_2O}$

  • Reactants: $H=2, O=2$

  • Products: $H=2, O=1$

• Balancing Steps:

  1. Oxygen is unbalanced. To balance the $O$ atoms, place a coefficient of $2$ in front of $H<em>2O$: ${H</em>2} + {O<em>2} \rightarrow {2H</em>2O}$

  2. Now recount:

     • Reactants: $H=2, O=2$

     • Products: $H=4, O=2$

  3. Hydrogen is now unbalanced. Place a coefficient of $2$ in front of $H<em>2$: ${2H</em>2} + {O<em>2} \rightarrow {2H</em>2O}$

• Balanced Equation: ${2H<em>2(g)} + {O</em>2(g)} \rightarrow {2H_2O(g)}$

  • Read as: Two moles of hydrogen gas plus one mole of oxygen gas yield two moles of steam.

Example 2: Haber Synthesis of Ammonia

• Unbalanced: ${H<em>2} + {N</em>2} \rightarrow {NH_3}$

  • Reactants: $H=2, N=2$

  • Products: $H=3, N=1$

• Balancing Steps:

  1. Nitrogen is unbalanced. Place a coefficient of $2$ in front of $NH<em>3$: ${H</em>2} + {N<em>2} \rightarrow {2NH</em>3}$

  2. Now recount:

     • Reactants: $H=2, N=2$

     • Products: $H=6, N=2$

  3. Hydrogen is now unbalanced. Place a coefficient of $3$ in front of $H<em>2$: ${3H</em>2} + {N<em>2} \rightarrow {2NH</em>3}$

• Balanced Equation: ${3H<em>2(g)} + {N</em>2(g)} \rightarrow {2NH_3(g)}$

  • This is a famous reaction, known as the Haber synthesis, used to produce ammonia for fertilizers and explosives.

  • External Factors: Chemical reactions can be affected by conditions such as temperature (e.g., $180 \; ^\circ\text{C}$) or catalysts (e.g., platinum, $Pt$), which may be indicated above or below the reaction arrow.

Types of Chemical Reactions: Combustion Reactions

• Definition: A combustion reaction is essentially a burning process. It always involves a substance reacting rapidly with oxygen, usually producing heat and light.

• Identification for Exams: The most important characteristic to identify a combustion reaction is the presence of $O_2$ as a reactant (usually with a carbon-containing compound).

• General Form:

  • Reactants: Carbon-containing compound (hydrocarbon) $+ O_2$

  • Products: Usually ${CO<em>2} + {H</em>2O}$

  • Additional Products: If the reactant contains nitrogen, $NO<em>2$ might be formed. If it contains sulfur, $SO</em>2$ might be formed.

• Balancing Hint: When balancing combustion reactions, always leave the oxygen ($O_2$) to be balanced last, as it is often an element by itself.

Example 1: Combustion of Ethene ($C<em>2H</em>4$)

• Unbalanced (Speaker starts with $C<em>2H</em>2$ but balances for $C<em>2H</em>4$): ${C<em>2H</em>4} + {O<em>2} \rightarrow {CO</em>2} + {H_2O}$

  • Balancing Steps:

    1. Balance Carbon: There are $2$ carbons on the left, so put $2$ in front of $CO<em>2$: ${C</em>2H<em>4} + {O</em>2} \rightarrow {2CO<em>2} + {H</em>2O}$

    2. Balance Hydrogen: There are $4$ hydrogens on the left, so put $2$ in front of $H<em>2O$: ${C</em>2H<em>4} + {O</em>2} \rightarrow {2CO<em>2} + {2H</em>2O}$

    3. Balance Oxygen (last): Count oxygen on the product side:

       • $2 imes O \text{ in } 2CO_2 = 4 \; O$ atoms

       • $2 imes O \text{ in } 2H_2O = 2 \; O$ atoms

       • Total oxygen atoms on product side = $4+2 = 6$

       • To get $6$ oxygen atoms on the reactant side from $O<em>2$, place a $3$ in front of $O</em>2$
         ${C<em>2H</em>4} + {3O<em>2} \rightarrow {2CO</em>2} + {2H_2O}$

• Balanced Equation: ${C<em>2H</em>4(g)} + {3O<em>2(g)} \rightarrow {2CO</em>2(g)} + {2H_2O(g)}$

Example 2: Combustion of Aniline ($C<em>6H</em>5NH_2$) - Complex Example

• Unbalanced: ${C<em>6H</em>5NH<em>2(l)} + {O</em>2(g)} \rightarrow {CO<em>2(g)} + {H</em>2O(l)} + {NO_2(g)}$

  • Balancing steps (with fractions, then clear them):

    1. Balance Carbon: $6$ carbons on left, so $6CO_2$

    2. Balance Nitrogen: $1$ nitrogen on left, so $1NO_2$

    3. Balance Hydrogen: $5+2 = 7$ hydrogens on left. To get $7$ on the right, use $\frac{7}{2}H<em>2O$ ${C</em>6H<em>5NH</em>2} + {O<em>2} \rightarrow {6CO</em>2} + {\frac{7}{2}H<em>2O} + {NO</em>2}$

    4. Balance Oxygen (last): Count oxygen on the product side:

       • $6 \times 2 \; O \text{ in } 6CO_2 = 12 \; O$

       • $\frac{7}{2} \times 1 \; O \text{ in } \frac{7}{2}H_2O = \frac{7}{2} \; O$

       • $1 \times 2 \; O \text{ in } NO_2 = 2 \; O$

       • Total oxygen: $12 + \frac{7}{2} + 2 = 14 + \frac{7}{2} = \frac{28}{2} + \frac{7}{2} = \frac{35}{2}$

       • To balance $O<em>2$ on the reactant side, we need $\frac{35}{2}$ oxygen atoms, meaning $\frac{35}{4}O</em>2$
         ${C<em>6H</em>5NH<em>2} + {\frac{35}{4}O</em>2} \rightarrow {6CO<em>2} + {\frac{7}{2}H</em>2O} + {NO_2}$

    5. Clear Fractions: Multiply all coefficients by $4$ to get whole numbers.
       ${4C<em>6H</em>5NH<em>2} + {35O</em>2} \rightarrow {24CO<em>2} + {14H</em>2O} + {4NO_2}$

  • Note: While such complex balancing problems can occur, students will not typically encounter ones requiring this many steps or fractions on exams. The purpose here is to demonstrate the method.

Precipitation Reactions and Solubility Rules

• Solubility: The ability of an ionic compound to dissolve in water.

• Dissociation: When a soluble ionic compound dissolves, it breaks apart (dissociates) into its individual ions. For example, solid $NaCl_{(s)}$ in water becomes separate $Na^+(aq)$ and $Cl^-(aq)$ ions. There is no $NaCl$ molecule in solution.

• Precipitation Reaction: A reaction that results in the formation of an insoluble solid (a precipitate) when two solutions are mixed.

• Identification for Exams: Precipitation reactions are always identified by the formation of a solid product in the chemical equation.

• Crucial for Exams: You must memorize the solubility rules, as they will not be provided on the test.

Solubility Rules (Must Know)

1. Always Soluble (Few Exceptions for Advanced Study):

   • All Group 1 salts (compounds containing $Li^+, Na^+, K^+, Rb^+, Cs^+$; excluding hydrogen for ionic compounds).

   • All salts containing nitrate (NO_3^-$).

   • All acetate salts (CH_3COO^-$).

   • All ammonium salts ($NH_4^+$).

2. Soluble, with Exceptions:

   • Chlorides (Cl^-$), Bromides (Br^-$), Iodides (I^-$): Soluble except when paired with silver (Ag^+$), lead ($Pb^{2+}$), or mercury(I) ($Hg_2^{2+}$).</p></li><li><p><strong>Sulfates ($SO_4^{2-}$): Soluble except when paired with calcium ($Ca^{2+}$), strontium ($Sr^{2+}$), or barium ($Ba^{2+}$) (Group 2 large ions).

3. Insoluble, with Exceptions:

   • Phosphates (PO4^{3-}$), Carbonates (CO3^{2-}$), Hydroxides (OH^-$), Sulfides (S^{2-}$), Oxides (O^{2-}$): Generally insoluble, unless they are paired with a Group 1 substance or ammonium (NH_4^+$). (Note: Calcium hydroxide is sparingly soluble).</p></li></ul></li></ol><h5 id="f7ed16bd-89ab-4290-8470-b2ab37494c39" data-toc-id="f7ed16bd-89ab-4290-8470-b2ab37494c39" collapsed="false" seolevelmigrated="true">Steps to Predict a Precipitation Reaction</h5><ol><li><p><strong>Identify Reactants</strong>: Write the chemical formulas of the reactants and their states (usually aqueous,$(aq) =$$ dissolved in water).

   • Dissociate Soluble Reactants: Based on solubility rules, show all ions that are present in solution from soluble reactants.

   • Exchange Partners: Determine the possible products by