Chapter 4: Compounds and Stoichiometry

Every compound has a chemical formula, but not every formula tells you the same story. There are two types you need to know cold for the MCAT, and confusing them is one of the most common mistakes on test day.

Empirical Formula - The Simplest Ratio

The empirical formula gives you the smallest whole-number ratio of atoms in a compound. It tells you the relative proportions, but not the actual count.

For example, glucose has the empirical formula CH₂O. This tells you that for every carbon atom, there are two hydrogen atoms and one oxygen atom. But it does not tell you how many of each atom are actually in one molecule of glucose.

Molecular Formula - The Actual Count

The molecular formula tells you the exact number of each type of atom in one molecule. Glucose’s molecular formula is C₆H₁₂O₆ - six carbons, twelve hydrogens, six oxygens. That is 6 times the empirical formula (6 x CH₂O).

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

Molecular formula = n x (Empirical formula)

where n = molecular weight / empirical formula weight.

How They Connect

Notice that glucose, acetic acid, and formaldehyde all share the same empirical formula. The empirical formula alone cannot distinguish between them - you need the molecular weight to determine n and therefore the molecular formula.

Determining the Empirical Formula from Percent Composition

This is a classic MCAT problem type. Here is the step-by-step process:

1. Assume 100 g of the compound (so percentages become grams directly)
2. Convert grams to moles for each element (divide by atomic mass)
3. Divide all mole values by the smallest mole value
4. Round to the nearest whole number (if you get values like 1.5 or 2.33, multiply all values by 2 or 3 to clear the fraction)

Example: A compound is 40.0% C, 6.7% H, and 53.3% O by mass. Find its empirical formula.

• C: 40.0 g / 12.0 g/mol = 3.33 mol
• H: 6.7 g / 1.0 g/mol = 6.7 mol
• O: 53.3 g / 16.0 g/mol = 3.33 mol

Divide by smallest (3.33): C = 1, H = 2, O = 1

Empirical formula: CH₂O

Determining the Molecular Formula

Once you have the empirical formula and the molecular weight (from mass spectrometry or other data), calculate n:

n = molecular weight / empirical formula weight

For the example above, if the molecular weight is 180 g/mol:

• Empirical formula weight of CH₂O = 12 + 2(1) + 16 = 30 g/mol
• n = 180 / 30 = 6
• Molecular formula: C₆H₁₂O₆ (glucose)

Naming compounds on the MCAT is not about memorizing thousands of names. It is about learning three sets of rules - one for ionic compounds, one for covalent compounds, and one for acids - and applying them consistently.

Ionic Compounds - Metal + Nonmetal

Ionic compounds form when a metal transfers electrons to a nonmetal. The naming rules are straightforward:

1. Name the cation (metal) first, then the anion (nonmetal) second
2. The anion gets the suffix -ide
3. If the metal can form more than one charge (transition metals), use Roman numerals in parentheses to indicate the charge

Polyatomic Ions - Know the Pattern

You need to recognize common polyatomic ions. Rather than memorizing each one independently, learn the patterns:

The -ate/-ite pattern: For oxyanions (polyatomic ions with oxygen), the -ate form has more oxygen atoms, and the -ite form has fewer:

Extended pattern for halogens: When there are four possible oxyanions, add per- (most O) and hypo- (least O):

Other essential polyatomic ions:

Covalent Compounds - Nonmetal + Nonmetal

Covalent compounds are named using Greek prefixes to indicate the number of each atom:

Rules:

• The first element gets a prefix only if there is more than one atom (never say “monocarbon”)
• The second element always gets a prefix AND the -ide suffix

Naming Acids

Acid naming follows a predictable pattern based on the anion:

Binary acids (no oxygen, H + nonmetal):

• Use the prefix hydro- and the suffix -ic acid
• HCl = hydrochloric acid, HBr = hydrobromic acid, HF = hydrofluoric acid

Oxyacids (H + polyatomic ion with oxygen):

• If the ion ends in -ate, the acid ends in -ic acid
• If the ion ends in -ite, the acid ends in -ous acid

Atoms are impossibly small. A single carbon atom has a mass of about $2 \times 10^{-23}$ grams. You cannot weigh one atom on any balance in any lab. But you can weigh $6.022 \times 10^{23}$ of them - and that clump happens to weigh exactly 12.01 grams, a number you can read right off the periodic table.

That number - $6.022 \times 10^{23}$ - is Avogadro’s number (Nₐ), and it defines the mole.

The Mole - Chemistry’s Counting Word

A “dozen” means 12. A “mole” means $6.022 \times 10^{23}$. That is all it is - a counting word for an absurdly large number.

One mole of any element contains exactly $6.022 \times 10^{23}$ atoms. One mole of any compound contains $6.022 \times 10^{23}$ molecules (or formula units for ionic compounds).

Molar Mass - The Bridge Between Grams and Moles

The molar mass (also called molecular weight or formula weight) is the mass of one mole of a substance, expressed in grams per mole (g/mol). For elements, it is the atomic mass from the periodic table. For compounds, it is the sum of all the atomic masses in the formula.

Examples:

• Molar mass of O₂ = 2(16.00) = 32.00 g/mol
• Molar mass of H₂O = 2(1.01) + 16.00 = 18.02 g/mol
• Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol
• Molar mass of CaCO₃ = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

The Mole Map - Your Navigation Tool

Every stoichiometry problem boils down to moving between three quantities: grams, moles, and number of particles. The mole sits at the center of this map.

The key insight: you must always pass through moles. You cannot go directly from grams of one substance to grams of another. The route is always grams → moles → ratio → moles → grams.

Working with Avogadro’s Number

Avogadro’s number connects the macroscopic (grams) to the microscopic (atoms, molecules). Here is how it works in practice:

Example: How many water molecules are in 36.04 g of water?

1. Convert to moles: 36.04 g / 18.02 g/mol = 2.00 mol
2. Convert to molecules: $2.00 \text{ mol} \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$ molecules

Example: How many oxygen atoms are in 36.04 g of water?

Each water molecule (H₂O) contains 1 oxygen atom. So the number of oxygen atoms equals the number of molecules: $1.204 \times 10^{24}$.

But each molecule also has 2 hydrogen atoms: $2 \times 1.204 \times 10^{24} = 2.409 \times 10^{24}$ hydrogen atoms.

Molar Volume at STP

For gases at standard temperature and pressure (STP: 0°C, 1 atm), one mole of any ideal gas occupies 22.4 liters. This gives you a third conversion pathway:

• Liters (at STP) → divide by 22.4 → moles
• Moles → multiply by 22.4 → liters (at STP)

This only applies to gases at STP. At other conditions, use the ideal gas law (PV = nRT) from Chapter 8.

Percent composition tells you what fraction of a compound’s mass comes from each element. It is the chemical equivalent of reading a nutrition label - instead of “20% protein, 50% carbohydrate,” you get “40% carbon, 6.7% hydrogen, 53.3% oxygen.”

Calculating Percent Composition

The formula is simple:

Example: Find the percent composition of water (H₂O).

• Molar mass of H₂O = 2(1.01) + 16.00 = 18.02 g/mol
• % H = [2(1.01) / 18.02] x 100 = 11.2%
• % O = [16.00 / 18.02] x 100 = 88.8%
• Check: 11.2 + 88.8 = 100.0% ✓

Working Backward - From Percent Composition to Empirical Formula

This is the reverse problem, and the MCAT loves it. Given percent composition data, determine the empirical formula:

1. Assume 100 g (percentages become grams)
2. Convert each to moles (divide by atomic mass)
3. Divide by the smallest mole value
4. Multiply to clear fractions if needed

Example: A compound is 26.7% C, 2.2% H, and 71.1% O.

• C: 26.7 g / 12.01 g/mol = 2.22 mol
• H: 2.2 g / 1.008 g/mol = 2.18 mol
• O: 71.1 g / 16.00 g/mol = 4.44 mol

Divide by smallest (2.18): C = 1.02, H = 1.00, O = 2.04

Round: C₁H₁O₂ → Empirical formula: CHO₂

Combustion Analysis - The MCAT’s Favorite Method

Combustion analysis is an experimental technique for determining the molecular formula of an unknown organic compound. You burn a sample in excess oxygen and measure the masses of CO₂ and H₂O produced.

The logic is beautifully simple:

• All the carbon in the original compound ends up in CO₂
• All the hydrogen in the original compound ends up in H₂O
• Any remaining mass is oxygen (if the compound contains oxygen)

Combustion Analysis Step-by-Step

Example: 0.255 g of an unknown compound containing only C, H, and O is burned. The combustion produces 0.561 g CO₂ and 0.306 g H₂O. The molecular weight is 180 g/mol. Find the molecular formula.

Step 1: Find moles of C from CO₂.

• Moles of CO₂ = 0.561 g / 44.01 g/mol = 0.01275 mol
• Moles of C = 0.01275 mol (1:1 ratio in CO₂)
• Mass of C = 0.01275 mol x 12.01 g/mol = 0.153 g

Step 2: Find moles of H from H₂O.

• Moles of H₂O = 0.306 g / 18.02 g/mol = 0.01698 mol
• Moles of H = 2 x 0.01698 = 0.03396 mol (2 H per H₂O)
• Mass of H = 0.03396 mol x 1.008 g/mol = 0.0342 g

Step 3: Find mass and moles of O by difference.

• Mass of O = 0.255 - 0.153 - 0.0342 = 0.0678 g
• Moles of O = 0.0678 g / 16.00 g/mol = 0.00424 mol

Step 4: Find the mole ratio.

• Divide by smallest (0.00424): C = 3.01, H = 8.01, O = 1.00
• Empirical formula: C₃H₈O

Step 5: Find n.

• Empirical formula weight = 3(12) + 8(1) + 16 = 60 g/mol
• n = 180 / 60 = 3
• Molecular formula: C₉H₂₄O₃

A chemical equation is a sentence written in the language of chemistry. Reactants go on the left, products on the right, and an arrow separates them. But the equation is only valid if it is balanced - meaning the same number of each type of atom appears on both sides.

This is not optional. It is the law of conservation of mass: atoms are neither created nor destroyed in a chemical reaction. What goes in must come out, just rearranged.

Conventions for Writing Chemical Equations

Before balancing, you need to know the notation:

Example: 2 H₂(g) + O₂(g) → 2 H₂O(l)

This tells you: two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of liquid water.

Stoichiometric Coefficients

The numbers in front of each formula are coefficients. They tell you the mole ratio - the recipe for the reaction.

In the equation above:

• 2 mol H₂ : 1 mol O₂ : 2 mol H₂O
• This ratio is fixed. You always need twice as many moles of H₂ as O₂.

How to Balance Equations - A Systematic Approach

There is no magic formula. But this systematic approach works reliably:

Step 1: Write the unbalanced equation with correct formulas.

Step 2: Balance elements that appear in only one reactant and one product first.

Step 3: Balance hydrogen and oxygen last (they often appear in multiple compounds).

Step 4: Use fractions if needed, then multiply everything by the denominator to clear them.

Step 5: Check your work - count every atom on both sides.

Worked Example

Balance: C₃H₈ + O₂ → CO₂ + H₂O (combustion of propane)

1. Start with carbon: 3 C on the left → put a 3 in front of CO₂

   • C₃H₈ + O₂ → 3 CO₂ + H₂O

2. Balance hydrogen: 8 H on the left → put a 4 in front of H₂O

   • C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

3. Balance oxygen: right side has 3(2) + 4(1) = 10 O atoms → put 5 in front of O₂

   • C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

4. Check: C: 3 = 3 ✓, H: 8 = 8 ✓, O: 10 = 10 ✓

Common Pitfalls

Never change subscripts to balance an equation. Subscripts define the compound. Changing H₂O to H₂O₂ does not “balance” anything - it changes water into hydrogen peroxide, a completely different substance. Only change coefficients.

Coefficients apply to the entire formula. A coefficient of 2 in front of Ca(OH)₂ means 2 calcium atoms, 4 oxygen atoms, and 4 hydrogen atoms - the coefficient multiplies every subscript in the formula.

Balancing Redox Equations

For redox reactions, balancing can be trickier because you need to balance both mass and charge. The half-reaction method is the most reliable approach (covered in detail in Chapter 11). For now, know the key idea: split the reaction into an oxidation half-reaction and a reduction half-reaction, balance each separately, then combine them so the electrons cancel.

Chemical reactions can be classified into a handful of categories based on the pattern of how reactants rearrange into products. Recognizing these patterns lets you predict products without memorizing every individual reaction.

Combination (Synthesis) Reactions

Two or more simple substances combine to form one more complex substance.

Pattern: A + B → AB

Examples:

• 2 Na + Cl₂ → 2 NaCl (metal + nonmetal → ionic compound)
• 2 Mg + O₂ → 2 MgO
• SO₃ + H₂O → H₂SO₄ (nonmetal oxide + water → acid)
• CaO + H₂O → Ca(OH)₂ (metal oxide + water → base)

Decomposition Reactions

One complex substance breaks down into two or more simpler substances. This is the reverse of combination.

Pattern: AB → A + B

Examples:

• 2 H₂O₂ → 2 H₂O + O₂ (hydrogen peroxide decomposes)
• CaCO₃ → CaO + CO₂ (limestone heated)
• 2 KClO₃ → 2 KCl + 3 O₂ (potassium chlorate decomposes)

Single Displacement (Replacement) Reactions

One element replaces another element in a compound. The displaced element is “kicked out” because the incoming element is more reactive.

Pattern: A + BC → AC + B

Examples:

• Zn + CuSO₄ → ZnSO₄ + Cu (zinc displaces copper)
• Fe + CuSO₄ → FeSO₄ + Cu (iron displaces copper)
• Cl₂ + 2 NaBr → 2 NaCl + Br₂ (chlorine displaces bromine)

The activity series determines whether a displacement reaction will occur. A more active metal displaces a less active metal from solution. If the incoming element is less active than the element in the compound, no reaction occurs.

Double Displacement (Metathesis) Reactions

Two compounds exchange partners. The cations trade anions.

Pattern: AB + CD → AD + CB

Examples:

• AgNO₃ + NaCl → AgCl(s) + NaNO₃ (precipitation)
• HCl + NaOH → NaCl + H₂O (neutralization)
• Na₂CO₃ + 2 HCl → 2 NaCl + H₂O + CO₂ (gas evolution)

Double displacement reactions are driven by the formation of:

1. A precipitate (insoluble solid)
2. Water (neutralization)
3. A gas (that escapes the solution)

If none of these forms, no net reaction occurs.

Combustion Reactions

A substance reacts with oxygen, typically producing heat and light. For organic compounds (containing C and H), the products are always CO₂ and H₂O.

Pattern: Hydrocarbon + O₂ → CO₂ + H₂O

Examples:

• CH₄ + 2 O₂ → CO₂ + 2 H₂O (methane combustion)
• 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O (ethane combustion)
• C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O (propane combustion)

If the compound also contains oxygen (like ethanol, C₂H₅OH), the products are still CO₂ and H₂O - the oxygen in the compound just reduces the amount of O₂ needed.

Disproportionation Reactions

In a disproportionation reaction, a single element is simultaneously oxidized and reduced. The same species acts as both the oxidizing agent and the reducing agent.

Example:

• 2 H₂O₂ → 2 H₂O + O₂

In hydrogen peroxide (H₂O₂), oxygen has an oxidation state of -1. In the products, oxygen in H₂O is -2 (reduced) and oxygen in O₂ is 0 (oxidized). The same element (-1 oxygen) both gained and lost electrons.

Another example:

• Cl₂ + 2 NaOH → NaCl + NaClO + H₂O

Chlorine starts at 0 and ends up as -1 (in NaCl, reduced) and +1 (in NaClO, oxidized).

Summary Table

When you are making sandwiches and you have 10 slices of bread but only 3 slices of cheese, the cheese limits your output. You can only make 3 sandwiches. The extra bread sits unused. That is the limiting reagent concept - and it is the single most tested stoichiometry topic on the MCAT.

The Limiting Reagent Determines the Product

The limiting reagent (or limiting reactant) is the reactant that runs out first. It determines the maximum amount of product that can form.

The excess reagent is whatever is left over after the limiting reagent is completely consumed.

How to Find the Limiting Reagent

There are two reliable methods. Use whichever feels more natural:

Method 1: Compare mole ratios

1. Convert all reactant amounts to moles
2. Divide each by its coefficient in the balanced equation
3. The reactant with the smallest result is the limiting reagent

Method 2: Calculate product from each reactant

1. Assume each reactant is limiting, one at a time
2. Calculate the moles of product each would produce
3. The reactant that produces the least product is the limiting reagent

Worked Example

Problem: Given 10.0 g of N₂ and 10.0 g of H₂, how many grams of NH₃ can be produced?

Balanced equation: N₂ + 3 H₂ → 2 NH₃

Step 1: Convert to moles.

• N₂: 10.0 g / 28.0 g/mol = 0.357 mol
• H₂: 10.0 g / 2.02 g/mol = 4.95 mol

Step 2: Divide by coefficients.

• N₂: 0.357 / 1 = 0.357
• H₂: 4.95 / 3 = 1.65

Step 3: Identify limiting reagent.

• N₂ gives the smaller value (0.357 < 1.65), so N₂ is limiting

Step 4: Calculate product from limiting reagent.

• Mole ratio: 1 mol N₂ → 2 mol NH₃
• 0.357 mol N₂ x (2 mol NH₃ / 1 mol N₂) = 0.714 mol NH₃
• Mass: 0.714 mol x 17.0 g/mol = 12.1 g NH₃

Theoretical Yield

The theoretical yield is the maximum amount of product that can form based on the limiting reagent, assuming the reaction goes to completion with no losses. It is the calculated “perfect” amount.

In the example above, the theoretical yield of NH₃ is 12.1 g. In practice, you would get less than this due to side reactions, incomplete reactions, and mechanical losses during purification.

How Much Excess Reagent Remains?

After the limiting reagent is fully consumed, calculate how much of the excess reagent was actually used, then subtract from the starting amount.

From the example above:

• N₂ used all 0.357 mol
• H₂ needed: 0.357 mol N₂ x (3 mol H₂ / 1 mol N₂) = 1.071 mol H₂
• H₂ remaining: 4.95 - 1.071 = 3.88 mol H₂ = 7.83 g H₂ left over

The BCA Table

A BCA table (Before, Change, After) is a systematic way to organize limiting reagent problems. It works identically to an ICE table (which you will use in equilibrium) but for reactions that go to completion.

The limiting reagent is the one that reaches zero in the “After” row.

In a perfect world, every reaction would produce exactly the amount of product predicted by stoichiometry. In reality, they never do. Some product sticks to the glassware. Some reactant undergoes a side reaction. Some product decomposes before you can collect it. The gap between what you should get and what you actually get is measured by percent yield.

Three Types of Yield

Why Percent Yield is Always Less Than 100%

Several factors reduce actual yield:

1. Incomplete reactions - not all reactant converts to product (equilibrium is reached before completion)
2. Side reactions - reactants form unwanted byproducts
3. Mechanical losses - product left on glassware, lost during filtration or transfer
4. Purification losses - product lost during recrystallization, distillation, or chromatography
5. Decomposition - some product breaks down before collection

Worked Example

Problem: A student reacts 5.00 g of zinc with excess hydrochloric acid. The balanced equation is:

Zn + 2 HCl → ZnCl₂ + H₂

The student collects 0.125 g of H₂. What is the percent yield?

Step 1: Calculate theoretical yield.

• Moles of Zn = 5.00 g / 65.38 g/mol = 0.0765 mol
• HCl is in excess, so Zn is the limiting reagent
• Mole ratio: 1 mol Zn → 1 mol H₂
• Theoretical moles of H₂ = 0.0765 mol
• Theoretical yield = 0.0765 mol x 2.016 g/mol = 0.154 g

Step 2: Calculate percent yield.

• % yield = (0.125 g / 0.154 g) x 100% = 81.2%

Working Backward from Percent Yield

The MCAT might give you a percent yield and ask you to find how much reactant you need to produce a desired amount of product.

Example: You need 50.0 g of product. The reaction has a 75% yield. How many grams of theoretical yield do you need?

Theoretical yield = actual yield / (% yield / 100) = 50.0 / 0.75 = 66.7 g

Then calculate the reactant needed for 66.7 g of product using standard stoichiometry.

What Percent Yield Greater Than 100% Means

If your calculated percent yield exceeds 100%, it does not mean you created matter from nothing. It means your product is impure. Common causes:

• Product was not fully dried (residual solvent adds mass)
• Product is contaminated with unreacted starting material or byproducts
• Product absorbed moisture from the air

This is a favorite MCAT distractor. If a question asks “a student obtained a 110% yield, what is the most likely explanation?” the answer is always contamination or incomplete drying - never that the reaction was unusually efficient.

Concentration tells you how much solute is dissolved in a given amount of solution or solvent. There are four concentration units you need for the MCAT, and each is defined slightly differently. Knowing which one to use - and why - is the key.

Molarity (M) - Moles per Liter of Solution

Molarity is the most commonly used concentration unit in chemistry. It measures the number of moles of solute per liter of total solution.

Example: Dissolve 58.44 g of NaCl (1 mole) in enough water to make 500 mL of solution.

• M = 1 mol / 0.500 L = 2.0 M NaCl

Key limitation: Molarity changes with temperature because volume expands and contracts when heated or cooled. This is why colligative property equations use molality instead.

Molality (m) - Moles per Kilogram of Solvent

Molality measures the number of moles of solute per kilogram of solvent (not total solution).

Mole Fraction (X) - Ratio of Moles

Mole fraction is the ratio of moles of one component to the total moles of all components.

Example: A solution contains 2 mol of ethanol and 8 mol of water.

• $X_{\text{ethanol}}$ = 2 / (2 + 8) = 0.20
• $X_{\text{water}}$ = 8 / (2 + 8) = 0.80
• Check: 0.20 + 0.80 = 1.00 ✓

Normality (N) - Equivalents per Liter

Normality measures the number of equivalents per liter of solution. An equivalent is the amount of substance that reacts with or supplies one mole of the relevant species (H⁺, OH⁻, or electrons, depending on the reaction type).

Examples:

• HCl donates 1 H⁺: N = M x 1 (normality = molarity)
• H₂SO₄ donates 2 H⁺: N = M x 2 (1 M H₂SO₄ = 2 N)
• H₃PO₄ donates 3 H⁺: N = M x 3 (1 M H₃PO₄ = 3 N)

Comparison Table

Dilution is simple: you add more solvent to a solution to decrease its concentration. The amount of solute does not change - you are just spreading it through more liquid. This one idea gives you the most useful equation in solution chemistry.

The Dilution Equation

Why it works: Molarity times volume equals moles (mol/L x L = mol). When you dilute, you do not add or remove solute - only solvent. So the moles before dilution equal the moles after dilution.

Worked Example

Problem: You have 50.0 mL of 6.0 M HCl. How much water must you add to make a 1.0 M solution?

Step 1: Identify variables.

• M1 = 6.0 M, V1 = 50.0 mL
• M2 = 1.0 M, V2 = ?

Step 2: Solve for V2.

• M1V1 = M2V2
• (6.0)(50.0) = (1.0)(V2)
• V2 = 300 mL

Step 3: Find water added.

• Water added = V2 - V1 = 300 - 50.0 = 250 mL

Note the question asked how much water to add, not the final volume. This is a common MCAT trap - read carefully.

When M1V1 = M2V2 Does NOT Apply

This equation only works when the moles of solute stay constant. It does not apply when:

1. A chemical reaction occurs - if mixing two solutions produces a reaction, you cannot use M1V1 = M2V2 for the reacting species
2. You mix two solutions of the same solute - this is a mixing problem, not a dilution. Use: $M_{\text{final}}$ = (M1V1 + M2V2) / (V1 + V2)
3. Solute is added or removed - the equation assumes only solvent changes

Serial Dilutions

A serial dilution is a sequence of dilutions, each using the product of the previous step as the starting solution. They are used to achieve very low concentrations efficiently.

Example: Starting with 1.0 M NaCl, perform three consecutive 1:10 dilutions.

• After dilution 1: 1.0 / 10 = 0.10 M
• After dilution 2: 0.10 / 10 = 0.010 M
• After dilution 3: 0.010 / 10 = 0.0010 M = $1.0 \times 10^{-3}$ M

Each 1:10 dilution divides the concentration by 10. Three such dilutions give a 1:1000 overall dilution factor.

Preparing Standard Solutions

Labs often prepare standard solutions by diluting a concentrated stock solution. The process:

1. Calculate the volume of stock solution needed using M1V1 = M2V2
2. Measure that volume with a pipette
3. Transfer to a volumetric flask
4. Add solvent until the solution reaches the calibration mark on the flask

Example: Prepare 250 mL of 0.10 M NaOH from a 2.0 M stock solution.

• V1 = M2V2 / M1 = (0.10)(250) / (2.0) = 12.5 mL of stock solution
• Add 12.5 mL of 2.0 M NaOH to a 250 mL volumetric flask, then fill to the mark with water

Solution stoichiometry is regular stoichiometry with one extra step: converting between volume and moles using molarity. If you can do grams-to-moles conversions and use mole ratios, you can do solution stoichiometry. The only new tool is the relationship M x V = moles.

The Key Relationship

Worked Example - Precipitation Reaction

Problem: How many mL of 0.200 M AgNO3 are needed to completely react with 35.0 mL of 0.100 M NaCl?

Balanced equation: AgNO3 + NaCl -> AgCl(s) + NaNO3

Step 1: Find moles of NaCl.

• moles NaCl = M x V = (0.100 mol/L)(0.0350 L) = 0.00350 mol

Step 2: Use mole ratio.

• 1 mol AgNO3 : 1 mol NaCl
• moles AgNO3 needed = 0.00350 mol

Step 3: Convert moles to volume.

• V = moles / M = 0.00350 / 0.200 = 0.0175 L = 17.5 mL

Titration Calculations

A titration is a controlled reaction between a solution of known concentration (the titrant, in the buret) and a solution of unknown concentration (the analyte, in the flask). The goal is to find the unknown concentration.

At the equivalence point, the titrant has supplied exactly enough moles to react completely with the analyte. No excess of either remains.

Titration Worked Example

Problem: A 25.0 mL sample of HCl of unknown concentration is titrated with 0.150 M NaOH. The equivalence point is reached after 32.0 mL of NaOH is added. What is the concentration of HCl?

Balanced equation: HCl + NaOH -> NaCl + H2O (1:1 ratio)

Step 1: Find moles of NaOH used.

• moles NaOH = (0.150)(0.0320) = 0.00480 mol

Step 2: Use mole ratio.

• 1 mol HCl : 1 mol NaOH
• moles HCl = 0.00480 mol

Step 3: Find molarity of HCl.

• M = moles / V = 0.00480 / 0.0250 = 0.192 M HCl

Non-1:1 Titrations

When the acid and base react in a ratio other than 1:1, you must account for stoichiometry.

Example: Titrating H2SO4 with NaOH.

Balanced equation: H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O

Here, 1 mole of H2SO4 reacts with 2 moles of NaOH. At the equivalence point:

moles NaOH = 2 x moles H2SO4

If 20.0 mL of 0.100 M H2SO4 is titrated with 0.100 M NaOH:

• moles H2SO4 = (0.100)(0.0200) = 0.00200 mol
• moles NaOH needed = 2 x 0.00200 = 0.00400 mol
• Volume NaOH = 0.00400 / 0.100 = 0.0400 L = 40.0 mL

Notice it took twice the volume of NaOH even though the molarities were equal - because each H2SO4 molecule requires two NaOH molecules.

Titration vs. Dilution - Do Not Confuse Them

Both use M x V, but for completely different reasons:

Back Titration

Sometimes the analyte does not react cleanly with an indicator or is insoluble. In a back titration, you add a known excess of one reagent, let it react with the analyte, and then titrate the leftover excess with a second standard solution.

Logic:

1. Moles of reagent added (known)
2. Moles of reagent that reacted with the analyte = moles added - moles left over
3. Use stoichiometry to find moles of analyte

This technique appears in MCAT passages about antacids (excess HCl + antacid, then titrate remaining HCl with NaOH) and calcium carbonate analysis.

Density and dimensional analysis are not glamorous topics, but they are the backbone of MCAT quantitative reasoning. Density connects mass to volume. Dimensional analysis connects everything else. Together, they let you solve multi-step problems by chaining conversion factors until your units work out.

Density

Key density values for the MCAT:

• Water: 1.00 g/mL (at 4 degrees C, where water is densest)
• Ice: 0.917 g/mL (less dense than liquid water - this is why ice floats)
• Most organic liquids: 0.7-0.9 g/mL (less dense than water)
• Most metals: much greater than 1.0 g/mL

Density as a Conversion Factor

Density is a fraction (g/mL), so you can use it as a conversion factor in two directions:

• Mass to volume: divide by density (or multiply by 1/d)
• Volume to mass: multiply by density

Example: What is the mass of 250 mL of ethanol (d = 0.789 g/mL)?

• mass = V x d = 250 mL x 0.789 g/mL = 197 g

Example: What volume does 500 g of mercury occupy (d = 13.6 g/mL)?

• V = m / d = 500 g / 13.6 g/mL = 36.8 mL

Specific Gravity

Specific gravity is commonly used in clinical settings. Urine specific gravity (normally 1.005-1.030) indicates how concentrated the urine is. Blood has a specific gravity of about 1.06. These values appear in MCAT passages.

Dimensional Analysis (Factor-Label Method)

Dimensional analysis is a systematic method for converting between units by multiplying by conversion factors that equal 1.

The rules:

1. Write down your starting value with its units
2. Multiply by conversion factors arranged so that unwanted units cancel
3. Continue until you reach the desired units
4. If the units work out, the math is set up correctly

Worked Example - Multi-Step Conversion

Problem: Convert 5.0 kg of aluminum to the number of atoms. (Molar mass of Al = 27.0 g/mol)

Step 1: Start with given: 5.0 kg

Step 2: Chain conversion factors:

$5.0 \text{ kg} \times (1000 \text{ g} / 1 \text{ kg}) \times (1 \text{ mol} / 27.0 \text{ g}) \times (6.022 \times 10^{23} \text{ atoms} / 1 \text{ mol})$

Step 3: Cancel units and calculate:

• 5.0 x 1000 = 5000 g
• 5000 / 27.0 = 185.2 mol
• $185.2 \times 6.022 \times 10^{23} =$ $1.11 \times 10^{26}$ atoms

Every unit cancels except “atoms” - which is exactly what we wanted.

Common Conversion Factors for the MCAT

Combining Density with Dimensional Analysis

Many MCAT problems require chaining density with other conversions. Density acts as just another conversion factor in the chain.

Example: How many molecules of ethanol are in 100 mL of pure ethanol? (d = 0.789 g/mL, MW = 46.07 g/mol)

$100 \text{ mL} \times (0.789 \text{ g} / 1 \text{ mL}) \times (1 \text{ mol} / 46.07 \text{ g}) \times (6.022 \times 10^{23} \text{ molecules} / 1 \text{ mol}) =$ $1.03 \times 10^{24}$ molecules

Each step converts one unit to the next: mL -> g -> mol -> molecules.

Density of Solutions

For solutions, density connects the total mass of the solution to its volume. This lets you convert between concentration units.

Example: Converting molarity to molality using density.

A 2.00 M NaCl solution has a density of 1.08 g/mL. Find the molality.

1. Take 1.00 L of solution as a basis
2. Moles of NaCl = 2.00 mol
3. Mass of solution = 1000 mL x 1.08 g/mL = 1080 g
4. Mass of NaCl = 2.00 mol x 58.44 g/mol = 116.9 g
5. Mass of solvent = 1080 - 116.9 = 963.1 g = 0.9631 kg
6. Molality = 2.00 mol / 0.9631 kg = 2.08 m

This type of conversion requires density as the bridge between volume-based (molarity) and mass-based (molality) concentration units.