Gen Chem Ch. 8 – Moles and Molecules (P. 2)

What is a mole?

The amount of substance containing the same number of entities as there are atoms in 12 g of carbon-12 (6.022 × 10²³ particles).

What is Avogadro's number?

6.022 × 10²³ mol⁻¹; it relates microscopic particles to macroscopic moles.

How do you convert between moles and particles?

particles = moles × 6.022 × 10²³; moles = particles ÷ 6.022 × 10²³.

How many atoms are in 0.300 mol Na?

0.300 × 6.022 × 10²³ = 1.81 × 10²³ atoms Na.

Why is the mole important?

It links measurable mass to atomic-scale counts, allowing equations to be quantitative.

Define molar mass (MM).

The mass of one mole of a substance (in g mol⁻¹); for compounds it's the sum of all atomic masses.

Find the molar mass of NaCl.

22.99 + 35.45 = 58.44 g mol⁻¹.

How do you convert between mass and moles?

moles = mass ÷ MM; mass = moles × MM.

How many moles are in 7.9 g Ca?

7.9 ÷ 40.08 = 0.197 mol Ca.

What is the mass of 1.2 × 10²³ Na atoms?

(1.2 × 10²³ / 6.022 × 10²³) × 22.99 = 4.58 g Na.

What is a stoichiometric ratio?

The fixed whole-number ratio of moles of reactants and products from a balanced equation.

How are mole ratios used?

As conversion factors between moles of different substances using equation coefficients.

From 2 N₂ + 5 O₂ → 2 N₂O₅, what is the ratio O₂:N₂O₅?

5:2.

If 0.864 mol P are present, how many mol Ca₃(PO₄)₂?

0.864 × (1 / 2) = 0.432 mol Ca₃(PO₄)₂.

What is an empirical formula?

Simplest whole-number ratio of elements in a compound (e.g. CH₂O).

What is a molecular formula?

Actual numbers of each atom (e.g. C₆H₁₂O₆).

Steps to find an empirical formula from % composition.

Assume 100 g; convert %→g; convert g→mol; divide by smallest; adjust to integers.

Example 43.64 % P and 56.36 % O empirical formula.

P₂O₅.

How do you find a molecular formula?

n = molecular mass ÷ empirical mass; multiply subscripts by n.

Example empirical P₂O₅ MM = 283.9 g mol⁻¹.

n ≈ 2 → P₄O₁₀.

What is percent composition by mass?

(mass element / mass compound) × 100 %.

Example Na₂CO₃ composition.

43.38 % Na, 11.33 % C, 45.29 % O.

Why is % composition useful?

It helps identify unknown compounds and check purity.

Example 8.657 g compound → 5.217 g C, 0.962 g H, 2.478 g O % composition.

C 60.26 %, H 11.11 %, O 28.62 %.

What is combustion analysis?

A compound containing C and H (± O) is burned in O₂; CO₂ and H₂O masses reveal C and H moles; O by difference.

Example 5.217 g sample → 7.406 g CO₂ & 4.512 g H₂O empirical formula.

CH₃O.

Guidelines for balancing equations.

Balance complex substance first; elements in two compounds next; H₂ & O₂ last; use whole numbers.

Why must equations be balanced?

Law of conservation of mass.

Balance KClO₃ → KCl + O₂.

2 KClO₃ → 2 KCl + 3 O₂.

Balance Ba(OH)₂ + Na₂SO₄ → BaSO₄ + NaOH.

Ba(OH)₂ + Na₂SO₄ → BaSO₄ + 2 NaOH.

Define limiting reactant.

Reactant that is completely consumed and limits product formation.

Define excess reactant.

Reactant present in greater amount than needed.

Steps to identify limiting reactant.

Balance → convert masses to moles → use mole ratios → compare needed vs available.

Example 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O (30.0 g NH₃, 40.0 g O₂).

O₂ limiting; produces ≈ 30.0 g NO.

Define theoretical yield.

Maximum possible product from limiting reactant.

Define actual yield.

Measured amount of product obtained.

Define percent yield.

(actual / theoretical) × 100 %.

Example 36 predicted cookies, 24 actual.

67 %.

Example NH₃ + CuO → Cu (18.1 g NH₃, 90.4 g CuO). Actual 58.3 g Cu.

% yield = 80.7 %.

CHAPTER 9 - MOLARITY & DILUTIONS

What is a solution?

A homogeneous mixture where components are evenly distributed at the molecular level.

Define solvent.

Component present in largest amount; dissolves the solute.

Define solute.

Substance present in smaller amount that is dissolved in the solvent.

What is an aqueous solution?

A solution with water as the solvent.

Why are solutions important in chemistry?

Most reactions occur in solution because reactants can move and collide freely.

Differentiate between dilute and concentrated solutions.

Dilute = low solute/solvent ratio; concentrated = high ratio.

Which has more solute: 0.1 M or 1.0 M NaCl?

1.0 M.

Define molarity (M).

M = mol solute / L solution.

Example 0.100 M NaCl.

0.100 mol NaCl in 1.00 L solution.

How does molarity act as a conversion factor?

mol = M × V; V = mol / M.

Example 11.5 g NaOH in 1.50 L.

0.2875 mol / 1.50 L = 0.192 M.

Example needed mL of 0.250 M NaCl for 0.100 mol.

0.400 L = 400 mL.

Why is molarity useful?

Relates moles of solute to volume for stoichiometric calculations.

Steps to prepare a solution of known molarity.

1 Weigh solute 2 Transfer to flask 3 Add water and dissolve 4 Fill to mark 5 Invert to mix.

Why use a volumetric flask?

Ensures precise final volume and accurate molarity.

Example prepare 250.0 mL 0.100 M Sr(NO₃)₂.

0.0250 mol × 211.63 = 5.29 g.

Why invert flask?

To mix completely for uniform concentration.

What is a dilution?

Lowering solution concentration by adding solvent; moles of solute unchanged.

Give the dilution equation.

M₁ V₁ = M₂ V₂.

Example prepare 1.00 L 2.00 M H₂SO₄ from 16.0 M stock.

V₁ = 0.125 L = 125 mL.

Example 25 mL 6.0 M HCl → 500 mL.

M₂ = 0.30 M.

What stays constant during a dilution?

Number of moles of solute.

What is solution stoichiometry?

Quantitative relationship between volumes, concentrations, and reactant/product amounts in solution reactions.

Steps for solution stoichiometry problems.

Balance eq → convert V→mol → use mole ratios → convert to desired quantity.

Example NaOH neutralizing 25 mL 0.100 M H₂SO₄ using 0.250 M NaOH.

20.0 mL NaOH required.

Why is solution stoichiometry useful?

Enables determining unknown concentrations (titrations) and reactant requirements.

Summarize the relationship among moles, molarity, and volume.

n = M × V.

Key link between Ch 8 and Ch 9.

Ch 8: mass-mole-particle conversions; Ch 9: mole-molarity-volume relationships in solution.