Conversions, Molarity, and Dilution

Overview

John Flood builds a unit-cancellation roadmap for converting among particles, moles, and mass, using examples such as copper(II) sulfate pentahydrate (249.6 g/mol), methane molecules, and copper iodide. He extends the same quantitative reasoning to percent composition, empirical and molecular formulas, electrolytes, molarity, dilution, and ppm/ppb, emphasizing that explicit units determine which conversion factors to use.

Key takeaways

Explicit units determine whether a conversion factor should be inverted: arrange units to cancel first, then calculate, rather than memorizing when to multiply or divide.
Molar mass is the bridge between mass and amount: CuSO₄·5H₂O has a formula mass near 249.6 amu and a molar mass near 249.6 g/mol.
Compound subscripts provide mole ratios: one mole of Fe₂O₃ corresponds to three moles of oxide ions, while one mole of Na₂CO₃ corresponds to two moles of sodium ions.
Empirical formulas come from converting element masses—or percentages treated as masses in a 100 g sample—to moles, then dividing all mole amounts by the smallest.
A molecular formula is an integer multiple of its empirical formula; a molar mass of 178 g/mol compared with an empirical mass of 89 g/mol requires doubling every subscript.
Molarity is moles per liter, and dilution preserves the number of solute moles, which is why M₁V₁ = M₂V₂.

Chapters

0:00 STEM Center Resources and the Start of Unit 5

John Flood encourages students to use the STEM center as a study space and ask its tutors questions while working there.
Unit 5 focuses mainly on calculations, and Flood revisits formula mass because it supports later conversions.
Students are reminded that earlier skills, including chemical formulas, remain necessary foundations for new calculations.

2:00 Formula Mass Depends on Knowing the Compound Formula

The worksheet asks for the mass of 0.1 mol of copper(II) sulfate pentahydrate, requiring its formula before any formula-mass calculation.
Flood uses the problem to review how compound names and ionic formulas connect across units.
He recommends targeted practice with flashcards, Quizlet, and teaching formulas to someone else.

4:00 Building CuSO₄·5H₂O from Ion Charges and the Hydrate Name

Copper(II) means Cu²⁺, sulfate is SO₄²⁻, and pentahydrate adds five water molecules.
Because Cu²⁺ and SO₄²⁻ balance in a 1:1 ratio, the ionic portion is CuSO₄ rather than Cu₂(SO₄)₂.
Flood contrasts sulfate with sulfite and notes that recognizing common polyatomic ions is an assumed skill in later problems.

10:00 Calculate the Formula Mass of Copper(II) Sulfate Pentahydrate

The calculation adds one Cu, one S, four O atoms, and five H₂O units using periodic-table atomic masses.
Using approximately 63.546 for Cu, 32.065 for S, 15.994 for O, and 18 for each water gives about 249.6 amu.
The formula mass counts the atoms specified by the subscripts, including all five waters of hydration.

15:00 Formula Mass in amu Versus Molar Mass in g/mol

The same numerical sum can be reported as formula mass in atomic mass units or molar mass in grams per mole.
Flood identifies molar mass—not formula mass—as the conversion factor between moles and measurable mass.
For CuSO₄·5H₂O, the approximate molar mass is 249.6 g/mol.

17:00 The Roadmap from Particles to Moles to Mass

Moles connect particle counts to mass, with Avogadro’s number linking individual entities to moles.
Flood compares a mole to a dozen: both are counting units, but a mole contains about 6.022 × 10²³ entities.
Molar mass then connects the amount in moles to the mass measured on a balance.

22:00 Convert 0.1 Mole of Hydrate to Grams by Cancelling Units

Starting with 0.1 mol CuSO₄·5H₂O, students multiply by 249.6 g per mole so the mole units cancel.
The result is about 24.96 g, reported as 25 g to two significant figures.
Flood recommends checking magnitude: less than one mole should produce less than the 249.6 g mass of one mole.

24:00 Use Avogadro’s Number to Convert Methane Molecules to Moles

For 1.5 × 10²³ methane molecules, the conversion factor places molecules in the denominator and moles in the numerator.
Dividing by 6.022 × 10²³ molecules per mole gives approximately 0.25 mol CH₄ to two significant figures.
The powers of ten cancel because both the particle count and Avogadro’s number are on the 10²³ scale.

31:00 Use Subscripts as Mole Ratios in Fe₂O₃

The formula Fe₂O₃ shows that one formula unit contains three oxide ions, so five moles of Fe₂O₃ correspond to 15 moles of oxide ions.
Flood compares a compound to a car and its subscripts to parts: one car has four tires, just as one Fe₂O₃ unit has three oxide ions.
The compound name iron(III) oxide follows from oxide’s 2− charge and the Fe₂O₃ ratio.

42:00 Solve a Two-Step Conversion from Water to Oxygen Mass

To find oxygen mass in one mole of water, first use the 1:1 mole ratio between H₂O and oxygen atoms.
Then convert oxygen moles to grams using approximately 16.00 g O per mole O.
The resulting oxygen contribution is about 16 g per mole of water; students may combine the factors or calculate one step at a time.

50:00 Start the Sodium Carbonate Problem with the Given Particle Count

Flood guides students to identify the supplied quantity as sodium ions, a particle count, rather than grams or moles.
Avogadro’s number converts sodium ions to moles of sodium ions; molar mass is not the appropriate first factor.
The formula Na₂CO₃ supplies the next relationship: two sodium ions per formula unit of sodium carbonate.

56:00 Convert Sodium Ions into Moles of Sodium Carbonate

The worked setup converts 1.204 × 10²⁴ sodium ions to moles, then uses the 2:1 Na⁺-to-Na₂CO₃ ratio.
About 2 mol Na⁺ corresponds to 1 mol Na₂CO₃, so the example yields approximately 1 mol of sodium carbonate.
Flood stresses that the known quantity must remain the starting value; assuming one mole of the requested compound would skip the problem.

1:00:00 Convert Iodide Mass to Copper(I) Iodide Mass

The example converts 63.5 g iodide to moles using iodine’s molar mass, about 126.9 g/mol.
For CuI, iodide and copper(I) iodide have a 1:1 mole ratio; the product’s molar mass is approximately 190.45 g/mol.
The combined unit-cancellation setup gives about 95.3 g CuI, and the placement of grams depends on which units must cancel.

1:08:00 Why Chemical Mass Conversions Pass Through Moles

Different substances have different molar masses, so grams of one species cannot be directly equated to grams of another.
Moles express the particle ratios in formulas, such as one Cu to one I in CuI or two H to one O in water.
A conversion between species therefore uses a mass-to-mole step, a mole ratio, and usually a mole-to-mass step.

1:12:00 Calculate Percent Composition from Sample Masses

For a 174 g sample containing 108 g carbon, 18 g hydrogen, and 48 g oxygen, each element’s mass is divided by the total sample mass.
Multiplying the carbon fraction by 100 gives approximately 62% carbon by mass.
For water, oxygen contributes about 16 of the formula’s 18 mass units, or roughly 89% by mass.

1:16:00 Find an Ionic Empirical Formula from Iron and Chlorine Masses

The example starts with 11.17 g Fe and 21.27 g Cl and converts each mass to moles using its molar mass.
The amounts are approximately 0.20 mol Fe and 0.60 mol Cl, giving a 1:3 ratio after division by the smaller amount.
The resulting empirical formula is FeCl₃, named iron(III) chloride rather than iron trichloride.

1:22:00 Turn Percent Composition into an Empirical Formula

For a composition of 40.4% C, 7.9% H, 15.7% N, and 36.0% O, assume a convenient 100 g sample.
Convert the resulting gram amounts to moles, obtaining approximately 3.37 mol C, 7.9 mol H, 1.12 mol N, and 2.25 mol O.
Dividing each amount by the smallest, about 1.12, gives the empirical formula C₃H₇NO₂.

1:28:00 Use Molar Mass to Scale an Empirical Formula

The empirical formula C₃H₇NO₂ has a formula mass of about 89 g/mol, matching the stated molar mass in the worked example.
When a sample’s molar mass is 178 g/mol instead, dividing 178 by 89 gives a factor of 2.
Multiplying every empirical subscript by 2 gives the molecular formula C₆H₁₄N₂O₄.

1:33:00 Distinguish Soluble Compounds, Electrolytes, and Nonelectrolytes

A soluble substance forms a homogeneous solution, but solubility alone does not mean the solution conducts electricity.
Electrolytes form ions in solution: examples include hydrochloric acid producing hydronium and chloride ions and FeCl₃ producing Fe³⁺ and chloride ions.
Sugar dissolves in water but remains molecular, so it is a nonelectrolyte and does not conduct electricity.

1:37:00 Calculate Molarity from NaOH Mass and Solution Volume

Molarity measures moles of solute per liter of solution; it describes concentration even when the actual sample is not one liter.
Four grams of NaOH corresponds to 0.1 mol using its 40 g/mol molar mass, so in 1 L the concentration is 0.1 M.
Because NaOH contains one sodium per formula unit, the sodium-ion concentration is also 0.1 mol/L.

1:40:00 Dilution Preserves Moles; ppm and ppb Scale Ratios

The dilution equation M₁V₁ = M₂V₂ works because molarity multiplied by volume gives moles, which remain constant when solvent is added.
Percent, parts per million, and parts per billion multiply a ratio by 100, 1,000,000, and 1,000,000,000 respectively; ppm and ppb differ by a factor of 1,000.
For dilute aqueous solutions, ppm can be usefully expressed as approximately milligrams per liter; Flood closes by directing students to the unit test and scheduled midterm.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, John Flood Chemistry.