Chapter 2 The Mole: The Link Between the Macroscopic and the Atomic Worlds of Chemistry In this chapter we explore the quantitative aspect of chemistry, also known as stoichiometry. The concept of the mole is a key part of stoichiometry.

The Mole as the Bridge Between the Macroscopic and Atomic Scales Central concept Connects microscopic and macroscopic. Defined as the number of atoms in exactly 12 grams of 12C. (For a lot of chemistry, it doesn’t matter what the numerical value of a mole equals!)

The Mole as the Bridge Between the Macroscopic and Atomic Scales 30.97376 grams of 31P has the same number of atoms as 12.0000 grams of 12C. Both quantities are referred to as one mole.

The Mole as a Collection of Atoms The numerical value of the number of atoms in 12 grams of 12C is called Avogadro’s Number.

Converting Grams into Moles and Number of Atoms Grams  Moles Moles → Grams Moles → Number of Atoms Number of Atoms  Moles

The Mole as a Collection of Molecules Need Avogadro’s number. Need molar mass. Mass corresponding to one mole. For elements, same as atomic weight in grams unless the element is represented as a molecular element such as Cl2.

The Mole as a Collection of Molecules

The Mole as a Collection of Molecules Also known as Molar Mass. For compounds, use chemical formula. Illustrates the importance of chemical formula: Symbols Subscripts

The Mole as a Collection of Molecules Calculate the molecular mass of formaldehyde, H2CO. Which atoms are present? How many of each of these atoms?

The Mole as a Collection of Molecules 2 H @ 1.0079 amu/H = 2.0158 1 C @ 12.011 amu/C = 12.011 1 O @ 15.999 amu/O = 15.999 Molecular Mass of formaldehyde is 30.025 grams/mole.

The Mole as a Collection of Molecules With the molecular mass, many questions can be answered: How many grams are in one mole of NH3? How many moles are in 75 grams of H2? What’s the average mass of one molecule of CO2?

The Mole as a Collection of Molecules Correctly calculating molecular mass is an important skill in chemistry.

Percent Mass Calculated with same data used to calculate molecular mass. In the formaldehyde example, 12.011 grams of the 30.025 grams/mole come from carbon.

Determining the Formula of a Compound From the % mass of the elements in a compound, we can determine the empirical formula of the compound. If we know the molecular mass, we can determine the molecular formula from the empirical formula.

Two Views of Chemical Equations: Molecules Versus Moles What does this equation mean? 2Mg(s) + O2(g) → 2MgO(s) • Two atoms of Mg will react with one molecule of oxygen to produce 2 molecules of MgO. • Also means 48 grams of Mg will react with 32 grams of oxygen to produce 80 grams of MgO.

Two Views of Chemical Equations: Molecules Versus Moles First meaning is microscopic. Second is macroscopic.

Mole Ratios and Chemical Equations Chemical equations predict. The numbers used to balance the equation are called stoichiometric coefficients. These coefficients represent mole ratios.

Mole Ratios and Chemical Equations 2Mg(s) + O2(g) → 2MgO(s) “Two moles of magnesium will react with 1 mole of oxygen to produce 2 moles of magnesium oxide.”

Mole Ratios and Chemical Equations How many grams of oxygen are required to completely react with 10.00 grams of magnesium? 2Mg(s) + O2(g) → 2MgO(s)

Mole Ratios and Chemical Equations 10.00 grams Mg = 0.4114 moles 1 mole of O2 will react with 2 moles of Mg. 0.2057 moles of O2 will react with 0.4114 moles Mg. 1 mole of O2 = 31.9988 grams 0.2057 moles O2 = 6.582 grams

Mole Ratios and Chemical Equations How many grams of magnesium oxide will be formed?

Mole Ratios and Chemical Equations How many grams of magnesium oxide will be formed? 16.582 grams

Stoichiometry The calculations just shown are an example of stoichiometry. Stoichiometry is what makes chemistry a quantitative science.

Stoichiometry Imagine that the previous problem had been stated as follows: “How much magnesium oxide can be produced with 20 grams of magnesium and 10 grams of oxygen?”

Stoichiometry Is the correct answer 30 grams?

Stoichiometry Is the correct answer 30 grams? No. Why not?

Stoichiometry Is the correct answer 30 grams? No. Why not? The mole ratio of Mg to O2 is not 2 to 1 in this problem.

The Stoichiometry of the Breathalyzer

The Nuts and Bolts of Limiting Reagents The reaction between 20 grams of Mg and 10 grams of O2 is an example of a Limiting Reagent problem. Reactions can be either stoichiometric or have a limiting reagent.

The Nuts and Bolts of Limiting Reagents The mathematics of a limiting reagent problem extend beyond chemistry. - Text example with nuts and bolts, section 2.11. - Imagine other examples such as recipes and assembly lines.

The Nuts and Bolts of Limiting Reagents In a limiting reagent problem, the most important task is to determine which reagent is the limiting reagent. All subsequent calculations are based on this determination.

Density Mass of a sample divided by its volume. Common unit is grams/cm3. Applies to all forms of matter. Most importantly, density is independent of the sample size. Can be used to identify a substance.

Solute, Solvent, and Solution A lot of chemistry is done with solutions. Important terms associated with solutions: Solvent Solute Solution

Solute, Solvent, and Solution Solutions are homogeneous. Solutions are mixtures. The composition of a solution must be specified.

Solute, Solvent, and Solution Compare these statements: “I heated 50 mL of ethanol to 35 °C and held it at that temperature for 1 hour.” “I heated 50 mL of an aqueous solution of sodium nitrate to 35 °C and held it at that temperature for 1 hour.”

Solute, Solvent, and Solution Why is the second one ambiguous?

Solute, Solvent, and Solution Why is the second one ambiguous? The composition of the solution was not specified. This leads to the next topic: Concentration.

Concentration Most useful with solutions. Expressed as amount of solute per amount of solvent or amount of solute per amount of solution. Important to know the difference between solvent and solution.

Molarity as a Way of Counting Particles in Solution Most common concentration unit in chemistry is Molarity. Equal to moles of solute/Liter of solution. Symbol M used to express units mol/L.

Molarity as a Way of Counting Particles in Solution Important Equation: M × V = n (Molarity of a solution) × (volume of solution expressed in Liters) = number of moles of solute in said volume

Dilution Calculations Use M × V = n Recall Molarity is moles solute/Liter solution.

Solution Stoichiometry Balanced chemical reactions involve moles of substances, n. Solution information must be converted to number of moles, n. M × V = n Important skill given the extent of solution work in chemistry.

Solution Stoichiometry Net ionic equation sometimes used Unreactive common ions (sometimes called spectator ions) are subtracted from balanced chemical equation. H2C2O4(aq) +2Na+(aq) + 2OH-(aq) → 2Na+(aq) + C2O4-2(aq) + 2H2O(l) H2C2O4(aq) + 2OH-(aq) → C2O4-2(aq) + 2H2O(l)