Mole: base unit for an amount of substance A mole contains Avogadro s number (N A ) of particles (atoms, molecules, ions, formula units )

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1 Mole: base unit for an amount of substance A mole contains Avogadro s number (N A ) of particles (atoms, molecules, ions, formula units ) N A mol -1 1 mol substance contains N A Molar mass (g/mol) numerically equals Elements Molecular substances Ionic compounds Na, Al, C, Si H, O, HCl, C H 6 NaCl, Al O Na atoms H molecules NaCl formula units Atomic mass Molecular mass Formula mass Mass percent composition Empirical formula Molecular mass Molecular formula Example.1 Burning a g sample of a carbon-hydrogen-oxygen compound in oxygen yields g CO and g H O. Determine the mass percent composition.?g C g CO 1 mol CO g CO 1 mol C 1 mol CO g C 1 mol C 1 mol HO mol H g H?g H g H O g H O 1 mol H O 1 mol H g C Mass % C 100% 5.0% C g sample g H Mass % H 100% 11.19% H g sample Mass % O % 5.0% 11.19% 5.51% O g C Molar mass of CO Factor relating mol C to mol CO Molar mass of C g H

2 Determine the empirical formula Step 1: A g sample contains 5.0 g C, g H, and 5.51 g O. Step : convert masses to moles 1mol C? mol C 5.0 g C 1.011g C 4.48 mol C 1mol H? mol H g H mol H g H 1mol O? mol O 5.51g O g O.0 mol O Step : tentative formula C 4.48 H O.0 Step 4: C 4.48/.0 H 11.10/.0 O.0/.0 C H O C H 5 O Empirical formula C H 5 O Empirical formula mass 45.0 u If the experimentally determined molecular mass is 90 u: molecular mass 90 u Integral factor empirical formula mass 45.0 u The molecular formula is (C H 5 O), or C 4 H 10 O.

3 Stoichiometry of Chemical Reactions.7 Writing and Balancing Chemical Equations For the reaction of carbon with a plentiful source of oxygen: Reactants Product C(s) + O (g) CO (g) Physical forms (g) gas; (l) liquid; (s) solid; (aq) aqueous solution Interpretation of a chemical equation a qualitative description of the reaction Solid carbon and gaseous oxygen react to form gaseous carbon dioxide. microscopic interpretation One carbon atom reacts with one oxygen molecule to form one molecule of carbon dioxide. macroscopic interpretation 1 mol (1.01 g) of carbon reacts with 1 mol (.00 g) of oxygen gas to produce 1 mol (44.01 g) of carbon dioxide. H (g) + O (g) H O(l) (not balanced) The law of conservation of mass Stoichiometric coefficients: used to balance the equation; indicating the combining ratios. H (g) + O (g) H O(l) (O balanced, H not balanced)

4 H (g) + O (g) H O(l) (not balanced) The law of conservation of mass Stoichiometric coefficients: to balance the equation; indicating the combining ratios. H (g) + O (g) H O(l) ( balanced) For each element, the same number of atoms must appear on each side of the arrow. We can balance an equation only by adjusting coefficients! Figure.6 in the textbook.

5 Balancing by inspection If an element is present in just one compound on each side, balance it first. If an element is present as the free element, balance it last. Often certain groupings of atoms (such as polyatomic ions) proceed unchanged through a reaction; balance them as a unit. At times, balance by first using a fractional coefficient(s) and then clearing the fractions by multiplying by a common multiplier. Example.1 Balance the equation Fe + O Fe O (not balanced) Fe + O Fe O (Fe balanced, O not balanced) Fe + O Fe O (balanced, fractional coefficient) 4 Fe + O Fe O (balanced) Another Example C H 6 O + O CO + H O (not balanced) Balance O last; start with C (or H) C H 6 O + O CO + H O (C balanced, H and O not balanced) Balance H: C H 6 O + O CO + H O (C, H balanced, O not balanced) Balance O: C H 6 O + O CO + H O (balanced)

6 Example.15 H PO 4 + NaCN HCN + Na PO 4 Note PO 4 and CN groups remain unchanged in the reaction. Balance H: Balance CN: H PO 4 + NaCN HCN + Na PO 4 H PO 4 + NaCN HCN + Na PO 4 (balanced) Na and PO 4 also balanced..8 Reaction Stoichiometry CO + H CH OH Stoichiometrically equivalent: 1 mol CO mol H ~ 1 mol CH OH Equivalence fraction 1 mol CO mol H 1 mol CO 1 mol CH OH mol H 1 mol CHOH 1 mol CO 1 mol CO Example.17 When mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is C H O CO + 4 H O 5 mol O Equivalence 1 mol C H 8 ~ 5 mol O 1 mol CH8 1 mol CH 5 mol O 8 5 mol O? mol O mol CH mol O 1 mol CH8

7 Calculations involving mass Grams of A 1 molar mass of Moles of A mol B mol A Moles of B A molar mass of B Example.18 NO + H O HNO + NO How many grams of nitric acid are produced for every g NO that reacts? 1 mol NO? mol NO g NO.174 mol NO g NO mol HNO? mol HNO.174 mol NO mol HNO mol NO Grams of B 6.01 g HNO? g HNO mol HNO 91.1 g HNO 1 mol HNO Combine steps: 1 mol NO mol HNO 6.01 g HNO? g HNO g NO 91.1 g HNO g NO mol NO 1 mol HNO.9 Limiting Reactants C H 8 (g) + 5 O (g) CO (g) +4 H O (l) 1 mol C H 8 and 5 mol O (in stoichiometric proportions) All the reactants are totally consumed if the reaction goes to completion. 1 mol C H 8 and 15 mol O C H 8 totally consumed, some unreacted O remains (10 mol; excess reactant) Limiting reactant: the reactant that is completely consumed in a reaction it limits the amount of products formed C H 8 is the limiting reactant; O is in excess

8 Determine the limiting reactant Example.0 Magnesium nitride can be found by the reaction of magnesium metal with nitrogen gas: Mg (s) + N (g) Mg N (s) How many grams of Mg N (s) can be made in the reaction of 5.00 g of Mg and g of N? Assuming Mg is the limiting reactant: 1 mol Mg 1 mol Mg N? mol Mg N 5.00 g Mg 4.05 g Mg mol Mg mol MgN Assuming N is the limiting reactant:? mol Mg 1 mol N 1 mol Mg N N g N mol MgN 8.01 g N 1 mol N Determine the limiting reactant Example.0 Magnesium nitride can be found by the reaction of magnesium metal with nitrogen gas: Mg (s) + N (g) Mg N (s) How many grams of can be made in the reaction of 5.00 g of Mg and g of N? Assuming Mg is the limiting reactant: 1 mol Mg 1 mol Mg N? mol Mg 4.05 g Mg mol Mg N 5.00 g Mg mol MgN Assuming N is the limiting reactant: N is in excess 1 mol N 1 mol Mg N? mol Mg N g N mol MgN 8.01 g N 1 mol N grams of N consumed: 1 mol N 8.01 g N? g N mol MgN 1.45 g N 1 mol MgN 1 mol N Mass of excess N : g N (initially) 1.45 g N (consumed) 1.55 g N

9 .11 Solutions and Solution Stoichiometry Solution is a homogeneous mixture of two or more substances. Its composition is uniform throughout. Solute(s): the substance(s) being dissolved Solvent: the substance doing the dissolve. Most of the solutions we encounter are aqueous solutions; water is the solvent. Some organic solvents: Hexane: dissolves grease Ethanol: solvent for many medicines Concentration: refers to the amount of solute dissolved in a certain quantity of solution or solvent a dilute solution vs. a concentrated solution Molar concentration, or molarity moles of solute Molarity (M) liters of solution Dissolving.50 mol NaCl in enough water to produce.00 L of solution:.50 mol NaCl Molarity (M) 1.75 M NaCl 1.75 molar NaCl.00 L solution 6.68 M NaOH 6.68 mol NaOH ~ 1 L solution Molarity conversion factors: 6.68 mol NaOH 1L soln 1L soln 6.68 mol NaOH

10 Example.4 We want to prepare a 6.68 M NaOH solution. (a) How many moles of NaOH are required to prepare L of this solution? (b) How many liters of this solution can we prepare with.5 kg NaOH? 6.68 mol NaOH (a)? mol NaOH L soln.4 mol NaOH 1L soln kg NaOH g NaOH mol NaOH L soln 1000 g NaOH 1mol NaOH 1L soln (b)? L soln.5 kg NaOH 8.79 L soln 1kg NaOH g NaOH 6.68 mol NaOH Dilutions of solutions adding solvent to a concentrated solution to get a dilute solution solution concentration changes Amount of solute does not change M conc Vconc Mdil Vdil Dilution equation Example.6 How many milliliters of a.00 M CuSO 4 stock solution are needed to prepare 0.50 L of M CuSO 4? V dil M M V dil conc conc?.00 M Vconc M 0.50 L M 0.50 L 1000 ml V conc L 50.0 ml.00 M 1L

11 Solutions in Chemical Reactions Stoichiometric factors (mole ratio) derived from chemical equation Molarity conversion factors Example.7 CaCO (s) + HCl(aq) CaCl (aq) + H O(l) + CO (g) How many grams of CaCO are consumed in a reaction with 5 ml of.5 M HCl? 1L soln.5 mol HCl 1mol CaCO? mol CaCO 1000 ml soln 1L soln mol HCl 5 ml soln 0.66 mol CaCO 100.1g CaCO? g CaCO 0.66 mol CaCO 6.6 g CaCO 1mol CaCO

The Mole. Relative Atomic Mass Ar

The Mole. Relative Atomic Mass Ar STOICHIOMETRY The Mole Relative Atomic Mass Ar Relative Molecular Mass Mr Defined as mass of one atom of the element when compared with 1/12 of an atom of carbon-12 Some Ar values are not whole numbers