Conservation of Mass and Stoichiometry Cartoon courtesy of NearingZero.net Conservation of Mass and Stoichiometry

Ions Cation: A positive ion Mg2+, NH4+ Anion: A negative ion Cl-, SO42- Ionic Bonding: Force of attraction between oppositely charged ions.

Predicting Ionic Charges Group 1: Lose 1 electron to form 1+ ions H+ Li+ Na+ K+

Predicting Ionic Charges Group 2: Loses 2 electrons to form 2+ ions Be2+ Mg2+ Ca2+ Sr2+ Ba2+

Predicting Ionic Charges Loses 3 electrons to form 3+ ions Group 13: B3+ Al3+ Ga3+

Predicting Ionic Charges Lose 4 electrons or gain 4 electrons? Group 14: Neither! Group 13 elements rarely form ions.

Predicting Ionic Charges Nitride Gains 3 electrons to form 3- ions Group 15: P3- Phosphide As3- Arsenide

Predicting Ionic Charges Oxide Gains 2 electrons to form 2- ions Group 16: S2- Sulfide Se2- Selenide

Predicting Ionic Charges F1- Fluoride Br1- Bromide Group 17: Gains 1 electron to form 1- ions Cl1- Chloride I1- Iodide

Predicting Ionic Charges Stable Noble gases do not form ions! Group 18:

Predicting Ionic Charges Many transition elements have more than one possible oxidation state. Groups 3 - 12: Iron(II) = Fe2+ Iron(III) = Fe3+

Predicting Ionic Charges Some transition elements have only one possible oxidation state. Groups 3 - 12: Zinc = Zn2+ Silver = Ag+

Writing Ionic Compound Formulas Example: Barium nitrate 1. Write the formulas for the cation and anion, including CHARGES! ( ) 2. Check to see if charges are balanced. Ba2+ NO3- 2 Not balanced! 3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Writing Ionic Compound Formulas Example: Ammonium sulfate 1. Write the formulas for the cation and anion, including CHARGES! ( ) NH4+ SO42- 2. Check to see if charges are balanced. 2 3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion. Not balanced!

Writing Ionic Compound Formulas Example: Iron(III) chloride 1. Write the formulas for the cation and anion, including CHARGES! Fe3+ Cl- 2. Check to see if charges are balanced. 3 3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion. Not balanced!

Writing Ionic Compound Formulas Example: Aluminum sulfide 1. Write the formulas for the cation and anion, including CHARGES! 2. Check to see if charges are balanced. Al3+ S2- 2 3 3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion. Not balanced!

Writing Ionic Compound Formulas Example: Magnesium carbonate 1. Write the formulas for the cation and anion, including CHARGES! Mg2+ CO32- 2. Check to see if charges are balanced. They are balanced!

Writing Ionic Compound Formulas Example: Zinc hydroxide 1. Write the formulas for the cation and anion, including CHARGES! ( ) 2. Check to see if charges are balanced. Zn2+ OH- 2 3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion. Not balanced!

Writing Ionic Compound Formulas Example: Aluminum phosphate 1. Write the formulas for the cation and anion, including CHARGES! 2. Check to see if charges are balanced. Al3+ PO43- They ARE balanced!

Naming Ionic Compounds 1. Cation first, then anion 2. Monatomic cation = name of the element Ca2+ = calcium ion 3. Monatomic anion = root + -ide Cl- = chloride CaCl2 = calcium chloride

Naming Ionic Compounds (continued) Metals with multiple oxidation states - some metal forms more than one cation - use Roman numeral in name PbCl2 Pb2+ is cation PbCl2 = lead(II) chloride

Naming Binary Compounds - Compounds between two nonmetals - First element in the formula is named first. - Second element is named as if it were an anion. - Use prefixes - Only use mono on second element - P2O5 = diphosphorus pentoxide CO2 = carbon dioxide CO = carbon monoxide N2O = dinitrogen monoxide

Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

Formulas molecular formula = (empirical formula)n [n = integer] Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11

Empirical Formula Determination Base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. Divide each value of moles by the smallest of the values. Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

Synthesis (Composition) Reactions Two or more substances combine to form a new compound. A + X  AX Reaction of elements with oxygen and sulfur Reactions of metals with Halogens Synthesis Reactions with Oxides There are others not covered here!

Decomposition Reactions A single compound undergoes a reaction that produces two or more simpler substances AX  A + X Decomposition of: Binary compounds H2O(l )  2H2(g) + O2(g) Metal carbonates CaCO3(s)  CaO(s) + CO2(g) Metal hydroxides Ca(OH)2(s)  CaO(s) + H2O(g) Metal chlorates 2KClO3(s)  2KCl(s) + 3O2(g) Oxyacids H2CO3(aq)  CO2(g) + H2O(l ) Peroxide 2H2O2 (l)  2H2O(l) + O2 (g)

Single Replacement Reactions A + BX  AX + B BX + Y  BY + X Replacement of: Metals by another metal Hydrogen in water by a metal Hydrogen in an acid by a metal Halogens by more active halogens

The Activity Series of the Metals Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron Nickel Lead Hydrogen Bismuth Copper Mercury Silver Platinum Gold Metals can replace other metals provided that they are above the metal that they are trying to replace. Metals above hydrogen can replace hydrogen in acids. Metals from sodium upward can replace hydrogen in water (cold) Metals between iron upward to magnesium can replace hydrogen in hot water (steam)

The Activity Series of the Halogens Fluorine Chlorine Bromine Iodine Halogens can replace other halogens in compounds, provided that they are above the halogen that they are trying to replace. 2NaCl(s) + F2(g)  2NaF(s) + Cl2(g) MgCl2(s) + Br2(g)  ??? No Reaction

Double Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX A+ + X- + B+ + Y-  A+ + Y- + B+ + X- One of the compounds formed is usually a precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

Combustion Reactions A substance combines with oxygen, releasing a large amount of energy in the form of light and heat. Reactive elements combine with oxygen P4(s) + 5O2(g)  P4O10(s) (This is also a synthesis reaction) The burning of natural gas, wood, gasoline C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

Balancing Chemical reactions “Satisfies the Law of conservation of Matter” Formulas must be written correctly Balance Polyatomic ions first Balance elements appearing only once on each side Balance Hydrogen and oxygen last

Stoichiometry “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.” Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Review: Atomic Masses Carbon = 98.89% 12C 1.11% 13C <0.01% 14C Elements occur in nature as mixtures of isotopes Carbon = 98.89% 12C 1.11% 13C <0.01% 14C Carbon’s atomic mass = 12.01 amu

Review: The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of anything = 6.022 ´ 1023 units of that thing

The Mole

Using Compound Masses

Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = 44.01 grams per mole H2O = 18.02 grams per mole Ca(OH)2 = 74.10 grams per mole

Review: Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Mole Relations

Calculating Masses of Reactants and Products Balance the equation. Convert mass to moles. Set up mole ratios. Use mole ratios to calculate moles of desired substituent. Convert moles to grams, if necessary.

Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2 Al2O3 a. Every reaction needs a yield sign! b. What are the reactants? c. What are the products? d. What are the balanced coefficients?

Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2  2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 = ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3

Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

Limiting Reagents - Combustion

Solving a Stoichiometry Problem Balance the equation. Convert masses to moles. Determine which reactant is limiting. Use moles of limiting reactant and mole ratios to find moles of desired product. Convert from moles to grams.