Changing Matter • Matter can be changed two ways: • Physically • Physical reaction • Physical change • Chemically • Chemical reaction • Chemical change
Physical Changes • DoNOT CHANGE THE TYPE OF MATTER • Nothing new or different is formed • Could be a change in: • Mass • Volume • Density • Change in state • Color • Shape Size
Examples of Physical Changes • Boiling, vaporization…. Any state change • Dissolving • Breaking • Making a mixture • 2 or more types of matter (substances) mixed together • Not in specific amounts • Can be separated physically
Chemical Changes • Atoms have electrons arranged in energy levels or energy shells • Electrons in the last (outermost) shell are called valence electrons • Valence electrons let atoms bond with other atoms • Ionic bonding • Gaining or losing electrons • Covalent bonding • Sharing electrons
Chemical Changes • Atoms that bond form molecules • May be the same type (nonmetals) of atom or, • Different types (metal + nonmetal) of atoms • Different types compounds
Chemical Changes • Molecules can bond and “unbond” • Atoms can re-arranged in different combinations • For example: CaCO3 (1 atom Ca, 1 atom C, 3 atoms O) Add heat to re-arranged the atoms: CaO CO2
Chemical Changes • Evidence of a chemical reaction • Formation of gas • Formation of precipitate • Change in color • Change in energy • Endothermic • Absorbs heat energy (gets cold) • Exothermic • Releases heat energy (gets hot)
Chemical Changes • Chemical reactions can be represented by equations CaCO3 CaO + CO2 Reactants Products
Chemical Changes • Atoms are re-arranged, NOT created or destroyed • Law of Conservation of Matter • Law of Conservation of Mass
Chemical Changes • Matter is conserved type of atoms does not change • Nothing is created or destroyed • Mass is conserved amount of atoms cannot change • Nothing is created or destroyed
Chemical Changes • To show conservation of mass Balance equations • Make sure there are the same number of each type of atom in the products and in the reactants
Balancing Equations • The equation for the burning of methane gas in oxygen is: • CH4 + 2 O2 → CO2 + 2 H2O Subscript Shows # of atoms Coefficient Shows # of molecules
Balancing Equations • No subscript or coefficient is understood to be 1 CH4 + 2 O2 → CO2 + 2 H2O = C1H4 + 2 O2 → C1O2 + 2 H2O1 1 C 1 C 4 H 4 H 4 O 4 O
Periodic Properties Post Lab • Periodic Trends affect properties because of Zeff • Ionization energy • Energy required to remove an electron • Electronegativity • Ability to attract electrons • Atomic Radius • Distance of the valence electrons to the nucleus
Other Periodic Trends • Groups have similar properties • Valence electrons • Members of the same representative family have the same number of valence electrons • Reactivity • Because they have the same number of electrons, they react similarly. • CaCl2, MgCl2, SrCl2, etc. • Density = mass/volume • % error = I actual – theoretical I theoretical Sn = 7.265 Pb = 11.34 Si = 2.33 Ge = 5.323
Solubility Trends • Common in Double Replacement Rxns
Double Replacement Rxns • Two compounds react to form two new compounds • Metal replaces metal • Remember basic formula writing rules Ex: Ca(NO3)2 + Na2CO3 CaCO3 + NaNO3 2 ppt
Some Types of Chemical Reactions Two or more elements or compounds combine to make a more complex substance A + B → AB Compounds break down into simpler substances AB → A + B Occurs when one element replaces another one in a compound AB + C → AC + B AB + CD → AC + BD Occurs when different atoms in two different compounds trade places
Identifying Chemical Reactions S = Synthesis D = Decomposition SR = Single Replacement DR = Double Replacement ____ P + O2 → P4O10 ____ Mg + O2 → MgO ____ HgO → Hg + O2 ____ Al2O3 → Al + O2 ____ Cl2 + NaBr → NaCl + Br2 ____ H2 + N2 → NH3
MOLAR MASS • Molar Mass is shown below the element symbol on the Periodic Table. Units: grams mole • Use molar mass to convert between: Number of Moles Mass (grams) Number of Molecules or Atoms
Molar Mass Examples • sodium bicarbonate • sucrose • NaHCO3 • 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol • C12H22O11 • 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol
Multiply moles by Molar Mass to get Divide the mass (in grams) by Molar Mass to get the number of moles. MASS in GRAMS MOLAR MASS grams/mol Multiply the number of moles by Avogadro’s Constant to get the number of atoms or molecules. MOLES Multiply the moles by the Molar Mass to get the mass (in grams). AVOGADRO’S CONSTANT 6.022x1023 particles/mol MOLECULESATOMS Divide the number of molecules or atoms by Avogadro’s Constant to get the number of moles.
Molar Conversion Examples • How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C
Molar Conversion Examples • How many molecules are in 2.50 moles of C12H22O11? 6.02 1023 molecules 1 mol 2.50 mol = 1.51 1024 molecules C12H22O11
Molar Conversion Examples • Find the mass of 2.1 1024 molecules of NaHCO3. 2.1 1024 molecules 84.01 g 1 mol 1 mol 6.02 1023 molecules = 290 g NaHCO3
Sample Problems – Moles and Atoms • Determine the number of atoms present in 2.50 moles of strontium. • Convert 5.01 x 1024 atoms of strontium to moles of strontium.
Sample Problems – Moles and Mass • Determine the mass in grams of 2.50 moles of strontium. • Determine the number of moles represented by 943.5 grams of strontium.
Sample Problems – Moles, Atoms, Mass • Determine the mass in grams of (exactly) 5 atoms of strontium. • Determine the number of atoms represented by 43.5 grams of strontium.
Percentage Composition • the percentage by mass of each element in a compound
127.10 g Cu 159.17 g Cu2S 32.07 g S 159.17 g Cu2S Percentage Composition • Find the % composition of Cu2S. 79.852% Cu %Cu = 100 = 20.15% S %S = 100 =
28 g 36 g 8.0 g 36 g Percentage Composition • Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. 100 = 78% Fe %Fe = 100 = 22% O %O =
Percentage Composition • How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.852% Cu (38.0 g Cu2S)(0.79852) = 30.3 g Cu
36.04 g 147.02 g Percentage Composition • Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 24.51% H2O %H2O = 100 =
CH3 Empirical Formula • Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts
Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.
1.85 mol 1.85 mol Empirical Formula • Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O
N2O5 Empirical Formula N1O2.5 Need to make the subscripts whole numbers multiply by 2
C2H6 Molecular Formula • “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? molecular formula
Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.
28.1 g/mol 14.03 g/mol Molecular Formula • The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? empirical mass = 14.03 g/mol = 2.00 (CH2)2 C2H4
Using the Basics… The empirical formula of a compound is found to be P2O5. The molar mass of the compound is 284 grams/mole. What is the molecular formula for the compound?
Proportional Relationships • Stoichiometry • mass relationships between substances in a chemical reaction • based on the mole ratio • Mole Ratio • indicated by coefficients in a balanced equation 2 Mg + O2 2 MgO
3 mol 2 mol 1 mol N2 + 3H2 2NH3
1 mol 3 mol 2 mol N2 + 3H2 2NH3
Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio - moles moles • Molar mass - moles grams • Molarity - moles liters soln • Molar volume - moles liters gas • Mole ratio - moles moles Core step in all stoichiometry problems!! 4. Check answer.
Standard Temperature & Pressure 0°C and 1 atm Molar Volume at STP 1 mol of a gas=22.4 L at STP
Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02 1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION
Phosphoric Acid • Phosphoric acid (H3PO4) is one of the most widely produced industrial chemicals in the world. • Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca5(PO4)3F, with sulfuric acid (H2SO4). Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4
Mole Ratio Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4). Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1Moles starting substance: 10.0 mol H2SO4 Step 2The conversion needed is moles H2SO4 moles H3PO4