Lecture (1) Chem 101 2024 PDF

 What is chemistry? ◼Chemistry - study of matter and how it changes Also interested in the energy associated with the change  States of Matter ◼Defined by how a container is filled: 1.Solid – fixed shape Doesn’tconform to a container nor fills it Atoms/molecules: rigid...

 What is chemistry? ◼Chemistry - study of matter and how it changes Also interested in the energy associated with the change  States of Matter ◼Defined by how a container is filled: 1.Solid – fixed shape Doesn’tconform to a container nor fills it Atoms/molecules: rigid & organized 2. Liquid – no specific shape Does conform to a container, somewhat fills it Atoms/molecules: close together & disorganized 3. Gas – no shape Does conform to container, completely fills it Atoms/molecules: far apart & disorganized  Types of Matters 1. Pure substances: have distinct properties & a fixed composition a. Atoms ( smallest building blocks of matter) or elements b. Molecules – 2 or more atoms jointed together  1. Pure Substances A. Elements: can’t decompose into anything smaller Listed on Periodic Table (118 are known) -Carbon =C -Oxygen = O -Cobalt = Co  Pure Substances(Elements) Elements are made up of 1 type of individual atom Copper:  Pure Substances (Compounds) B. Compounds: molecules of 2 or more different elements Combined in a fixed ratio Water (H2O) vs Peroxide (H2O2) 2. Mixtures 10  What is a Solution? ◼ Solution (Soln) – uniform mixture of 2 or more substances Soln = solute + solvent Solute - usually present in a smaller amounts It is being dissolved Solvent - usually present in a larger amounts In aqueous solutions it is H2O 11 A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass? 17  Example ◼ What is the mass of one atom of calcium? Solution: 1 mol Ca = 6.022 × 1023 atoms 1 mol Ca = 40 g Ca 1 atom Ca = 40/ 6.022 × 1023 = 6.66 × 10-23 Mass of 1 atom Ca = atomic weight / NA 19 20 21 22 23 24 25 Structural formula: is the formula that represent the chemical bonds between the atoms in the molecule. 26 ►Acetic acid: E. F. M. F. S. F. O H CH2O C2H4O2 C H C O H H Proplem A sample of a brown- colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the simplest formula of the compound? 1 mole of N 14.0 ? 2.34 Number of moles of N = 1 x 2.34 / 14.0 = 0.167 mole 1 mole of O 16.0 ? 5.34 Number of moles of O = 5.34 / 16.0 = 0.334 mole N: O 0.167 : 0.334 N: 2 NO2  Proplem : Colored gas is used in rocket engines , with empirical formula is NO2 and has a molecular weight of 92.0 gm. What is the molecular formula ? Ratio (x)= Mol. Wt. / E. F. Wt. Molecular weight of empirical formula= 14 + 2 x 16= 46 g/mol Ratio = 92 / 46 = 2 Molecular Formula (M.F.) = x * empirical formula M. F. = 2 x NO2= N2O4 30 31 Stoichiometry  Stoichiometry & Solutions ◼ Stoichiometry – mole relationships between 2 DIFFERENT chemicals in a balanced equation ◼ Steps for Stoichiometry: 1. Correctly balance the equation. 2. Convert your given chemical into mols (mol:mol ratio) 3. Convert to mols of desired chemical 4. Convert to correct units 33  Mole ratio Reaction between magnesium and oxygen to form magnesium oxide. 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1 2  Mole to Mole conversions  How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2Al2O3 → 4Al + 3O2 2mole : 3mole 3.34 : ? No. of moles = 3 x 3.34 / 2 = 5.01 mole of O2 A can of butane lighter fluid contains 1.20 moles of butane (C4H10). Calculate the number of moles of carbon dioxide given off when this butane is burned. Equation of reaction 2C4H10 + 13O2 8CO2 + 10H2O Mole ratio 2C4H10 : 8CO2 C4H10 4CO2 1 4 1.2 : ? By cross-multiplication X = 1.2 x 4 / 1 = 4.8 moles of CO2 given off Problem : 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? [k = 39, Cl = 35.5, O = 16] 2 KClO3 2 KCl + 3 O2 2 : 3 1.50 : X X = 3 x 1.50 / 2 = 2.25 mole Convert to mass 2.25 mol x 32.0 g/mol = 72.0 grams  We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required? [k = 39, Cl = 35.5, O = 16] Soln: 2 KClO3 2 KCl + 3 O2 2 : 2 X :2.75 X = 2 X 2.75 / 2 = 2.75 mol In mass: 2.75mol X 122.55 g/mol = 337 grams  Note: The mass of the reactants must equal the mass of the products. 2H2 + O2 → 2H2O 2x2 g H2 + 32O2 = 36.00 g H2O 36.00 grams reactant = 36.00 grams product  Practice 2C2H2 + 5 O2 → 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? (9.6 mol) How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol) If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol) Assignment 1A ◼ If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO4 → Fe2(SO4)3 + 3Cu 2x56 : 3x63.5 10.1: ? Number of grams of Cu = 3 x 63.55 x 10.1 / 2 x 56 = 17.24 g Cu ◼ Calculate mass of iron in a 10.0 g sample of rust, Fe2O3 1 mol Fe2O3 → 2 mol Fe 2 x 56 + 3 x 16 : 2 x 56 160 g/mol : 112 g/mol 10 g : X X = (10 × 112) / 160 = 7.0 g Fe  Sample Problem ◼ How many mols of arsenic(III) sulfide will form when 0.3000L of 2.12 M sodium sulfide reacts with 19.57g of arsenic(III) chloride? 2 AsCl3 (aq) +3Na2S(aq)→ As2S3 (s) + 6 NaCl (aq) 19.57g v = 0.3000L mols = ? M = 2.12mol/L LIMITING REACTANT!? info about BOTH reactants & asked about product 43  4.16: Moles Produced? 2 AsCl3 (aq) +3Na2S(aq)→ As2S3 (s) + 6 NaCl (aq) A. 0.05398 19.57g v = 0.3000L mols = ? B. 0.158 M = 2.12mol/L C. 0.212 D. 0.266 19.57g AsCl3  1mol AsCl3 1mol As 2 S3    = 0.05398mols As2 S3  181.27 gAsCl3 2mols AsCl3 0.3000L Na2 S 2.12mols Na2 S 1mol As 2 S3    = 0.212mols As2 S3  1L 3mols Na2 S 4.14: What start with? A. Volume B. Molarity 4.15: where put 2.12? A. Top B. Bottom 44 Assignment 2A  What is a Solution? ◼ Solution (Soln) – uniform mixture of 2 or more substances Soln = solute + solvent Solute - usually present in a smaller amounts It is being dissolved Solvent - usually present in a larger amounts In aqueous solutions it is H2O 46 47  Concentration ◼ Molarity (M) – mols of solute per liter of solution mols M= L ◼ Many ways of expressing molarity unit: A Solution of 0.901 mols NaCl per liter of water is: 0.901 ________ mols/L or M 0.901 ______ or 0.901 _______ molar 48  Sample Problem ◼ Calculate the molarity of 1.42g of Al(NO3)3 in 15.4mL of water. mols M= L  1molAl ( NO3 ) 3  1.42 gAl ( NO3 ) 3   = 0.00667molAl ( NO3 ) 3  213.01gAl ( NO3 ) 3  15.4mL = 0.0154 L 0.00667mols M= = 0.433M 0.0154L 49  Limiting Reactant Is the reactant that disappear firstly from the reaction. Proplem Zinc & sulfur react to form zinc sulfide, a substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is: Zn + S → ZnS 1. How many grams of ZnS can be formed when 12.0 gm of Zn are allowed to react with 6.5 gm of S ? 2.How much of which elements will remain unreacted ? Solution No. of mole of Zn = 1 x 12 / 64.4 = 0. 183 mole No. of moles of S = 1 x 6.5 / 32.1 = 0. 202 mole Zn + S → ZnS 1 : 1 1 0.183 0.183 0.183 1 mole ZnS → 97.5 gm ZnS 0.183 mole → ? No. of gm of ZnS = 0.183 x 97.5 / 1 = 17.8 gm  Zn is the limiting reactant because it is disappearing firstly from the reaction. # of unreacted moles of S = Total number of moles-Number of reacted moles 0.202 – 0.183= 0.019 moles No. of grams of unreacted = # of unreacted moles of S*At.Wt.= 0.019 x 32.1 = 0.16 gm  ►What is Yield?  Yield is the amount of product made in a chemical reaction.  There are three types: 1. Actual yield (Experimental)- what you actually get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield = [Actual / Theoretical] x 100 Zn + S → ZnS Theoretical yield= 17.8 g Actual yield = 15 g Percent yield= (15/17.8) *100