Summary Foundations of Chemistry Sheet

MOLE FRACTION (w/w%) Mole fraction of carbon tetrachloride (CCl4) if 3.47 moles dissolved in solution w 8.54 moles of benzene (C6H6)? MOLES à MOLALITY 15.1g NaCl = 0.258 moles. Solvent = 970g. What molality? =0.266mol/kg **molality should be close to molar mass/moles Total molality of ions in solution If you dissolve 2g of Ca(NO3)2 in 750g of water, what is the total molality of ions in solution? (Assume total dissociation) - Find molar mass - convert mass Ca(NO3)2 into moles - divide that by mass of water in kgs 2g / 164g/mol = 0.0122 mol / 0.750 kg = 0.0163 molal Since Ca(NO3)2 produces 3 ions on dissolving, the molality of ions will be x3: Molality of ions = 3 x 0.0163 = 0.0489 molal Mole fraction of 25.7g CH2O dissolved in 3.25 moles CCl4? 1. convert mass (g) to moles of CH2O TITRATION FORMULA half equivalence point = to pKa of weak acid Conc. of 25mL NaOH if 35mL of 1.25M HCl is needed to titrate MOLARITY (mol/L) Molarity if 1.47 moles sugar dissolved in 2.31L solution? equivalence pt? 1. balanced equation means 1:1 use equation above How many g NaCl needed to make 1.5L aqueous 0.500M solution? 1. solve for moles NaCl first then convert to grams after 2. convert moles to grams using molar mass 10g KCL is dissolved in 250mL distilled water, calc molar concentration (aka molarity) 1. find molar mass of KCl from periodic table 0!1 M= = Psolution = vapour pressure (VP) of solution Xsolvent = mole fraction of solvent in solution P0solvent = VP of pure solvent at standard conditions What is VP of solution with 3.5 moles of glucose in 10 moles of water? (VP of pure water at 100oC is 101.325kPa) 2. solve for Raoult’s law: Psolution = 0.741 x 101.325kPa= 75kPa What is VP when 25.5g of glucose is dissolved in 212g water? 1. must convert (g)glucose and (g)water to moles 3. solve for Raoult’s law: Psolution= 0.988 x 101.325kPa=100kPa EFG : H&I&JG &I&&KLMNOPQ)*+ DILUTION FORMULA If 234mL of 2.51M aq solution is diluted by adding water to give final vol of 365mL,what is M of final solution? 1. use equation above: 2.51M x 234mL = M2 x 356mL 2. rearrange… #"$0Y&4&#D3;5 = M2 = 1.65M **Kf =freezing pt depression constant for H2O=1.86K kg/mol ETf = 1.86K kg/mol x m of solution = …K (this is decrease E6 Tf = 100oC - E 1.6 g naphthalene (C10H8) is dissolved in 20g of benzene. Freezing point of benzene is 5.5oC, and freezing point of solution is 2.8oC. What is the molal freezing point depression constant, Kf of benzene? V" EFG = (Kf of pure solvent) x (Kf of solution) = 5.5 – 2.8 = 2.7oC D$X;5 What vol water must be added to 345mL of 1.75M solution in order to dilute to 1.25M? 1. use equation above: 1.75M x 345mL = 1.25M x V2 0"-$Y&4&D3$;5 2. rearrange… = V2 = 483mL 0"#$Y 3. How much needs to be added!! 483mL–345mL = 138mL 2. calc m concentration of the solution ( MASS FRACTION/PERCENT (no units) ) ß mass fraction of solute only. Can do mass fraction of solvent moles naphthalene = ( pH FORMULA. pH +pOH = 14 pH = -log[H+] [H+] = 10-pH pOH = -log[OH-] [OH-] = 10-pOH Calc pH of HCL in water as concentration of 0.0050mol/L? 1. determine strong/weak acid or strong/weak base 2. HCl strong so: 3. 1:1 ratio so pH= -log[0.0050] = 2.3 CALC SOLUBILITY PRODUCT FROM CONCENTRATION Ksp = ? Ca(OH)2 4.9g in 6L H2O 2. Balance equation Ca(OH)2 e Ca2+ + 2OH-. Ksp = [Ca2+] x [OH-]2 3. Stoichiometry for solution. Ca2+ only come from solute [Ca2+] = c. OH- come from H2O as well [OH-] = 2c. c x (2c)2 = 4c3 = 4 x molar concentration (mol L-1)3 = Ksp 3. calc Kf of solution (2.7oC ÷ 0.625m) = 6.3oC = -6.3oC Calculate change in freezing point: DT= Kf x i x msolute MOLAR SOLUBILITY (IGNORE SOLIDS) Calc. molar solubility of PbCl2 (Ksp= 1.6x10-5) in a 0.1M solution of KCl. Common ions??? Initial solution contains K+ and ClPbCl2 is a salt and sparingly soluble. PbCl2 e Pb2+ + 2ClInitial conc (M) 1 0 0.1M Change (M) -x +x +2x Equilibrium conc (M) 0 x (0.1+2x) Ksp= [Pb2+]1 x [Cl-]2 à1.6x10-5= X1 x (0.1M x 2x)2 millionaire rule à 1.6x10-5 = X1 x (0.1M)2 à1.6x10-5 = X x (0.01M) …solve for X 1.6x10-5 Ö 0.01M = X = 0.0016M If Ksp = [3x]3 x [x] à 27x3 x [x] à 27x4 à rearrange à or CONDUCTIVITY Calculate molar conductivity from given data: 1. write balanced stoich equation to found out how many of given data for ions are needed for total electrolyte 2. cancel any common species!! 3. calculate conductivity from data & stoich (mS m2 mol-1) MASS à MOLES 15.1g NaCl how many moles? Calc molar mass à Nam (22.990g/mol) + Clm (35.45g/mol) = 58.44g/mol 0$"01 moles = = = 0.25838mol 2 6 Calculate pressure/volume work Vol change = DV = Vfinal - Vinitial (convert from L àm3) L Ö 1000 = m3 i.e. 10LÖ1000 = 0.01m3 Work = -Psurroundings x DV (convertà) atm x 101.325 = kPa i.e. 6atm x 101.325 = 607.95kPa Calculate heat change. Temp change =D t = tfinal - tinitial q = n(no moles) x c(cpm) x Dt IF cpm is given OR q = m(mass) x c(cps) x Dt IF cps is given Cpm = molar heat capacity J K-1 mol-1. Cps = specific heat capacity J K-1 kg-1 (or g-1 convert!) Newman Projections…eclipsed + staggered Organic Acids + Bases: Staggered: lower energy, more stable à Resonance: more resonance = more delocalisation of Eclipsed: higher energy, less stable à electrons = greater stability = lower reactivity Put higher atomic no closer for ^^^ energy Aromatic rings: increase acidity Amides/Amino Acids/Proteins: Order of acidity: amine, alcohol, carboxylic acid&& -Classification of an amide by how many H &&&½&Base w resonance: stronger base groups attached to the N &&&½ Acid w resonance: weaker acid Functional groups of an amino acid: amine + Electronegativity: ^^^ = more acidic carboxylic acid (bonded by peptide bonds) Atomic size: ^^^ = more acidic -Protein functions: enzymes, structural Orbital Hybridization: sp3 à sp2 àsp …increasing acidity Can be simple, conjugated, globular, fibrous Lipids à any molecule that is fat soluble, can be extracted by non-polar solvents Structure: Carboxylic acids with long hydrocarbon chain. Most common is triglycerol à glycerol + 3 fatty acids Carbohydrates: Compound where intramolecular reaction can occur between an aldehyde + alcohol Structure: sugars = polyhydroxy aldehydes and ketones à lots of OH groups +++ TYPE 1:1 1:1 EXAMPLE AgBr Hg2Cl2 CATION c 2c ANION C 2c Ksp C2 16c4 3 1: 2 2:1 Li2CO3 Ca(OH)2 2c c c 2c 4c 2:2 BaSO4 c c C2 1:3 3:1 Li3PO4 Fe(OH)3 3c C c 3c 27c4 2:3 3:2 Mg3(PO4)2 Y2(CO3)3 3c 2c 2c 3c 108c5 3:3 AlPO4 c C C2 C ¶•¨ž · h · ´ •¨ž ¸ •¨ž ¹ ¶•¨ž · h •¨ž ¸º •¨ž · V»µ ¼ ¶•¨ž A chemist wants to dilute 14.0 M HCl to make 600.0 mL of 3.5 M HCl solution. How will he do this? The Chemist would add 14 M HCl to 150 mL water slowly with constant stirring to make 600 mL of 3.5M HCl. Formula used is C1V1 = C2V2 Saponification rx of a triglyceride (ester) = glycerol + the salts of fatty acids à the salts are soap molecules. In why sugars can be both chain & ring form à water they form micelles due to IMFs à can arrangement of 5/6 membered chain allows encapsulate non-polar subs to make water-soluble reversible intramolecular cyclisation Reactions: Substitution: 1 bond broken @ carbon while another 1 forms Addition: break C=C multiple bond, form 2 new sigma bonds Saponification: formation of soap molecules, base catalysed hydrolysis of esters Esterification: formation of an ester by à carboxylic acid + alcohol = ester Condensation: 2 molecules combine + a small stable molecule eliminated OIL RIG: oxidation is losing, reduction is gaining Nucleic Acids Nucleosides = nitrogen base + pentose sugar Nucleotides = nitrogen base, pentose sugar + phosphate group Ribose v Deoxyribose = has an Oxygen in it Adenine and guanine = purine (2 rings) Thymine, cytosine, uracil = pyrimidine (1 ring) T-A. G-C