B3053 CH4701 2C Wed 10 am CH4701 2B Wed 3 pm CH4711 2A Th 10 am CH4721 2A Th 1 pm CH4701 2A Fri 10 am

Example What mass of NaOH and HCl are require to make g of NaCl? NaOH + HCl ——>NaCl + H 2 O(1) Atomic masses: Na a.m.u., Cl a.m.u., O a.m.u. and H a.m.u. 1 molecule NaOH + 1 molecule HCl ——> 1 molecule NaCl + 1 molecule H 2 O 1 mole NaOH + 1 mole HCl ——>1 mole NaCl + 1 mole H 2 O x mole NaOH + x mole HCl ——>x mole NaCl + x mole H 2 O where x is the number of moles present in g NaCl Molecular mass of NaCl = = a.m.u. 1 mole of NaCl has a mass of g

X = 1 x 29.22/58.44 = 0.5 mole Calculate the mass of 0.5 mole of NaOH and of HCl Mass of 0.5 mole of NaOH = 0.5 x ( ) = 20.00g Mass of 0.5 mole HCl = 0.5 x ( ) = g Answer: g NaCl and g HCl.

Q Exam The alcohol content of a beer is 5% by weight. How many moles of ethanol (C 2 H 5 OH) are there in 500 g of beer? (a)5.4(b) 0.54(c) 0.44(d) 5 (b)none of the above Mass of alcohol = 0.05 x 500 g = 25 g C 2 H 5 OH mass of one molecule = (2 x 12) + (5 x 1) = 46 amu Mass of one mole of ethanol = 46 g mol -1 # of moles = 25/46 g/ g mol -1 = 0.54 moles

Early Chemical Concepts and Their Present Day Uses

Basic idea of the atom: Democritus (400 B.C.) Alchemists (Middle Ages) Lavoisier ( ). Viewed the notion of atoms from a philosophical point of view, i.e. did not try to prove their existence. Robert Boyle ( ) Disproved Aristotelian approach (based on a priore theories). Introduced the “Scientific Method” as the underlying philosophy of science.

Observation The Scientific Method

Observation Hypothesis/Theory Prediction Experiment Do results agree with theory? yes Scientific Law Modify theory no The Scientific Method

John Dalton ( ) Used the Scientific Method to develop a basic atomic theory: 1. Matter is made up of atoms which are indivisible and indestructible. 2. Atoms of a given element are identical both in mass and in chemical proportions. 3. Atoms of different elements have different masses and different chemical properties. 4. Atoms of different elements combine in simple whole numbers to form compounds. 5. When a compound is decomposed, the recovered atoms are unchanged and can form the same or new compounds.

Gay-Lussac ( ) “Gases react in simple whole number units of volume, and where the products are gases they also have simple whole number units of volume.” Thus: 2 volumes H volume O 2 2 volumes H 2 O 3 volumes H volume N 2 2 volumes NH 3 Note: this relationship is only true if pressure and temperature are constant throughout.

Avogadro ( ) Avogadro’s Hypothesis (1811): "Equal volumes of gases under the same conditions of temperature and pressure contain the same number of particles (i.e. atoms or molecules)." Volume of 1 mole of any gaseous pure substance at STP (273.15K, kPa) = dm 3 = Gas Molar Volume Thus for any gas at STP: 1 mole (= atomic or molecular mass in grams) occupies dm 3. Examples:1 mol of H 2 (M.M.= 2.02) has a mass of 2.02 g and at STP occupies 22.41dm mol of CO (M.M.= 28.01) has a mass of g and at STP occupies dm 3.

Avogadro's Hypothesis implies gas density is proportional to molecular mass. Allowed the early chemists to deduce molecular mass from gas density measurements.

Avogadro's Hypothesis implies gas density is proportional to molecular mass. Allowed the early chemists to deduce molecular mass from gas density measurements. Cannizzaro (1858) Put all of the foregoing hypotheses together into a method for determining accurate atomic and molecular masses, but only for gaseous compounds. Postulated that over a large number of compounds of a particular element, at least one will have only one atom of the element per molecule. Mass of the element in 22.4 L of this compound is the atomic mass.

CompoundMass of Mass of contained gas/ g hydrogen/ g Hydrogen (H 2 )22 Methane (CH 4 ) 164 Ethane (C 2 H 6 )306 Water (H 2 O)182 Hydrogen sulfide (H 2 S)342 Hydrogen cyanide (HCN)271 Hydrogen chloride (HCl)361 Ammonia (NH 3 )173 Pyridine (C 5 H 5 N)795 Minimum mass of hydrogen = 1 g Atomic mass of hydrogen = 1 a.m.u.

Dulong and Petit (1819) Found a correlation between atomic mass and specific heat of solid elements: Atomic mass x specific heat  25 JK -1 mo1 -1 where atomic mass is in grams per mole and specific heat in JK -1 g -1. ElementExperimental values of heat capacity Al24.3 J K -1 mol -1 Fe25.2 Ni26.0 Ag25.5 Au25.2 Pb26.8 For Bi, heat capacity = J => atomic mass of 211 a.m.u. measured atomic mass = 209 a.m.u.

Determination of Empirical and Molecular Formulae Empirical formula: the relative number of atoms of each element in a molecule or formula unit Molecular formula: the actual number of atoms of each element in a molecule. For ethane: molecular formula = C 2 H 6, empirical formula = CH 3. Elemental composition data Empirical formula Molecular formula Molecular mass

Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element. 3. Divide each value in point 2 by the smallest of them. 4. If necessary, multiply all the numbers by the smallest integer possible to obtain approximate whole number values for each element. 5. Write these whole numbers as subscripts of the element symbols to give an empirical formula. 6. If the molecular weight is known, determine how many empirical formula weights are required to obtain the molecular weight. Use this factor to multiply the number of atoms of each element in the empirical formula to give the molecular formula.

A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is g mol Mass % of carbon = 3.27g/ 8.00 g x 100 = 40.9% Mass % of hydrogen = g/ 8.00 g x 100 = 4.58% Mass % of oxygen = 4.36 g/ 8.00 g x 100 = 54.5% In 100 g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. # of moles C = 40.9 g /12.01 g mol -1 = 3.41 mol # of moles H = 4.58 g /1.008 g mol -1 = 4.54 mol # of moles O = 54.5 g /16.00 g mol -1 = 3.41 mol C: 4.54 H : 3.41 O divide by 3.41 =>1 C: 1.33 H : 1.33 O 4. Multiply by 3 3 C: 4 H : 3 O

5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x x x = g mol # of empirical formula units per molecule = / g g -1 mol mol -1 = 2 Molecular formula is C 6 H 8 O 6 The molar mass of oxalic acid is 90 g mol -1 and its empirical formula is CHO 2. What is its molecular formula? g mol Empirical formula is CHO 2. Molar mass of empirical formula: x = g mol # of empirical formula units per molecule = 90/ g g -1 mol mol -1 = 2 Molecular formula is C 2 H 2 O 4

Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = Multiply by 2 to obtain approx. whole nos.: N2, H8, Cr2, O7 5. Write as subscripts: N 2 H 8 Cr 2 O 7

Q 14, Germanium has five naturally occurring isotopes: 70 Ge, 20.5%, a.m.u.; 72 Ge, 27.4%, a.m.u.; 73 Ge, 7.8%, a.m.u.; 74 Ge, 36.5%, a.m.u., 76 Ge, 7.8%, a.m.u. What is the atomic weight of germanium? (a)73.42 (b) (c) (d) (e) = ( x x x x x 7.8) / 100 = a.m.u.

Atoms and Molecules- Summary: 1. Atomic structure, ions, isotopes. 2. Atomic number, mass no., atomic mass, chemical atomic weight, molecular weight. 3. Mass- no. of particles for practical use, the mole, Avogadro’s number. 4. Dalton atomic theory and early ideas on atoms and molecules. 5. Volume relationships in gas phase reactions (Gay-Lussac). 6. Volume - no. of particle - mass relationships in gases (Avogadro’s hypothesis). 7. Determination of empirical and molecular formulae.