Quantitative Composition of Compounds. Preparation for College Chemistry Luis Avila Columbia University Department of Chemistry

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1 Quantitative Composition of Compounds Preparation for College Chemistry Luis Avila Columbia University Department of Chemistry

2 Depending upon Bonding type Compounds Molecular (Covalent bonds) Ionic (Coulombic forces) Molecules Cations Anions

3 Claude Berthollet: Proportions by mass of elements in a compound VARY OVER A CERTAIN RANGE Joseph Proust: Proportions by mass of elements in a compound ARE FIXED. VARIATIONS ARE DUE TO IMPURITIES. Careful experimentation lead Proust to demonstrate THE LAW OF DEFINITE PROPORTIONS (CONSTANT COMPOSITION): The proportions by mass of the elements in a compound ARE FIXED, and do not depend on its mode of preparation. H 2(g) + Cl 2(g) 2HCl (g) H 2 SO 4(l) + 2NaCl (s) 2HCl (g) + Na 2 SO 4(aq)

4 Certain SOLIDS are exceptions of the Law of Constant Composition: NON STOICHIOMETRIC COMPOUNDS (BERTHOLLIDES) Wüstite, an iron oxide whose simplest formula is FeO, with 77.73%Fe. Its composition truly ranges from Fe 0.95 O (76.8% Fe) to Fe 0.85 O (74.8% Fe) depending of the method of preparation. All gaseous compounds OBEY THE LAW OF DEFINITE PROPORTIONS.

5 The composition of a compound is shown by its CHEMICAL FORMULA. Let s take the elements C and O: C + O 2 A CHEMICAL ANALYSIS: (1.000 g C and g O) C + O 2 B (1.000 g C and g O) For a FIXED mass of C the ratio of O in A and B is: : or 1: 2 If A is CO then B = CO 2 If A is CO 2 then B is C 2 O 4 We are unable to say which one is the right formula, but we know the ratio C : O is the QUOTIENT OF INTEGERS.

6 Molecules Composition Types of Formulas

7 Composition Held together by covalent bonds Usually made up of nonmetal atoms

8 Types of Formulas Empirical CH 3 Molecular C 2 H 6 H H Structural H C C H H H

9 Atomic and Formula Masses Masses of Individual Atoms Meaning of Atomic Masses Formula Mass

10 Masses of Individual Atoms The atomic masses of H, Cl, and Ni are H = amu Cl = amu Ni = amu Therefore 1.008g H, 35.45g Cl, and 58.69g Ni all have the same number of atoms: N A N A = Avogadro s number = x 10 23

11 Meaning of Atomic Masses Give relative masses of atoms based on C 12 scale The Most common isotope of carbon is assigned an atomic mass of 12 amu. The amu is defined as 1/12 of the mass of one neutral carbon atom 1amu = 1dalton = 1 Ê Á 12g 12 6 C 12 mol 12 6 C 1mol 12 6 C Ë atoms 6 12 C ˆ = g/ atom 6 12 C

12 Masses of Individual Atoms Mass of H atom: 1 H atom x 1.008g H x atoms = x g Number of atoms in one gram of nickel: x g Ni x 23 atoms Ni = x atoms 58.69g Ni

13 Formula Mass The formula for water is H 2 O. What is its molar mass? 2H = 2(1.008 g/mol) = g/mol 1O = 1(16.00 g/mol) = g/mol g/mol = molar mass of water

14 The Mole Meaning Molar Mass Mole - Mass Conversions

15 Meaning 1 mol = x items Cl 2 HCl H Cl x x x x molecules molecules atoms atoms g Cl g HCl 1.008g H 35.45g Cl 1 mol Cl 2 1 mol HCl 1 at-gr H 1 at-gr Cl 1 molar mass Cl 2 1 molar mass 1 molar mass 1 molar mass Cl HCl H

16 Molar Mass Generalizing from the previous examples, the molar mass, M, is numerically equal to the formula mass formula mass molar mass CaCl amu g/mol C 6 H 12 O amu g/mol

17 Mole-Mass Conversions Calculate mass in grams of 13.2 mol CaCl 2 mass = 13.2 mol CaCl g CaCl 2 2 x = 1.47 x 10 3 g 1 mol CaCl 2 Calculate number of moles in 16.4g C 6 H 12 O 6 moles = 16.4g C 6 H 12 O 6 x 1 mol C 6 H 12 O g C 6 H 12 O 6 = 9.10 x 10-2 mol

18 Calculating Composition % Composition from Formula % Composition from Experimental Data Empirical Formula from % Composition Molecular Formula from Empirical Formula

19 Mass % from Formula Percent composition of potassium dichromate, K 2 Cr 2 O 7? molar mass K 2 Cr 2 O 7 = ( )g/mol = g/mol %K = x 100 = 26.58% %Cr = x 100 = 35.35% %O = x 100 = 38.07% Note that percents must add to 100

20 % Composition from Experimental Data Aluminum chloride is formed by reacting g aluminum with g chlorine. What is the % composition of the compound? q Calculate mass of compound formed gal gcl = galcl 3 q Divide mass of each element by total mass of compound and multiply by 100. Ê gal Á Ë g AlCl 3 ˆ 100 = 20.16%Al Ê Á Ë gcl ˆ 100 = 79.84%Cl g AlCl 3

21 Empirical Formula from % Composition Empirical formula of compound containing 26.6% K, 35.4% Cr, 38.0% O work with 100g sample:26.6 g K, 35.4 g Cr, 38.0 g O moles K = 26.6g x 1 mol 39.10g = mol K moles Cr = 35.4g x 1 mol 52.00g = mol Cr

22 Empirical Formula from % Composition moles O = 38.0g x 1 mol 16.00g = 2.38 mol O Note that 2.38 / = 3.50 = 7 / 2 Empirical formula: K 2 Cr 2 O 7 Potassium Dichromate?

23 Empirical Formula from Analytical Data A sample of acetic acid (C, H, O atoms) weighing g burns to give g CO 2 and g H 2 O. Empirical formula? Solution: find mass of C in sample (from CO 2 ) find mass of H in sample (from H 2 O) find mass of O by difference

24 Empirical Formula from Analytical Data mass C : gco 2 1mol CO g CO 2 1molC 1mol CO gC 1molC = 0.394gC mass H = g H 2 O x 2.02g H 18.02g H 2 O = g H mass O = 1.00g 0.394g 0.067g = 0.539g O

25 Simplest Formula from Analytical Data moles C = 0.394g C x 1 mol C = mol C 12.01g C moles H = g H x moles O = 0.533g O x 1 mol H 1.008g H 1 mol O 16.00g O = mol H = mol O Empirical formula is CH 2 O

26 Molecular Formula from Empirical Formula Must know molar mass n = molar mass mass of empirical formula = number of empirical formula units Calculate empirical and molecular formulas of a compound that contains 80%C, 20%H, and has a molar mass of g/mol. C : Ê Á 80.0 gc Ë 1 mol C atoms g ˆ = mol C

27 Molecular Formula from Empirical Formula H : Ê Á 20.0 gh Ë 1 mol H atoms g ˆ = mol H Divide each value by smaller number of moles H : = 2.97 ~ 3.00 C : = 1.00 Empirical Formula: CH 3 n = molar mass mass of empirical formula = 30.0g 15.01g ~ 2 Molecular Formula: (CH 3 ) 2 = C 2 H 6

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