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

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

3. Careful experimentation lead Proust to demonstrate H 2(g) + Cl 2(g) 2HCl (g) H 2 SO 4(l) + 2NaCl (s) 2HCl (g) + Na 2 SO 4(aq) Proportions by mass of elements in a compound VARY OVER A CERTAIN RANGE Proportions by mass of elements in a compound ARE FIXED. VARIATIONS ARE DUE TO IMPURITIES. Claude Berthollet: Joseph Proust: 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.”

4. Wüstite, an iron oxide whose simplest formula is FeO, with 77.73%Fe. All gaseous compounds OBEY THE LAW OF DEFINITE PROPORTIONS. Certain SOLIDS are exceptions of the Law of Constant Composition: NON STOICHIOMETRIC COMPOUNDS (BERTHOLLIDES) 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.

5. The composition of a compound is shown by its CHEMICAL FORMULA. C + O 2 A B CHEMICAL ANALYSIS: If A is CO then B = CO 2 For a FIXED mass of C the ratio of O in A and B is: If A is CO 2 then B is C 2 O 4 Let’s take the elements C and O: (1.000 g C and 1.333 g O) (1.000 g C and 2.667 g O) 1.333 : 2.667 or 1: 2 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 Types of Formulas Composition

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

8. Types of Formulas Empirical Molecular Structural CH 3 C2H6C2H6

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

10. Masses of Individual Atoms The atomic masses of H, Cl, and Ni are H = 1.008 amu Cl = 35.45 amu Ni = 58.69 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 = 6.022 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

12. Mass of H atom: 1 H atom x= 1.674 x 10 –24 g Number of atoms in one gram of nickel: 1.00g Ni x= 1.026 x 10 22 atoms Masses of Individual Atoms 1.008g H 6.022 x 10 23 atoms 6.022 x 10 23 atoms Ni 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) = 2.016 g/mol 1O = 1(16.00 g/mol) = 16.00 g/mol 18.02 g/mol = molar mass of water

14. The Mole Meaning Molar Mass Mole - Mass Conversions

15. Meaning 1 mol = 6.022 x 10 23 items Cl 2 HCl H Cl 6.022 x 10 23 molecules 6.022 x 10 23 atoms 70.90 g Cl 2 36.46g HCl 1.008g H35.45g Cl 1 mol Cl 2 1 mol HCl1 at-gr H1 at-gr Cl 1 molar mass Cl 2 1 molar mass HCl1 molar mass H 1 molar mass Cl

16. Molar Mass Generalizing from the previous examples, the molar mass, M, is numerically equal to the formula mass

17. Mole-Mass Conversions 110.98g CaCl 2 1 mol CaCl 2 mass = 13.2 mol CaCl 2 x = 1.47 x 10 3 g Calculate mass in grams of 13.2 mol CaCl 2 Calculate number of moles in 16.4g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180.18g C 6 H 12 O 6 moles = 16.4g C 6 H 12 O 6 x = 9.10 x 10 -2 mol

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

19. Mass % from Formula Percent composition of potassium dichromate, K 2 Cr 2 O 7 ? molar mass K 2 Cr 2 O 7 = (78.20 + 104.00 + 112.00)g/mol = 294.20g/mol 78.20 294.20 %K = x 100 = 26.58% 112.00 294.20 %O = x 100 = 38.07% Note that percents must add to 100 104.00 294.20 %Cr = x 100 = 35.35%

20. % Composition from Experimental Data q Calculate mass of compound formed q Divide mass of each element by total mass of compound and multiply by 100. Aluminum chloride is formed by reacting 13.43 g aluminum with 53.18 g chlorine. What is the % composition of the compound?

21. Empirical Formula from % Composition Empirical formula of compound containing 26.6% K, 35.4% Cr, 38.0% O moles K = 26.6g x = 0.680 mol K 1 mol 39.10g moles Cr = 35.4g x = 0.681 mol Cr 1 mol 52.00g work with 100g sample:26.6 g K, 35.4 g Cr, 38.0 g O

22. Empirical Formula from % Composition moles O = 38.0g x = 2.38 mol O 1 mol 16.00g Note that 2.38 / 0.680 = 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 1.000 g burns to give 1.446 g CO 2 and 0.6001 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 2.02g H 18.02g H 2 O mass H = 0.6001g H 2 O x= 0.0673g H mass O = 1.00g – 0.394g – 0.067g = 0.539g O

25. Simplest Formula from Analytical Data 1 mol C 12.01g C moles C = 0.394g C x = 0.0328 mol C 1 mol H 1.008g H moles H = 0.0673g H x= 0.0668 mol H 1 mol O 16.00g O moles O = 0.533g O x = 0.0333 mol O Empirical formula is CH 2 O

26. Molecular Formula from Empirical Formula Must know molar mass Calculate empirical and molecular formulas of a compound that contains 80%C, 20%H, and has a molar mass of 30.00 g/mol.

27. Molecular Formula from Empirical Formula Divide each value by smaller number of moles Empirical Formula: CH 3 Molecular Formula: (CH 3 ) 2 = C 2 H 6