1. Composition and formulae Of moles and men

2. Learning objectives  Count atoms in formula  Define the mole  Determine numbers of atoms or molecules in molar quantities  Determine molar mass from chemical formula  Determine moles from mass of substance  Perform calculations of:  Percent composition  Empirical formula  Molecular formula

3. Molecules or moles  The numbers (coefficients) in chemical equations can refer to molecules  But for practical applications, we need a more useful number: we cannot count molecules

4. The Mole  The mole is a unit of quantity used in chemistry to measure the number of atoms or molecules  DEFINITION:  The number of atoms in exactly 12 g of 12 C  A mole of anything always has the same number of particles: atoms, molecules or potatoes – 6.02 x 10 23 – Avogadro’s number

5. Atomic and molecular mass  Two scales:  Atomic mass unit scale  Mass of individual atom or molecule in atomic mass units (amu)  Molar mass scale  Mass of mole of atoms or molecules in grams  Confusing?...

6. The Good News  Molar mass in grams of element has same numerical value as mass of atom in amu  Atomic mass of carbon = 12 amu  Molar mass of carbon = 12 g  Formula mass of H 2 O molecule = 18 amu  Molar mass of H 2 O = 18 g

7. Examples  How many moles of Li are in 6.94 g if atomic mass of Li is 6.94 amu?  1.00 mol  What is the molar mass of NH 3 if atomic mass of H = 1 amu and N = 14 amu  17 g/mol

8. Particle – mole conversions

9. Gram – mole conversions

10. Particle – gram conversions

11. Significance of formula unit  Ionic compounds do not contain molecules. Simplest formula is the formula unit  Covalent compounds, the molecular formula is the formula unit

12. Percent composition and empirical formula  Chemical analysis gives the mass % of each element in the compound  Molar masses give the number of moles  Obtain mole ratios  Determine empirical formula

13. Determining percent composition  Percent composition is obtained from the actual masses. Example: Sample contained 0.4205 g of C and 0.0795 g of H. Total mass = 0.5000 g (0.4205 + 0.0795) Therefore: in 100 g there are: (84.10 %) (84.10 %) (15.90 %) (15.90 %) Percent composition: 84.10 % C, 15.90 % H

14. Percent composition from formula  What is percent composition of C 5 H 10 O 2 ?  1 mol C 5 H 10 O 2 contains 5 mol C, 10 mol H and 2 mol O atoms  Mass of each element  Total mass = 102.13 g

15. Convert masses into percents  Percent composition: 58.80 % C + 9.870 % H + 31.33 % O = 100.00%

16. Empirical formula from percent composition: 84.1 % C, 15.9 % H 1.Convert percents into moles 84.10 g of C ≡ 7.00 mol C 15.9 g of H ≡ 15.8 mol H 2.Determine mole ratio Mole ratio H:C =  Simplest formula (decimal form): C 1 H 2.26  Make smallest integers by multiplying C 4 H 9 May require rounding. Errors in real data cause problems  Do percent composition and empirical formula exercises

17. Empirical formula with more than two elements  Percent composition of vitamin C is:  40.9 % C, 4.58 % H, 54.5 % O 1.Convert into moles 2.Determine mole ratios 3.Find lowest whole numbers

18. Inaccuracy can lead to ambiguous or incorrect formulas  What if H:C is 2.20 rather than 2.26? An error of only 3 %  Formula becomes C 5 H 11 rather than C 4 H 9  What if H:C is 2.30 rather than 2.26? An error of only 2 %  Formula becomes C 3 H 7  Sometimes chemical intuition is required: we know there is FeO, Fe 3 O 4 and Fe 2 O 3 ; so a formula FeO 3 would indicate an error

19. Practice empirical formula problem  A compound contains 62.1 % C, 5.21 % H, 12.1 % N and 20.7 % O. What is the empirical formula?

20. Empirical and molecular formula  Percent composition gives the empirical (simplest) formula. It says nothing about the molecular formula.  Molecular formula describes number of atoms in the molecule  May be much larger than the empirical formula in the case of molecular covalent compounds  For ionic compounds empirical formula = “molecular” formula

21. Elements and compounds can have molecular formula different from simplest formula Substance Empirical formula Molecular formula Substance Empirical formula Molecular formula SulphurS S8S8S8S8PhosphorousP P4P4P4P4 BenzeneCH C6H6C6H6C6H6C6H6AcetyleneCH C2H2C2H2C2H2C2H2 Ethylene CH 2 C2H4C2H4C2H4C2H4Cyclohexane C 6 H 12

22. Determination of molecular formula  Require: 1.Empirical formula from percent composition analysis 2.Molar mass from some other source  Number of empirical formula units in molecule:  There are n (A a B b C c ) in molecule:  Molecular formula is A na B nb C nc

23. Molecular formula of vitamin C  Empirical formula of vitamin C is C 3 H 4 O 3  Molar mass vitamin C is 176.12 g/mol  Mass of empirical formula = 88.06 g/mol  (3 x 12.01 + 4 x 1.008 + 3 x 16.00)  Number of formula units per molecule =  Molecular formula = 2(C 3 H 4 O 3 ) = C 6 H 8 O 6

24. Practice molecular formula problem  Ibuprofen contains 75.69 % C, 8.80 % H and 15.51 % O. What is the molecular formula if molar mass is 206 g/mol?

25. Molarity  Concentration is usually expressed in terms of molarity:  Moles of solute/liters of solution (M)  Moles of solute = molarity x volume of solution

26. Example  What is molarity of 50 ml solution containing 2.355 g H 2 SO 4 ?  Molar mass H 2 SO 4 = 98.1 g/mol  Moles H 2 SO 4 =.0240 mol  Volume of solution = 50/1000 =.050 L  Concentration = moles/volume =.0240/.050 = 0.480 M =.0240/.050 = 0.480 M

27. Dilution  More dilute solutions are prepared from concentrated ones by addition of solvent M 1 V 1 = M 2 V 2 Molarity of new solution M 2 = M 1 V 1 /V 2 Molarity of new solution M 2 = M 1 V 1 /V 2 To dilute by factor of ten, increase volume by factor of ten