Composition and formulae. Of moles and men

Transcription

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 Counting particles: 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 Avogadro s number

5 Mole conversions

6 Weighing molecules and moles Two scales: Atomic mass unit scale The mass of an individual atom or molecule in atomic mass units (amu) Molar mass scale The mass of a mole of atoms or molecules in grams Confusing?

7 The Good News The mass of a single atom or molecule in amu has same numerical value as molar mass in grams The atomic mass of carbon is 12 amu The molar mass of carbon is 12 g/mol The same is true for molecules and compounds The formula mass of H 2 O is 18 amu (1+1+16) The molar mass of H 2 O is 18 g/mol

8 Calculations with molar mass Moles = mass molar mass How many moles are in g of lithium if the atomic mass of Li is 6.94 amu? 2.00

9 Particle mole conversions Moles (mol) = No particles 1 mol x 10 ( ) 23 1 No particles = Moles (mol) x 6.02 x 10 ( ) mol

10 Gram mole conversions Mass (g) Moles(mol) = Molar mass (g/mol) Mass(g) = Moles(mol) x Molar mass (g/mol)

11 Particle gram conversions Particles = Mass (g) = Mass (g) 23 x 6.02 x 10 (/mol) Molar mass (g/mol) Particles x Molar mass (g/mol) x 10 (/mol)

12 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

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

14 Determining percent composition Percent composition is obtained from the actual masses. Example: Sample contained g of C and g of H. Total mass = g ( ) Therefore: in 100 g there are: 100 g (84.10 %).5000 g 100 g.5000 g x = g C x = g H (15.90 %) Percent composition: % C, % H

15 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 (multiply subscript by molar mass) g C 5 mol C = g C 1 mol C g H 10 mol H = g H 1 mol H g O 2 mol O = g O 1 mol O Total mass of one mole = g

16 Convert mass of elements into percents g C % C = x100 = % C g C H O g H % H = x100 = % H g C H O g O % O = x100 = % O g C H O Percent composition: % C % H % O = % (Check: adds to 100 %)

17 Empirical formula from percent composition: 84.1 % C, 15.9 % H 1. Convert percents into moles (divide by molar mass) 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 g C g/mol 15.9 g H g/mol 15.8 mol H 2.26 : mol C C 4 H 9 May require rounding. Errors in real data cause problems Do percent composition and empirical formula exercises

18 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

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 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

21 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

22 Elements and compounds can have molecular formula different from simplest formula Substance Empirical formula Molecular formula Substance Empirical formula Molecular formula Sulphur S S 8 Phosphorous P P 4 Benzene CH C 6 H 6 Acetylene CH C 2 H 2 Ethylene CH 2 C 2 H 4 Cyclohexane CH 2 C 6 H 12

23 Which substances have same empirical and molecular formula?

24 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: Molar mass Empirical formula mass n There are n (A a B b C c ) in molecule: Molecular formula is A na B nb C nc

25 Molecular formula of vitamin C Empirical formula of vitamin C is C 3 H 4 O 3 Molar mass vitamin C is g/mol Mass of empirical formula = g/mol (3 x x x 16.00) Number of formula units per molecule = Molar mass vitamin C n 2 Empirical formula mass vitamin C Molecular formula = 2(C 3 H 4 O 3 ) = C 6 H 8 O 6

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

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

28 Example What is molarity of 50 ml solution containing 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 = M

29 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 To dilute by factor of ten, increase volume by factor of ten