Composition and formulae Of moles and men

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

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

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

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?...

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

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

Particle – mole conversions

Gram – mole conversions

Particle – gram conversions

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

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

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: (84.10 %) (84.10 %) (15.90 %) (15.90 %) Percent composition: % C, % H

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 = g

Convert masses into percents  Percent composition: % C % H % O = %

Empirical formula from percent composition: 84.1 % C, 15.9 % H 1.Convert percents into moles 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

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

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

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?

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

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

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

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 =  Molecular formula = 2(C 3 H 4 O 3 ) = C 6 H 8 O 6

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

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

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 =.0240/.050 = M

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