The Mole & Formulas Dr. Ron Rusay Spring 2008 © Copyright 2008 R.J. Rusay

Stoichiometry Mole - Mass Relationships Chemical Reactions The Mole % Composition: Determining the Formula of an Unknown Compound Writing and Balancing Chemical Equations Calculating the amounts of Reactant and Product Limiting Reactant

The Mole The number of carbon atoms in exactly 12 grams of pure 12 C. The number equalsThe number of carbon atoms in exactly 12 grams of pure 12 C. The number equals 6.02   1 mole of anything = 6.02  units 6.02  “units” of anything: atoms, people, stars, $s, etc., etc. = 6.02  “units” of anything: atoms, people, stars, $s, etc., etc. = 1 mole

Avogadro’s Number Avogadro’s number equals 1 mole ….which equals  “units”  “units”

Counting by Weighing 12 red 7g each = 84g 12 yellow each=48g 55.85g Fe = x atoms Fe 32.07g S = x atoms S

CaCO g Oxygen g Copper g Water g Relative Masses of 1 Mole

Atomic and Molecular Weights Mass Measurements 1 H weighs x g and 16 O x g. 1 H weighs x g and 16 O x g. DEFINITION: mass of 12 C = exactly 12 amu. DEFINITION: mass of 12 C = exactly 12 amu. Using atomic mass units: Using atomic mass units: 1 amu = x g1 amu = x g 1 g = x amu1 g = x amu

QUESTION

ANSWER  –22  23  –22 5) g Section 3.3 The Mole (p. 82) A mole of copper atoms has a mass of g. The mass of 1 copper atoms is g/( ) = g.

Atomic and Molecular Weights Formula Weight a.k.a. Molecular WeightFormula Weight a.k.a. Molecular Weight Formula weights (FW): sum of Atomic Weights (AW) for atoms in formula.Formula weights (FW): sum of Atomic Weights (AW) for atoms in formula. FW (H 2 SO 4 ) = 2AW(H) + AW(S) + 4AW(O)FW (H 2 SO 4 ) = 2AW(H) + AW(S) + 4AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0)= 2(1.0 amu) + (32.0 amu) + 4(16.0) = 98.0 amu= 98.0 amu

Atomic and Molecular Weights Molecular weight (MW) is the weight of the molecular formula in amu.Molecular weight (MW) is the weight of the molecular formula in amu. MW of sugar (C 6 H 12 O 6 ) = ?MW of sugar (C 6 H 12 O 6 ) = ? MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180 amu= 180 amu

Molar Mass A substance’s molar mass (equal to the formula weight: atomic or molecular weight in grams) is the mass in grams of one mole of the element or compound.A substance’s molar mass (equal to the formula weight: atomic or molecular weight in grams) is the mass in grams of one mole of the element or compound. ð C = grams per mole (g/mol) ð CO 2 = ?? ð grams per mole (g/mol) (16.00) = 44.01

QUESTION

ANSWER )  3)46.07 Section 3.4 Molar Mass(p. 86 The molar mass is the sum of masses of all the atoms in the molecule = 46.07

QUESTION

ANSWER 3 ) 3)CHCl Section 3.4 Molar Mass(p. 86 The molar mass has units of g/mol. (12.8 g/0.256 mol) = 50.0 g/mol. The molecule with the closest molar mass is CH 3 Cl.

QUESTION

ANSWER  2)278 g Section 3.4 Molar Mass (p. 86) The molar mass of sodium hydroxide, NaOH, is g/mol g/mol g/mol = g/mol. Convert to grams: 6.94 mol (40.00 g/mol) = 278 g.

Percent Composition Mass percent of an element:Mass percent of an element: For iron in (Fe 2 O 3 ), iron (III) oxide = ?For iron in (Fe 2 O 3 ), iron (III) oxide = ?

QUESTION

ANSWER 88 5)All of these Section 3.5 Percent Composition of Compounds (p. 89) The ratio of C to H in CH is 1:1. This is the same ratio found for each of the compounds, so all have the same percent composition by mass.

QUESTION Morphine, derived from opium plants, has the potential for use and abuse. It’s formula is C 17 H 19 NO 3. What percent, by mass, is the carbon in this compound? % % % 4.This cannot be solved until the mass of the sample is given.

ANSWER Choice 3 is correct. First determine the molar mass of the compound, then divide that into the total mass of carbon present,and finally, multiply that by 10. (17  12) / ((17  12) + (19  1) + (1  14) + 3  16)) =  100 = 71.6 % Section 3.4: Percent Composition of Compounds

QUESTION

ANSWER 27     3)4.04 g Section 3.5 Percent Composition of Compounds (p. 89) The molar mass of KCrO is = The mass fraction of potassium is ( )/242.2 = g = 4.04 g.

Formulas: Dalton’s Law Dalton’s law of multiple proportions:Dalton’s law of multiple proportions: When two elements form different compounds, the mass ratio of the elements in one compound is related to the mass ratio in the other by a small whole number.

Formulas: Multiple Proportions

Formulas & Multiple Proportions Components of acid rain, SO 2 (g) and SO 3 (g) Compound A contains:Compound A contains: g Sulfur & g Oxygen Compound B contains:Compound B contains: g Sulfur & g Oxygen Mass ratio A: 2 to 3; Mass ratio B: 1 to 1Mass ratio A: 2 to 3; Mass ratio B: 1 to 1 Adjusting for atomic mass differences: AW sulfur is 2x the AW oxygen; the atom ratios therefore are S 1 O 3 and S 1 O 2 respectivelyAdjusting for atomic mass differences: AW sulfur is 2x the AW oxygen; the atom ratios therefore are S 1 O 3 and S 1 O 2 respectively

Formulas & Molecular Representations ð molecular formula = C 6 H 6 Benzene ð empirical formula = CH = C 6/6 H 6/6 ð molecular formula = (empirical formula) n [ n = integer] (CH) 6 Other representations: Lewis Dot formulas, structural formulas, 2-D, 3-D Other representations: Lewis Dot formulas, structural formulas, 2-D, 3-D

Formulas & Molecular Representations

Empirical Formulas from Analyses

Empirical Formula Determination 1.Use percent analysis.1.Use percent analysis. Let 100 % = 100 grams of compound. 2.Determine the moles of each element.2.Determine the moles of each element. (Element % = grams of element.) (Element % = grams of element.) 3.Divide each value of moles by the smallest of the mole values.3.Divide each value of moles by the smallest of the mole values. 4.Multiply each number by an integer to obtain all whole numbers.4.Multiply each number by an integer to obtain all whole numbers.

QUESTION The dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo? 1.C 6 H 4 NO 2.C 8 H 3 NO 3.C 8 H 5 NO 4.I know this should be whole numbers for each atom, but I do not know how to accomplish that.

ANSWER Choice 3 is the smallest whole number ratio of the atoms that make up a molecule of indigo. The percentage must be converted to a mass, then the mass is converted to moles of the atoms and finally, the smallest is divided into the others to obtain the proper ratio. Section 3.5: Determining the Formula of a Compound

Empirical & Molecular Formula Determination Quinine: C 74.05%, H 7.46%, N 8.63%, O 9.86% 74.05/12.01, 7.46/1.008, 8.63/14.01, 9.86/ /12.01, 7.46/1.008, 8.63/14.01, 9.86/16.00 C H 7.40 N O Empirical Formula: C 10 H 12 N 1 O 1Empirical Formula: C 10 H 12 N 1 O 1 Empirical Formula Weight = ? Molecular Weight = Molecular Weight = Molecular Formula = 2x empirical formula Molecular Formula = C 20 H 24 N 2 O 2Molecular Formula = C 20 H 24 N 2 O 2

A Mass Spectrometer Recordds a mass spectrum

A mass spectrum records only positively charged fragments m/z = mass to charge ratio of the fragment

A Carbon atom is at each angle. Each C has 4 bonds (lines + Hs ). Hs are not always drawn in & must be added. C = 20 H = 24 N = 2 O = 2 From the structures, determine the molecular formula of quinine. C 20 H 24 N 2 O 2

QUESTION

ANSWER 2) C 8 H 8 Section 3.6 Determining the Formula of a Compound (p. 91) The mass of CH is divides into about 8 times. Therefore there are 8 CH groups in this compound. C 8 H 8 is the molecular formula.