STOICHIOMETRY The Mole Atomic and Molecular Masses Chemical Formula Stoichiometry

ATOMIC MASS Atomic mass standard is based on the atom Carbon-12 or 12 C (6 p +, 6 n 0, 6 e - ). One C-12 atom weighs exactly 12 amu 1 amu = atomic mass unit = 1.661E-27 kg

ATOMIC WEIGHT Atomic weight (mass) of an element is defined as a weighted average over all naturally occurring isotopes of the element; this is the number on bottom on each element box on the Periodic Table.

Calculate the Atomic Weight of Boron Boron has two naturally occurring isotopes. 10 B has a mass of M 1 = amu and abundance of A 1 = 19.78%. 11 B has a mass of M 2 = amu and abundance of A 2 = 80.22%. At. Wt = Σ (M i x A i )/100 = [ x x 80.22]/100 = amu (see PT)

MOLE The number of C-12 atoms in exactly 12 grams of pure C E+23 items Avogadro’s Number, N A mole of an element has a mass equal to its average atomic weight (mass). 1 mol of naturally occurring Boron has a mass of g.

MOLECULE and MOLAR MASS Molecule = arrangement of atoms chemically bonded together. Molar Mass = Sum of atomic masses of constituent atoms in one molecule (amu) or one mole of molecules (gram). Use atomic and molar masses to the 1/100 place.

CHEMICAL FORMULA Qualitative description of the constituent elements in a molecule or ion. –C 12 H 22 O 11 contains C, H and O –SO 4 2- contains S and O Quantitative description of the relative numbers (subscripts) of atoms of each element. –Can be used to determine % composition or mass %.

TYPES OF CHEMICAL FORMULAS Chemical - shows type and number of atoms (shorthand notation) Structural - shows chemical bonds (Fig 2.16) Ball and Stick - shows spatial arrangement, 3D (Fig 3.7 and 2.18) Space filling - shows space atoms fill, 3D (Fig 2.17, also p 95)

Figure 3.7 The Two Forms of Dichloroethane

Computer-Generated Molecule of Caffeine

CONVERSIONS grams to molesDivide by Molar Mass * * = Atomic or Molar Mass or Weight moles to gramsx by Molar Mass g/mol to g/unit** Divide by N=6.022E23 moles to #unitsx by N #units to molesDivide by N ** = Atoms or Molecules

Chapter 3 Problems 30, 40, 42, 44, 46, 48, 50

DETERMINATION OF A CHEMICAL FORMULA A chemical formula can be determined from the –Mass of each element in the formula –% Mass of each element in the formula (% Composition) –Number of moles of each element in the formula –Elemental analysis by combustion

CHEMICAL FORMULAS EMPIRICAL - includes all atoms in molecule in correct smallest integer ratios MOLECULAR - includes all atoms in molecule in actual numbers and correct ratios; can be determined from the empirical formula and molar mass.

Chapter 3 Problems 18, 64, 68, 80

CHEMICAL REACTION A chemical reaction involves rearrangements of atoms; breaking initial chemical bonds (in the reactants) and making new chemical bonds (in the products). R1 + R2  P1 + P2 + P3 Methane burns in oxygen to form carbon dioxide and water

CHEMICAL EQUATION Shorthand symbolic notation for a chemical reaction –CH 4 (g) + O 2 (g)  H 2 O(ℓ) + CO 2 (g) Note that this reaction is NOT BALANCED Qualitative aspect –identity of reactants [R] and products [P]; use study of nomenclature to write equations –Identify the state of matter for each [R] and [P] –identify reaction type

CHEMICAL EQUATION (2) Quantitative aspect –how much reactant is consumed and how much product is formed –coefficients must be consistent with the Law of Conservation of Mass; atoms are neither created nor destroyed in a chemical reaction. –i.e. chemical equation must be balanced CH 4 (g) + 2O 2 (g)  2H 2 O(ℓ) + CO 2 (g) Note that this reaction is BALANCED

STOICHIOMETRY Quantitative relationships in a chemical reaction based on a BALANCED chemical equation. Relationships between R(eactant)1 and R2 or R1 and P(roduct)2 or P1 and P2

C(s) + 2H 2 (g)  CH 4 (g) Formation of methane One atom of solid carbon reacts with two molecules of gaseous hydrogen to produce one molecule of gaseous methane. One mole of solid carbon reacts with two moles of hydrogen gas to produce one mole of methane gas g of C reacts with 4.0 g of H 2 to produce 16.0 g of CH 4. Note conservation of mass: 12+4 = 16

STOICHIOMETRIC COEFFICIENTS We will use mole interpretation for stoichiometric coefficients (SC), the coefficients in front of Rs and Ps. I.e., SCs represent # of moles of each R and P Provide quantitative (i.e. mole) relationships between R and P. Can be used to determine amount of mass of each R and P (using mol to g conversion)

MOLE RATIOS A mole ratio is a ratio of Stoichiometric Coefficients from a balanced chemical eqn. These ratios are conversion factors from amt of R1 to amt of R2, amt of P2 to amt of R1, etc

C(s) + 2H 2 (g)  CH 4 (g) 1 mol2 mol  1 mol 12.0 g 4.0 g  16.0 g How many g of carbon are needed to react with 10.0 g of hydrogen? How much CH 4 is formed g-H 2  mol-H 2  mol-C  g-C [10.0g H 2 /2.0g H 2 /mol]x[1 molC/2 mol H 2 ] x[12.0g C/mol] = 30.0 g C [10.0g H 2 /2.0g H 2 /mol]*[1 molCH 4 /2 mol H 2 ] * [16.0g CH 4 /mol] = 40.0 g CH 4 Is mass conserved?

Calculating Mass of Reactants and Products

Chapter 3 Problems 25, 82, 84, 94

REACTION YIELD In the previous example, say that only 32.0 g of CH 4 were produced due to side reactions and waste. We define the percent or reaction yield as [actual yield/theoretical yield]x100 This gives % yield = [32.0/40.0] x 100 = 80.0%

Stoichiometry and Limiting Reagent 4 wheels + 1 steering wheel + 2 airbags  one 4- passenger car Given 57 wheels, 13 steering wheels and 24 airbags, how many passengers (max) can ride to opening day of the Twins game? % mass of each atom in the cmp, find the EF. 12 cars can be made  48 passengers What is left over? 9 wheels, 1 steering wheels and 0 air bags

LIMITING REACTANT Find the actual moles of each reactant. Use the balanced chem eqn to determine how many mol of R2 is required to react completely with R1. Do you have enough R2? If not, R2 = limiting reactant = LR and R1 = reactant in excess = XS. Always use the LR to solve the stoichiometric problem to find the amount of product formed. Calculate the amount of XS left over. Calculate the grams of methane formed when 18.5 g carbon and 2.9 g hydrogen react.

C(s) + 2H 2 (g)  CH 4 (g) Formation of methane One mole of solid carbon reacts with two moles of hydrogen gas to produce one mole of methane gas. Calculate the grams of methane formed when 18.5 f C and 2.9g H 2 react. g C  mol C  mol H 2 needed g H 2  mol H 2 on hand

C(s) + 2H 2 (g)  CH 4 (g) Formation of methane Compare mol H 2 needed and mol H 2 on hand mol H 2 needed and 1.44 mol H 2 on hand. What is LR? What is XS? C = XS and H 2 = LR Use LR = H 2 to find g methane g H 2  mol H 2  mol CH 4  12. g CH 4

Solving a Stoichiom. Problem Involving Masses of Reactants and Products

Chapter 3 Problems 98, 104