STOICHIOMETRY

The MoleAtomic and Molecular Masses

Chemical FormulaStoichiometry

ATOMIC MASS

• Atomic mass standard is based on the atom Carbon-12 or 12C (6 p+, 6 n0, 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. 10B has a mass of M1 = 10.0129 amu and abundance of A1 = 19.78%. 11B has a mass of M2 = 11.0093 amu and abundance of A2 = 80.22%.

• At. Wt = Σ (Mi x Ai)/100 = [10.0129 x 19.78 + 11.0093 x 80.22]/100 =

10.81 amu (see PT)

MOLE

• The number of C-12 atoms in exactly 12 grams of pure C-12

• 6.022E+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 10.81 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.– C12H22O11 contains C, H and O

– SO42- 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 Moles Divide by Molar Mass*

* = Atomic or Molar Mass• Moles to Grams x by Molar Mass• Grams to amu Divide by N• amu to Grams x by N• Moles to #Units** x by N• #Units to Moles Divide by N

** = Atoms or Molecules

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.

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– CH4(g) + O2 (g) H2O(ℓ) + CO2(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

• CH4(g) + 2O2 (g) 2H2O(ℓ) + CO2(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) + 2H2(g) CH4(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.

• 12.0 g of C reacts with 4.0 g of H2 to produce 16.0 g of CH4. 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) + 2H2(g) CH4(g)1 mol 2 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 CH4is formed

• g-H2 mol-H2 mol-C g-C

• [10.0g H2 /2.0g H2/mol]x[1 molC/2 mol H2] x[12.0g C/mol] = 30.0 g C

• [10.0g H2 /2.0g H2/mol]*[1 molCH4/2 mol H2] * [16.0g CH4/mol] = 40.0 g CH4

• Is mass conserved?

Calculating Mass of Reactants and Products

REACTION YIELD

• In the previous example, say that only 32.0 g of CH4 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%

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.

Solving a Stoichiom. Problem Involving Masses of Reactants and Products