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Chapter 3Stoichiometry

3.1 Atomic Masses3.2 The Mole3.3 Molar Mass3.4 Percent Composition of Compounds3.5 Determining the Formula of a Compound3.6 Chemical Equations3.7 Balancing Chemical Equations3.8 Stoichiometric Calculations: Amounts of

Reactants and Products3.9 Calculations Involving a Limiting Reactant

Atomic Mass

• We measure mass, volume, pressure, etc.

• We want number of atoms (or number of moles)

• grams x (grams/mole)-1 = moles

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Atoms• Avogadro’s Number is the number of 12C atoms in exactly 12 grams of carbon

N0 = 6.02 X 1023

• The mass, in grams, of Avogadro's number of atoms of an element is numerically equal to the relative atomic mass of that element (relative to carbon)

• The mass of 6.02 X 1023 atoms of 12C = 12.00 g (defn)• The mass of 1 atom of 12C = 12.00 amu (12.00 Daltons)• The mass of 6.02 X 1023 atoms of C = 12.01 g• The mass of 6.02 X 1023 atoms of 35Cl = 35.0 g• The mass of 6.02 X 1023 atoms of Cl = 35.4 g

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Molar Mass

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Moles

• The # of moles of chemical is its amount.• One mole of a substance equals the amount that contains

Avogadro's number of atoms or molecules.• One mole of an element or molecule has a Molecular Weight

(MWt) of that element or molecule, expressed in grams• For example, the Molecular Weight of Glucose (Glucose

C6H12O6) is MWt = 6(12.0 g/mol)

+ 12(1.00 g/mol)+ 6(16.0 g/mol) = 180 g/mol

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Isoamyl acetate have the formula C7H14O2. Calculate (a) how many moles and (b) how many molecules are contained in 0.250 grams of isoamyl acetate.

Strategy: Use the units. You are given grams. You need moles. The molecular weight is given in g / mol). Then you need molecules. Avo’s # is molecules / mole.

– Calculate MWt of C7H14O2

– Calculate the number of moles in 0.250 grams– Using Avogadro’s number to calculate the number of

molecules in the number of moles of C7H14O2

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Percentage Composition (g/g) from Empirical or Molecular Formula

Tetrodotoxin, a potent poison found in the ovaries and liver of the globefish, has the empirical formula C11H17N3O8. Calculate the mass percentages of the four element in this compound.

Strategy:– Calculate molar mass of C11H17N3O8, by

finding the mass contributed by each element.

– Assume you have one mole of the compound.

– Divide the mass of each element by the total mass of the compound.

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Tetrodotoxin has the empirical formula C11H17N3O8.

Calculate the mass percentages (g/g) of the four element in this compound.

Solution:– Calculate molar mass of C11H17N3O8, by finding the

mass contributed by each elementC : 11x12 = 132 g / molH : 1x17 = 17N : 3x14 = 42O : 8x16 = 126tot = 317 g / mol

– Calculate the mass of one mole of tetrodotoxin317 g / mol x 1 mol = 317 g

– Divide the mass of each element by the mass of the compound.Example C: 132 g / 317 g = 0.42 g/g (42 %)

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1 mmol = 1 millimole =1x10-3 mol1mg = 1 milligram =1x10-3 g1msec = 1 millisec =1x10-3 sec1 mol = 1 micromole =1x10-6 mol1g = 1 microgram =1x10-6 gAlso nano (10-9)

pico (10-12) femto (10-15)

An Angstrom (Å)= 10-15 meters = 0.1 nm

Milli, Micro, etc

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The Law of Conservation of Mass

• The mass/matter of a closed system is constant.

• Mass can be rearranged but not created or /destroyed.

• In a chemical reaction the mass of the reactants equals the mass of the products.

• Mass is conserved during a change of state (solid to liquid to gas)

(This law holds in this class, but maybe not in your physics class)

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Using Conservation of Mass to determine an Empirical Formula

Moderate heating of 97.4 mg of a compound containing nickel, carbon and oxygen and no other elements drives off all of the carbon and oxygen in the form of carbon monoxide (CO) and leaves 33.5 mg of metallic nickel (Ni) behind. Determine the empirical formula of the compound.

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Using Conservation of Mass to determine an Empirical Formula

1. Write the reaction. NixCyOy -> X Ni + Y CO

2. Use conservation of mass to find the mass of CO.

97.4 mg (mass tot) – 33.5 mg (mass Ni) = 63.9 g (mass CO)

3. Find the number of moles of CO and of Ni.

CO : 63.9 mg / (12.0+16.0 g/mol) = 2.28 mmolNi : 33.5 mg / 58.7 g / mol) = 0.57 mmol

4. Find the ratios of the moles by dividing each by the smallest one, i.e., normalize to the smallest.

2.28 mmol CO /.57 mmol Ni = 4

5. Y/X = 4: answer is NiC4O4

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Chemical Equations

• Chemical Reactions tell us three things• What atoms or molecules react together to form

what products.• How much reactant how much product.• The state of each species

aA (l) + bB (s) cC (s) + dD (g) Reactants Products

The Law of Conservation of Mass says that a chemical equation must have the same number of atoms of a given kind on each side (a chemical reaction cannot create or destroy carbon, or oxygen, or hydrogen, or etc.)

Chemical equations must be balanced!

H2 + O2 H2O (not balanced)

2H2 + O2 2H2O (balanced)

Chemical Equations

Chemical Equations4 Al(s) + 3 O2(g) → 2 Al2O3(s)This equation means

4 Al atoms + 3 O2 molecules ---produces--->

2 molecules of Al2O3

or

4 moles of Al + 3 moles of O2 ---produces--->

2 moles of Al2O3

Br2 (l) + ___ Al (s) → __Al2Br6 (s)

Balancing Chemical Equations

• The same atoms are present in a reaction at the beginning and at the end.

KClO3 (s) KCl (s) + O2 (g) no

KClO3 (s) KCl (s) + 3/2O2 (g) better

2KClO3 (s) 2KCl (s) + 3O2 (g) best

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Step 1: Set the stoichimetric coefficient of the most complicated molecule (with the largest number of different elements ) to 1.

Step 4: Eliminate fractional coefficients.

To Balance an Equation

Step 2: Balance as many atoms as possible in the second most complicated molecule. Ignore atoms that show up elsewhere in homonuclear species like O2 and H2.

Step 3: Balance the atoms in the in the homonuclear species (O2 and H2).

Step 5: Count the each atom type on each side of the equation.

Balancing Equations

CC33HH88(g) + (g) + OO22(g) (g) COCO22(g) + H(g) + H22OO (g)(g)

Combustion of Propane

Balancing Equations

Combustion of Propane

11CC33HH88(g) + (g) + 55OO22(g) (g) 33COCO22(g) + (g) + 44HH22OO (g)(g)

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Writing Balanced Chemical Equations

PbO2 + Pb + H2SO4 → PbSO4 + H2O

PbO2 + Pb + 2 H2SO4 → 2 PbSO4 + 2 H2O

2 Pb 2 Pb

10 O 10 O4 H 4 HBalanced

Problem: Suppose we have 1.45 grams of Pb in the presence of excess lead oxide and sulfuric acid. How many grams of Lead Sulfate are produced?

Check for balance

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3 Br2 (l) + 2 Al (s) → 1 Al2Br6 (s)

Limiting Reactants

excess limiting Given the amounts below

Limiting Reagent

1. Balance the reaction.

2. Convert reactant masses to moles [g(g/mol)-

1=mol]

3. Normalize the moles of each reactant by its stoichiometric coefficient.

4. Find the smallest normalized number of moles

5. Use the ratio of stoichiometric coefficients to find the moles of product.

6. Convert to mass (if necessary).

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Example Calculation Involving a Limiting Reactant

Suppose that 1.00 g of sodium and 1.00 g of chlorine react to form sodium chloride (NaCl). Which of these is limiting, and what is the mass of product.

2 Na + Cl2 → 2 NaCl

nNa = 1.00 g x (1 mol Na / 23.0 g Na) = 0.0435 mol Na

nCl2 = 1.00 g × (1 mol / 70.9 g Cl2) = 0.0141 mol Cl2

nNa/2 = 0.022 (normalized) nCl2/1 = 0.0141 (normalized)

Cl2 is the limiting reagent

0.0141 moles of Cl2 gives 0.0282 moles of NaCl. Use the molecular weight of NaCl to find the mass of NaCl produced.

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What mass (in grams) of xenon tetrafluoride would be required to react completely with 1.000 g of water?

XeF4 + 2 H2O → Xe + 4 HF + O2

=

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At one point in the purification of silicon, gaseous SiHCl3 reacts with gaseous H2 to give gaseous HCl and solid Si.

(a) Determine the chemical amount (in moles) of H2 required to react with 160.4 mol of SiHCl3.(b) Determine the chemical amount of HCl that is produced.

(c) Determine the mass (in grams) of Si that is produced.

SiHCl3 (g) + H2 (g) → 3 HCl (g) + Si (s)

(a)

(b)

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At one point in the purification of silicon, gaseous SiHCl3 reacts with gaseous H2 to give gaseous HCl and solid Si.

(a) Determine the chemical amount (in moles) of H2 required to react with 160.4 mol of SiHCl3.(b) Determine the chemical amount of HCl that is produced.

(c) Determine the mass (in grams) of Si that is produced.

SiHCl3 (g) + H2 (g) → 3 HCl (g) + Si (s)1 mol SiHCl3 1 mol H2 3 mol HCl 1 mol Si

(c)

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Isotopes of Cl:

031 150 msecEC/ECp,11.980S-31/P-3030.9924

032 298 msecEC/ECa/ECp,12.685S-32/Si-28/P-3131.985688

033 2.511 secEC,5.583 MeVS-3332.97745

034 1.5264 secEC,5.492 MeVS-3433.97376

035 75.77%

Stable

34.9688

036 3.01E+5 yrB-/EC,10.413Ar-36/S-3635.9683

037 24.23%

Stable

36.9659

038 37.24 minB-,4.917 MeVAr-3837.968010

039 55.6 minB-,3.442 MeVAr-3938.968008

040 1.35 minB-,7.480 MeVAr-4039.970413

041 38.4 secB-,5.730 MeVAr-4140.970649

042 6.8 secB-,9.430 MeVAr-4241.973172

043 3.3 secB-,7.950 MeVAr-4342.974202

044 0.43 secB-/B-n,3.920Ar-44/Ar-4343.978539

045 400 msecB-/B-n,10.800Ar-45/Ar-4444.979710

046 0.22 secB-/B-n,6.900Ar-46/Ar-4545.984111

047 200 nsecB-/B-n,14.700Ar-47/Ar-4646.987976

MASS abund.HalflifeParticle, EnergyDecay Product(s)Isotopic Mass

35Cl contains 17 protons and 18 neutrons

37Cl contains 17 protons and 20 neutrons

1 1 + 2 2 + ... + n nAverage Relative Atomic Mass = A P A P A P

Chlorine = 35.45