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Chapter 3

Mass Relationships

In Chemical Reactions

Chapter 3

• Measuring atomic and molecular masses

– Mass spectrometry

• The mole

– Scaling molecular mass to a size we can weigh

• Chemical formulas

• Experimentally determining molecular formulas

• Equations representing chemical reactions

• Mass relationships in chemical reactions

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Atomic Mass Units

• Units for the mass of atoms and molecules (amu)

• Defined so that mass of 126C is exactly 12

• Gives most of common elements an integral mass )or close)

• Not all, though

Mass Spectrometer

One type of Mass spec

• Most accurate way of determining masses

– Expose gas to beam of electrons

• Strips electrons from particles to form ions

• Breaks some molecules into fragments

– Pass beam of charged fragments through magnetic field

– Curvature of path depends on mass of fragments

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Chlorine Mass Spectrum

• Part of chlorine mass spectrum looks like below.

• X-axis units are in amu

• Why are there two peaks?

Mass Averages

• Chlorine has 2 isotopes

– 3517Cl: 75.8% 37

17Cl: 24.2%

– Different number of neutrons

• Elements have several isotopes in nature

• The atomic mass is an average of the isotopes

– For chlorine:

– (.758)(35 amu) + (.242)(37 amu) = 35.5 amu

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Atomic Mass

• Look at Periodic Table

– Which of the following probably has more than one common isotope?

• N

• F

• Al

• Ge

Molecular Mass

• Add atomic masses of atoms in molecule

• H2O: 2(1.008 amu) + 16.00 amu = 18.02 amu

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The Mole

• Problem: difficult to weigh one atom or molecule

– Need to scale up masses to a size we can deal with in the laboratory

• 1 mole = amount of substance that contains as many particles as exactly 12 g of C-12

• Number of particles in a mole = Avogadro’s Number

– 6.022 x 1023 1/mole

– Know this

Molar Mass

• Molar mass = mass of one mole in grams

– Units = g/mole Important

– H: 1.008 g/mole

• Molar mass has same value as molecular mass

– Units are different

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Molar Mass

• Use Factor-Label method to convert g <-> mole

• Question: How many moles in 46.0 g water

• Question: How many g in 0.76 moles of NaCl?

• Question: How many molecules in 100.0 g water?

Chemical Formulas

• Molecular Formula: exact number of atoms in a molecule

– Example: ethylene glycol

– C2H6O2

• Empirical Formula: simplest whole number ratio of atoms in molecule

– Ethylene glycol: CH3O

– Empirical formula is used in determining molecular formula

Black: Carbon Red: Oxygen Yellow: Hydrogen

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Empirical Formulas

• What are the empirical formulas of these compounds?

– N2O4

– C6H6O3

– C6H6O

Empirical Formulas

• Why do we care about empirical formulas?

– Experiments sometimes determine empirical formulas.

– First step in figuring out a molecular formula

• Experiment -> empirical formula

• Experiment -> molecular mass

• Put them together -> molecular formula

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Empirical -> Molecular Formula

• Suppose we know the empirical formula of a compound

– C3H7

• What is the molecular formula of the compound?

Empirical -> Molecular Formula • Mass spectrum of the compound

• Generally the peak with largest mass is the molecular mass – Ignore the C-13 isotope peak

• This example: Molecular mass = 86 amu

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Empirical -> Molecular Formula • Empirical formula = C3H7.

• Molecular mass = 86 amu

• Question: What is molecular formula of compound?

• Note: Only need approximate molecular mass

Percent Composition

• Example: H2O

• %H =

• %O =

• Percent compositions can be determined experimentally

• Can use percent composition to determine empirical formulas

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Determining Empirical Formulas

• Experimental % composition of compound is:

– 50.05% S

– 49.95% O

• What is the empirical formula of the compound?

Determining Molecular Formulas

• Molecular mass is around 64 g/mole.

• What is the molecular formula of the compound?

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Determining Molecular Formulas

• Compound contains only C, H, and O.

– In 2.034 g of compound, have the following:

• C: 1.031 g

• H: 0.0864 g

• O: 0.917 g

– The mass spectrum of compound indicates that its molar mass is around 140 g/mole.

– What are empirical and molecular formulas of compound?

Chemical Equations

• Express what occurs during a chemical reaction

• 2 Al(s) + 3 Br2(l) Al2Br6(s)

Reactants Product(s)

– Coefficients = # molecules or # moles

– Equation must be balanced • Same number elements on both sides

– Physical states in parentheses • s, l, g, aq

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Chemical Equations

• Question: Observe the video of heating mercury(II) oxide. – What is formula of mercury(II) oxide?

– One of the products is metallic mercury

– Other product of reaction supports combustion. What is this gas?

– What is the balanced equation for the reaction?

Chemical Equations

• Reaction of metallic magnesium in aqueous HCl.

– What is name of HCl?

– The gas liberated is hydrogen H2

– What is formula and name of the other product?

– What is the balanced equation for the reaction?

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Stoichiometry

• Relationship between amounts of reactants and products

• 2 H2 (g) + O2 (g) ----> 2 H2O (l)

• Question: How many moles of water will be produced from the reaction of 2.52 moles oxygen with excess hydrogen?

Stoichiometry

• 2 H2 (g) + O2 (g) ----> 2 H2O (l)

• Question: How many grams of oxygen are needed to react with 3.40 g of hydrogen?

• Key to doing these problems

–Convert grams to moles

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Stoichiometry

• 2 CH3OH (l) + 3 O2 (g) ----> 2 CO2 (g) + 4 H2O (l)

• How many grams of water will be produced by burning 25.0 mL methanol in excess oxygen?

– What do you need to look up to solve this problem?

– Use Wolfram Alpha

Combustion Analysis

• Burn known amount of organic compound in excess oxygen

• Measure masses of CO2 and H2O produced

CmHn + O2 mCO2 + (n/2)H2O

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Combustion Analysis

• Combining three topics covered in this chapter

– Stoichiometry • Use g of CO2 to determine moles of C in compound

• Use g of H2O to determine moles of H in compound

– Empirical formula determination • Moles C and moles H empirical formula

– Molecular formula determination • Mass spectrum molecular mass

• Empirical formula and molecular mass molecular formula

Combustion Analysis

• Equation above isn’t balanced

– All we care about is C-CO2 and H-H2O relationship

• 1 mole CO2 = 1 mole C in compound

• 1 mole H2O = 2 moles H in compound

CmHn + O2 mCO2 + (n/2)H2O

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Combustion Analysis

• Burn 0.820 g of an unknown hydrocarbon

• Obtain 2.70 g CO2 and 0.73 g H2O

• Mass spectrum:

• What is compound’s molecular formula?

CmHn + O2 mCO2 + (n/2)H2O

• Don’t always have exact amounts of reactants to completely use them all up

• A + 2B C

• Initially:

– 5 A molecules

– 6 B molecules

• How many C molecules will be produced?

• Will there be any A or B molecules remaining when reaction is over?

Limiting Reactant

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Limiting Reactant

• A + 2B C

• B molecules are used up first

• # of C molecules produced depends on initial # of B molecules

– 6 B molecules 3 C molecules

• B = Limiting reactant

• A is in excess

Initially: 5 A molecules 6 B molecules

Limiting Reactant

C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O 74.1 g/mole 102.9 g/mole

• What are names of NaBr, H2SO4, NaHSO4?

• Want to use up all of C4H9OH

• Start with 50.0 g of C4H9OH

• What is minimum mass of NaBr necessary to use all of C4H9OH ? (excess H2SO4)

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Limiting Reactant

• N2 (g) + 3 H2 (g) 2 NH3 (g) 28.0 g/mole 2.02 g/mole 17.0 g/mole

• React 10.0 g N2 with 10.0 g H2. How much NH3 will be produced?

• Need to decide which reactant is limiting – Calculate amount of product possible for both reactants

– Smallest amount identifies limiting reactant

• Use limiting reactant to calculate amount of product

Reaction Yield

• Theoretical yield = amount of product produced if all limiting reactant is consumed

– Best you can do

• Actual yield = amount of product you actually get

– Usually the case that actual yield is less than theoretical

• Percent yield = 100*(Actual yield)/(Theoretical yield)

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Percent Yield

• Previous problem

• Theoretical yield = 18.2 g NH3

• Suppose actual yield = 15.3 g NH3

• What is percent yield?

Reaction Yield

2 N2H4 (l) + N2O4 (g) → 3 N2 (g) + 4 H2O (g) 32.0 g/mole 92.0 g/mole 28.0 g/mole

Reaction of hydrazine with dinitrogen tetraoxide

What mass of N2 gas theoretically would result from the reaction of 150.0 g of hydrazine and 100.0 g of N2O4? If you actually obtain 66.4 g N2, what is percent yield?

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Primary Productivity

• Primary productivity: How quickly and to what extent sunlight is converted into organic material by plants during photosynthesis.

• Photosynthesis: Simple equation describes production of sugars

• 6CO2 + 6H2O C6H12O6 + 6O2

• Nutrients: Other chemicals that also are required:

– Examples: Phosphorus and nitrogen

light

Limiting Nutrient

• Nutrients seldom used up at same rate

– Eventually, one of them may be depleted

• What will happen to primary production once one of nutrients is depleted?

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Limiting Nutrient

• Limiting nutrient

– The nutrient in shortest supply in a particular ecosystem

– In essence the limiting reactant in primary production

• Once a nutrient is depleted, plant growth stops in the ecosystem

Limiting Nutrient

• Phosphorus: limiting nutrient in many lakes

• How could we design an experiment to test this?

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Eutrophication

• Large increase in primary production in a lake

– Usually because phosphorus has been added

– Mainly due to fertilizers in runoff

– Reason phosphates are illegal in detergents in many states