stoichiometry quantitative nature of chemical formulas and chemical reactions
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Stoichiometry Quantitative nature of chemical formulas and chemical reactions. Chapter 3 (Sections 3.3 - 3.7). Atomic Mass Scale. Atomic mass units (amu) are convenient units to use when dealing with extremely small masses of individual atoms 1 amu = 1.66054 x 10 -24 g - PowerPoint PPT PresentationTRANSCRIPT
StoichiometryQuantitative nature of chemical formulas
and chemical reactions
Chapter 3(Sections 3.3 - 3.7)
Atomic Mass Scale• Atomic mass units (amu) are
convenient units to use when dealing with extremely small masses of individual atoms
• 1 amu = 1.66054 x 10-24 g• 1 g = 6.02214 x 1023 amu• By definition, the mass of C-12 is
exactly 12 amu
Average Atomic Mass(Atomic Weight)
• We average the masses of isotopes using their masses and relative abundances to give the average atomic mass of an element.
• Naturally occurring C consists of 98.892% C-12 (12.00 amu) and 1.108% C-13 (13.00335 amu)
• The average mass of C is• (0.98892)(12.00 amu) + (0.01108)(13.00335)
= 12.01 amu • Atomic weights are listed on the periodic
table
Calculating % Abundance• Chlorine is made up of two isotopes,
Cl-35 atomic mass = 34.969 amu and Cl- 37 atomic mass = 36.966 amu. Given chlorine’s atomic weight of 35.453, what is the % abundance of each isotope?
34.969(x) + (36.966) (1-x) = 35.45334.969x + 36.966 – 36.966x = 35.453x = .7576
75.76% Cl-35 and 24.24% Cl-37
Formula & Molecular Weights
• Formula Weight is the sum of atomic weights for the atoms present in the chemical formula
• Molecular Weight is the sum of atomic weights of the atoms in a molecule as shown in the molecular formula
• Sample Exercise 3.5 page 80
Percentage Composition from Formulas
• Obtained by dividing the mass contributed by each element (number of atoms times atomic weight) by the formula weight of the compound and multiplying by 100
• Sample Exercise 3.6 page 80
The Mass Spectrometer page 81• Mass spectrometers are pieces of equipment designed to
measure atomic and molecular masses accurately.• The sample is converted to positive ions by collisions with a
stream of high-energy electrons upon entering the spectrometer.
• The charged sample is accelerated using an applied voltage.• The ions are then passed into an evacuated tube through a
magnetic field.• The magnetic field causes the ions to be deflected by
different amounts depending on their mass – more mass, less deflection.
• The ions are then detected.
The Mole• The mole is a convenient
measure of chemical quantities. • 1 mole of something is 6.0221421
x 1023 of that thing.• This number is called
Avagadro’s number.• Thus 1 mole of carbon atoms =
6.0221421 x 1023 carbon atoms
Molar Mass• mass in grams of 1 mole of a
substance• expressed in units of g/mol• formula weights are numerically
equal to the molar mass • Sample Exercise 3.8 page 84
Interconverting Masses, Moles, & Number of Particles
• To convert between grams and moles, we use the molar mass
• To convert between moles and particles (atoms, molecules, or ions) we use Avogadro’s number
• Sample Exercises 3.7, 3.9, 3.10, 3.11 pages 82-86
Empirical Formula• Gives the relative number of
atoms of each element in the substance
• Can be calculated from mass percent data
• Sample Exercise 3.12 page 87
Molecular Formula• Actual number of atoms of
each element in one molecule of the substance
• Whole number multiple of empirical formula
• Sample Exercise 3.13 page 88
Combustion Analysis• A sample containing C, H,
and O is combusted in excess oxygen to produce CO2 & H2O
• Can be used to determine empirical formula or percent composition
Quantitative Information from Balanced Equations
• Coefficients can be interpreted as the relative numbers of molecules or formula units in the reaction as well as the relative number of moles
• See Figure 3.14 page 92• Sample Exercises 3.14, 13.15
pages 92 & 93
Limiting Reactant(s)• The reactant(s) that is completely consumed in
a reaction• Limits or determines the amount of product
that will be formed• The other reactant(s) that is left over is called
the excess reactant • Sample Exercise 3.16 and 3.17 pages 95-96• How much of the excess reactant is left over?
Percent Yield (actual yield / theoretical yield) X 100
• Actual is the amount of product recovered in the lab
• Theoretical is the amount predicted from stoichiometry
• Sample Exercise 3.18 page 97