Atomic Structure and Properties The relationship between structures and their behaviors
Methods of Separation Filtration Distillation
Energy and Chemical Change • Law of the Conservation of Energy • High Potential Energy = Unstable • Low Potential Energy = Stable • Energy is released into the surroundings • Compounds with high chemical potential can be used to do work
Density: An Intensive Property • Density is independent of the amount of a substance • Density decreases at heat energy increases (for most substances) • Density is still mass over volume
Significant Figures • They’re back!
Conversion Steps Starting Unit Conversion Factors
Conversions with Density The mass of fuel in a jet must be calculated before each flight to ensure that the jet is not too heavy to fly. A 747 is fueled with 173,231 L of jet fuel. If the density of the fuel is 0.768 g/cm3, what is the mass of the fuel in kilograms?
Moles and Molar Mass • The mole is the bridge between mass, volume, and parts • 1 mole = 6.022 x 1023 parts (molecules, compounds, elements) • 1 mole = the molar mass of a substance • Calculated by adding the individual masses of each element • 1 mole = 22.4 L of a gas AT STANDARD TEMPERATURE AND PRESSURE (STP) • This is important because our conditions will now be changing • Mole conversions follow the same conversion format previously covered
Glossary • Molecular ion - The ion obtained by the loss of one electron from the molecule (m+) • Base peak-The most intense peak in the MS, assigned 100% intensity • Radical cation-positively charged species with an odd number of electrons • Fragment ions - Lighter cations (and radical cations) formed by the decomposition of the molecular ion. These often correspond to stable carbcations. • m/z-mass to charge ratio
Octane, m+ = 114 m-71 Base peak m-57 m-29 m-43 m+
Law of Definite Proportions • French chemist, Joseph Proust • All samples of a given compound, regardless of their source or how they were prepared, have the same proportions of their constituent elements • Mass ratio = Mole ratio • Water: = 8.0 or 8:1 Mass Ratio • Ammonia? • = 4.7 or 4.7:1 Mass Ratio
Law of Definite Proportions Two samples of carbon dioxide are decomposed into their constituent elements. One sample produces 25.6 g of oxygen and 9.60 g of carbon, and the other produces 21.6 g of oxygen and 8.10 g of carbon. Show that these results are consistent with the law of definite proportions. Sample 1 and Sample 2 have the same mass ratio!
Law of Multiple Proportions • John Dalton • When two elements form two different compounds, the masses of element B that combine with 1 g of element A can be expressed as a ratio of small whole numbers • Carbon and Oxygen:
Law of Multiple Proportions Nitrogen forms several compounds with oxygen, including nitrogen dioxide and dinitrogen monoxide. Nitrogen dioxide contains 2.28 g oxygen to every 1.00 g nitrogen, while dinitrogen monoxide contains 0.570 g oxygen to every 1.00 g nitrogen. Show that these results are consistent with the law of multiple proportions. The small whole number ratio is 4!
Percent Composition • Breaks down the amount of an element in a compound by percentage • Total mass of element / total mass of compound • Result will always be a decimal • Multiply by 100 to get the final percentage • Data can also be acquired from Mass Spec data
Practice What is the percentage of Iron in Iron (III) Carbonate?
Empirical formulas • Formulas are represented in smallest whole number ratios • Derived from percent composition of a compound • Remember 100g = 100 % for easier calculations • Ratios can also be used (Law of Definite Proportions)
Talk it Out… • Change percentage to grams assuming there are 100g of the sample • Divide by the molar mass of each element • Divide the moles of each element by the lowest number of moles of all the elements • Round to whole numbers to find ratios UNLESS there is .5, then multiply each result by 2 for whole number ratios
Practice A 60.00g sample of a tetraethyl lead, a gasoline additive, contains 38.43g of Lead, 17.83g Carbon, and 3.74g Hydrogen. What is the compounds empirical formula?
Molecular Formula • You must have the empirical formula first • You must calculate the mass of the empirical formula • Don’t lose track of your numbers! • Law of Multiple Proportions applies to the molecule
Calculate empirical formula • Calculate the mass of the empirical formula • Divide the given mass by the empirical mass, this will give you an element multiplier • Multiply your chemical quantities by the multiplier • Rewrite molecular formula with new chemical values
Practice A compound containing 50.05% Sulfur and 49.95% Oxygen has a molar mass of 256.28 g/mol. Determine the empirical and molecular formula of this compound.
Combustion Analysis • Using an unknown compound and burning it in pure oxygen. The resulting gases are collected and massed, which is then used to calculate an empirical formula
Steps for Solving Combustion Analysis Scenarios • Start with given masses • Convert the masses of CO2 and H2O to moles • Convert the moles of CO2 and H2O to moles of C and H using the ratio from the chemical formulas • ***** If the compound contains an element other than C and H, find the mass of the other element by subtracting the sum of the masses of C and H (Step 3) from the mass of the sample. Convert this mass to moles of the other element • Divide moles calculated by the smallest moles of the group • *****If you have whole numbers YA DONE SON! ***** If you have decimals, multiply by the denominator to achieve whole numbers
Let’s Work Together! Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.6004 g CO2 and 0.6551 g H2O. Find the empirical formula of the compound. C2H4O
A compound containing the elements C, H, N, and O is analyzed. When a 1.2359g sample is burned in excess oxygen, 2.241g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648g of H. 1. Determine the mass, in grams, of C in the 1.2359g sample of the compound 2. When the compound is analyzed for N content only, the percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359g sample of the compound. 3. Determine the mass, in grams, of O in the original 1.2359g sample of the compound. 4. Calculate the empirical formula of the compound A different compound, which has the empirical formula CH2Br , has a vapor density of 6.00 g L− 1 at 375 K and 0.983 atm. Using these data, determine the following. 5. The molar mass of the compound 6. The molecular formula of the compound
Atomic Structure and Electron Configuration • Positively charge nucleus (protons and neutrons) surrounded by negatively charged cloud of electrons • Core electrons are close to the nucleus • Valence are far from the nucleus • 1s22s22p63s23p64s23d104p65s24d9 = Ag • Pauli Exclusion Principle states no two electrons can occupy the same space • Electron Configuration can play a big role in being able to predict behavior
Aufbau Principle • Electrons will occupy the lowest energy orbitals first • Best illustrated by orbital diagrams • Hund’s rule is an extension of the Aufbau principle that electrons will fill orbitals singly first with parallel spins
Coulomb’s Law • The potential energy of two charged particles depends on their charges (q1 and q2) and on their separation (r) • Used to calculate the force between two charged particles Charge to Distance Ratio Potential Energy Coulomb Constant
Coulomb’s Law • For like charges – potential energy is positive and decreases with increasing distance • For opposite charges – potential energy is negative and becomes more negative as distance decrease • The magnitude of interaction – increases with an increase in charge difference
Photoelectron Spectroscopy • A Spectroscopic tool whereby the binding energies of electrons can be measured. • Uses high energy UV or X-ray radiation to remove electrons from neutral atoms. • By measuring the kinetic energy of the removed electron and knowing the energy of the radiation source the binding energy can be calculated.
Kinetic Energy Analyzer X-ray or UV Source Kinetic Energy Analyzer 6.26 0.52 Binding Energy (MJ/mol) 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+
Photoelectron Spectrum • The results are contained in a Photoelectron Spectrum • These are simulated examples as the actual spectra are quite complex due to interference. • They give two pieces of information • The ionization energy (or binding energy) in MJ/mol • The relative number of electrons that have a particular IE Relative Intensity = 2 6.26 MJ/mol Number of electrons Relative Intensity = 1 0.52 MJ/mol 10 MJ/mol 0 MJ/mol
Photoelectron Spectrum Relative Intensity = 2 6.26 MJ/mol Number of electrons Relative Intensity = 1 0.52 MJ/mol • The larger peak represents two electrons with a high binding energy. • The smaller peak represents one electron with a low binding energy. • This is the PES for Lithium. 10 MJ/mol 0 MJ/mol