Introductory Material AP Chemistry Introductory Material

Chemical Foundations Chapter 1 Observations Hypotheses Predictions Scientific Method Theory Or Model Predictions Experiment Modify Theory

Scientific Method You are given a computer and asked to make a graph. After booting the computer, opening excel and entering data, the screen goes blank. Oh Gees! Now What!

Units of Measurement Expect you to know Units used in science pico to giga And be able to convert Units used in science Kilograms, meters, seconds, kelvins, amps, moles

Significant Figures There is more than one convention! AP Chemistry allows for some variation If you are within one sig fig, it is OK We will follow this Rules are on Pg 23 of your book We will use these for every calculation You lose a point for incorrect sig figs on test

Calculations Adding and subtraction Answer has the same number of decimal places as the least precise measurement. 12.11 18.0 1.013 31.123 31.1 Multiplication and Division Answer has the same number of significant figures as the least precise measurement 4.56 x 1.4 = 6.38 corrected 6.4 pH The number to the left of the decimal is the exponent The number to the right of the decimal contains the correct number of sig figs. pH = 7.07 has 2 sig figs

Dimensional Analysis Do I really have to? It’s way easier! No, but it will cost you extra work explaining yourself Units written out in Dim Analysis are self explanatory It’s way easier! Way, way easier!! Just Do it!

Mercury poisoning is a debilitating disease that is often fatal Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes. If the amount of mercury in a polluted lake is 00.4000 micrograms Hg per milliliter, what is the total mass in kilograms of mercury in the lake. The lake has a surface area of 0100. mi2 and an average depth of 20.1 ft. (5280. ft in a mile, 12 in in a foot, 2.54 cm in an inch, 106 micrograms in a gram)

Classification of Matter What is a mixture? Name two types How can we separate hetero? Homo? If I say something is a pure substance, what does that mean? What is the difference between an element and a compound? What is an element made up of?

It’s the Law Explain the following laws: Conservation of Mass Definite proportion Multiple proportion Name four parts of Dalton’s Atomic Theory Atoms All atoms of same element are identical Same compound always has same elements in same proportions Atoms themselves do not change in chemical reactions

Famous Atomic Experiments Describe the Experiment JJ Thompson and CRT’s Used CRT to determine charge to mass ratio Discovered electron Rutherford’s Gold Foil Used alpha particles and gold foil Discovered a dense, positive nucleus Millakan’s Oil Droplet Discovered the charge of an electron Calculated the mass of an electron with JJ’s reults

Modern Theory Subatomic particles are? Nucleus is composed of ? Electron, neutron and proton Nucleus is composed of ? Neutron and proton Electrons are in “clouds” What does that mean?

F Symbol -1 19 How many protons? 9 How many neutrons? How many electrons?

Periodic Table Describe the following: Metals Non-metals Semi-metals Alkali Metals Alkali Earth Metals Transition Metals Halogens Noble gases

Modern periodic table * * + + H He Li Be B C N O F Ne Na Mg Al Si P S 1 2 13 14 15 16 17 18 I A II A III A IV A V A VI A VIIA 0 H He 1 2 3 4 5 6 7 Li Be B C N O F Ne 3 4 5 6 7 8 9 10 11 12 III B IVB V B VIB VIIB VIII B IB IIB Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn + Fr Ra Lr * Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No La Ac Er Fm Tm Md Dy Cf Ho Es +

Metals * * + + H B He Ne F O N C Li Be Na Mg Al As Si Kr Br Se Ar Cl S P Ge K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn + Fr Ra Lr * La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb + Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

Nonmetals * * + + H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba Ly Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn + Fr Ra Lr * La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb + Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

Semimetals or Metalloids H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn + Fr Ra Lr * La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb + Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

Bonds and Stuff Explain the following Ion Cation and anion Ionic Bond Covalent Bond Molecule Formula Unit Chemical Formula Structural Formula

Ions and Ionic Compounds Important: note that there are no easily identified NaCl molecules in the ionic lattice. Therefore, we cannot use molecular formulas to describe ionic substances.

Naming Compounds Memorize all polyatomic ions pg 63 and how to determine the rest Memorize names of elements s block, p block = all Transition metals Need to know common metals Charges on Al+3, Zn+2, Ag+1, Cd+2

Ions Cation - a positively charged ion. Mg Mg2+ + 2e- Ions are charged particles formed by the transfer of electrons between elements or combinations of elements. Cation - a positively charged ion. Mg Mg2+ + 2e- Anion - a negatively charged ion. F2 + 2e- 2F-

Writing Formulas Al3+ O2- 2 3 Al2O3 All compounds are electrically neutral. The sum of the positive and negative charges must add up to zero. Al3+ O2- 2 3 Use subscripts to indicate how many of each ion is used. Al2O3

Naming inorganic compounds When an element forms only one compound with a given anion. name the cation name the anion using the ending (-ide) for monatomic ions NaCl sodium chloride MgBr2 magnesium bromide Al2O3 aluminum oxide K3N potassium nitride

Naming ionic compounds Many metals form more than one compound with some anions. For these, Roman numerals are used in the name to indicate the charge on the metal. Cu1+ + O2- = Cu2O copper(I) oxide copper(I) oxide Cu2+ + O2- = CuO copper(II) oxide copper(II) oxide

Naming ionic compounds Since the charge of some metal ions can vary, look at everything else first. What ever is left is the charge on the metal! FeBr3 The three bromides are each 1- so iron must be 3+ for the compound to have zero net charge. Iron (III) bromide

Examples FeCl2 iron (II) chloride FeCl3 iron (III) chloride SnS AgCl ZnS iron (II) chloride iron (III) chloride tin (II) sulfide tin (IV) sulfide silver chloride zinc sulfide Note: Some transition metals have only one oxidation state, so Roman numbers are omitted.

Metals with multiple charges Transition metals. Here it is easier to list some of the common elements that only have a single oxidation state. All Group 3B are 3+ Zn and Cd are 2+ Ag is 1+

Oxidation numbers and the P.T. Some observed trends in compounds. Metals have positive oxidation numbers. Transition metals typically have more than one oxidation number. Nonmetals and semimetals have both positive and negative oxidation numbers. No element exists in a compound with an oxidation number greater than +8. The most negative oxidation numbers equals the group number - 8

Common oxidation numbers Li +1 Na Cs Rb K Fr Ba +2 Be Mg Sr Ca Ra B +3 C +4 -2 -4 N +5 +4 +3 +2 +1 -3 O -1 -2 F -1 Ne H +1 He Al +3 Si +4 -4 P +5 +3 -3 S +6 +4 +2 -2 Cl +7 +5 +3 +1 -1 Ar Sc 3+ Ti +4 +3 +2 V +5 +4 +3 +2 Cr +6 +3 +2 Mn +7 +6 +4 +3 +2 Fe +3 +2 Co +3 +2 Ni +2 Cu +2 +1 Zn +2 Ga +3 Ge +4 -4 As 5+ 3+ 3- Se 6+ 4+ 2- Br +5 +1 -1 Kr +4 +2 Y +3 Zr +4 Nb +5 +4 Mo +6 +4 +3 Tc +7 +6 +4 Ru +8 +6 +4 +3 Rh +4 +3 +2 Pd +4 +2 Ag +1 Cd +2 In +3 Sn +4 +2 Sb +5 +3 -3 Te +6 +4 -2 I +7 +5 +1 -1 Xe +6 +4 +2 Lu +3 Hf +4 Ta +5 W +6 +4 Re +7 +6 +4 Os +8 +6 Ir +4 +3 Pt +4 +2 Au +3 +1 Hg +2 +1 Tl +3 +1 Pb +4 +2 Bi +5 +3 Po +2 At -1 Rn Lr +3 Common oxidation numbers

Polyatomic ions NH4+ ammonium NO3- nitrate SO42- sulfate OH- hydroxide A special class of ions where a group of atoms tend to stay together. NH4+ ammonium NO3- nitrate SO42- sulfate OH- hydroxide O22- peroxide Your book contains a more complete list.

Polyatomic ions For compounds that contain 1 or 2 polyatomic ions, base the formulas upon the given ion name(s). ammonium chloride NH4Cl sodium hydroxide NaOH potassium permanganate KMnO4 ammonium sulfate (NH4)2SO4

Naming Inorganic Compounds Names and Formulas of Ionic Compounds Polyatomic anions containing oxygen with more than two members in the series are named as follows (in order of decreasing oxygen): per- …. -ate ClO41- …. -ate ClO31- …. -ite ClO21- hypo- …. -ite ClO1-

Oxidation number and nomenclature Polyatomic anions containing oxygen rely on a modification of the name of the other element to indicate the oxidation number. Anions per ________ate ________ate ________ite hypo ________ite Increased #oxygen and Oxidation number

Oxidation number and nomenclature Examples Cl oxidation number Formula Name +7 NaClO4 sodium perchlorate +5 NaClO3 sodium chlorate +3 NaClO2 sodium chlorite +1 NaClO sodium hypochlorite -1 NaCl sodium chloride Usually, the overall charges of all ions for a nonmetal are the same. Sometimes the -ates and -ites have a different charge than the -ide ions.

Slivka’s Square -ate has 4 Oxygens -ate has 3 Oxygens Polyatomic Ions 1 2 13 14 15 16 17 18 I A II A III A IV A VA VI A VIIA 0 H He -ate & -ite charges usually = -ide charge 1 2 3 4 5 6 7 Li Be B C N O F Ne 3 4 5 6 7 8 9 10 11 12 III B IVB V B VIB VIIB VIII B IB IIB Na Mg Slivka’s Square Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn + Fr Ra Lr * Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No La Ac Er Fm Tm Md Dy Cf Ho Es +

Polyatomic Ions “ate” 4 Oxygens .. Inside Slivka’s Square ex: SO42- sulfate 3 Oxygens .. Borders the outside of the square ex: NO31- nitrate

1 less Oxygen compared to the -ate ex: ClO21- chlorite SO32- sulfite Polyatomic Ions “ite” 1 less Oxygen compared to the -ate ex: ClO21- chlorite SO32- sulfite

“per” root name “ate” has 1 more O than the “ate” ex: IO41- periodate Polyatomic Ions “per” root name “ate” has 1 more O than the “ate” ex: IO41- periodate “hypo” root name “ite” has 2 less O than the “ate” ex: ClO1- hypochlorite

“Per”-“ate” 1 more O - “ate” - “ite” 1 less O “hypo”-“ite” 2 less O Polyatomic Ions “Per”-“ate” 1 more O - “ate” - “ite” 1 less O “hypo”-“ite” 2 less O (also notice oxidation # of nonmetal changes)

follow Group A patterns Polyatomic Ions Group B Elements follow Group A patterns 1 2 18 I A II A VIIIA CrO42- chromate MnO41- permanganate H 13 14 15 16 17 He III A IV A VA VI A VIIA 1 2 3 4 5 6 7 Li Be B C N O F Ne 3 4 5 6 7 8 9 10 11 12 III B IVB V B VIB VIIB VIII B IB IIB Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn + Fr Ra Lr * Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No La Ac Er Fm Tm Md Dy Cf Ho Es +

Other Methods of Naming - Latin For ionic compounds containing a metal and a nonmetal, the Latin root word for the metal is sometimes used with an -ous or -ic suffix. The -ous suffix indicates a lower oxidation state, -ic a higher one. Ex. ferrous = Fe2+ ferric = Fe3+

Latin Root Words plumbous = Pb2+ cuprous = Cu1+ plumbic = Pb4+ cupric = Cu2+ stannous = Sn2+ stannic = Sn4+ plumbous = Pb2+ plumbic = Pb4+ aurous = Au1+ auric = Au3+ Note that there is no pattern between the -ous and -ic suffixes and the actual charge of the ions.

Naming Covalent Compounds What is the difference between SO32- and SO3 polyatomic ion vs. neutral compound sulfite ion vs. sulfur (VI) oxide

Naming Covalent Compounds Some nonmetals can have more than one positive oxidation state when they share electrons to form molecules called covalent compounds. Therefore Roman numbers must be used. SCl4 sulfur (IV) chloride SCl sulfur (VI) chloride CO carbon (II) oxide CO2 carbon (IV) oxide

Naming Inorganic Compounds Names and Formulas of Binary Molecular Compounds Binary molecular compounds are composed of two nonmetallic elements. The element with the positive oxidation number (the one closest to the lower left corner on the periodic table) is usually written first. Exception: NH3. Greek prefixes are used to indicate the number of atoms in the molecule(subscripts). PCl5 is phosphorus pentachloride

Naming Inorganic Compounds Names and Formulas of Binary Molecular Compounds Roman numerals can be used to indicate the positive oxidation number, but sometimes prefixes more accurately describe the actual composition of the molecule. Example: sulfur (V) fluoride exists as disulfur decafluoride molecules. F F F F S S F F F F F F

Other Methods of Naming molecules For binary molecular compounds composed of two nonmetals, prefixes are sometimes used to indicate the number of atoms of each element present. Common prefixes mono = 1 di = 2 tri = 3 tetra = 4 penta = 5 hexa = 6 hepta = 7 octa = 8 deca = 10

Other Methods of Naming molecules name elements in the formula. use prefixes to indicate how many atoms there are of each type. N2O5 CO2 CO CCl4 dinitrogen pentoxide carbon dioxide carbon monoxide carbon tetrachloride The rule may be modified to improve how a name sounds. Example - use monoxide not monooxide.

-ide becomes hydro-….-ic acid -ate becomes -ic acid Other Naming -Acids Acids are substances that produce H+ ions in water solutions (aqueous). The names of acids are related to the names of the anions to which H+ is bonded: -ide becomes hydro-….-ic acid H2S is hydrosulfuric acid -ate becomes -ic acid H3PO4 is phosphoric acid -ite becomes -ous acid HNO2 is nitrous acid

Naming Inorganic Acids

Naming Inorganic Acids Salt Name Formula Acid Name Formula Sodium acetate NaC2H3O2 Acetic acid HC2H3O2 Sodium chloride NaCl Hydrochloric acid HCl Sodium hyponitrite NaNO Hyponitrous acid HNO Sodium phosphite Na3PO3 Phosphorous acid H3PO3 Sodium sulfate Na2SO4 Sulfuric acid H2SO4

Naming Inorganic Compounds Names and Formulas of Acids Acids contain hydrogen as the only cation. The names of acids are related to the names of the anions: -ide becomes hydro-….-ic acid; H2S is hydrosulfuric acid -ate becomes -ic acid; H3PO4 is phosphoric acid -ite becomes -ous acid. HNO2 is nitrous acid

Other Naming -Double & Triple Salts Polyatomic anions containing oxygen with additional hydrogens are named by adding hydrogen (or bi-) for one extra H, dihydrogen (for two extra H), etc., to the name as follows: CO32- is named carbonate, but HCO3- is hydrogen carbonate (or bicarbonate) H2PO4- dihydrogen phosphate anion. Note that these are not named as acids, since another cation is still needed to balance the charge.

Other Naming -Double & Triple Salts Two or three different positive ions can be attracted to the same negative ion to form a single compound. These are called double or triple salts. Each ion is named as it appears. NaHCO3 sodium hydrogen carbonate (or sodium bicarbonate) AlK(SO4)2 aluminum potassium sulfate

Other Naming - Hydrates pentahydrate 5H2O Other Naming - Hydrates Hydrated compounds physically trap water molecules as part of their structure. A prefix is used to indicate the relative number of water molecules present with the word hydrate added after the compound’s name. copper (II) sulfate pentahydrate CuSO45H2O

Other Naming - Historical Names Sometimes the names of compounds are based upon their historical significance or derivation. There are no patterns or rules for determining these names, so they would have to be memorized. For example, H2O is called water, not dihydrogen monoxide Check out http://www.dhmo.org

A quick review of nomenclature Is a metal present as the first element? No No Is a nonmetal the first element? Is hydrogen first element? Yes Yes Yes Can the metal have more than one oxidation state? Use Roman numerals or may use prefixes (mono, di, tri ...) Name as an acid No -ides become hydro- -ic acids -ates become -ic acids -ites become -ous acids Yes Roman numerals are not needed. Use Roman numerals or may use Latin name with -ous/-ic suffixes

A quick review of nomenclature Look up the name or formula Is it a binary compound? No Yes No Does it contain one of the 8 common ions? Does it have more or less O atoms than one of the -ate ions? No Use the -ide suffix for the negative ion Yes Yes Name the common polyatomic ion Use per- -ate 1 more O -ite 1 less O hypo- -ite 2 less O

Naming Compounds Summary Simple rules that will keep you out of trouble most of the time. Groups IA, 2A, 3A (except Tl) only have a single oxidation state that is the same as the group number - don’t use numbers. Most other metals and semimetals have multiple oxidation states - use numbers. If you are sure that a transition group element only has a single state, don’t use a number.

Average Atomic Mass What is the average atomic mass of carbon-12? Why is this a bad question? If I traveled to alpha centauri, would the average atomic mass of chlorine be 35.45? Can you calculate the AAM of the following: 1H = 99% 2H = 1%

Moles and Moles How many atoms in a mole? What does the “mole” do? How do you calculate molar mass? What is an empirical formula? What is a molecular formula?

Atomic masses Atoms are composed of protons, neutrons and electrons. Almost all of the mass of an atom comes from the protons and neutrons. All atoms of the same element will have the same number of protons. The number of neutrons may vary - isotopes. Most elements exist as a mixture of isotopes.

Isotopes Isotopes Atoms of the same element but having different masses. Each isotope has a different number of neutrons. Isotopes of hydrogen H H H Isotopes of carbon C C C 1 2 1 3 1 12 6 13 6 14 6

Isotopes Most elements occur in nature as a mixture of isotopes. Element Number of stable isotopes H 2 C 2 O 3 Fe 4 Sn 10 This is one reason why atomic masses(weights) are not whole numbers. They are based on averages.

Atomic masses As a reference point, we use the atomic mass unit (u), which is equal to 1/12th of the mass of a 12C atom. (One atomic mass unit (u) = 1.66 x 10-24 gram) Using this relative system, the mass of all other atoms can be assigned. Examples 7Li = 7.016 004 u 14N = 14.003 074 01 u 29Si = 28.976 4947 u

Average atomic masses One can calculate the average atomic weight of an element if the abundance of each isotope for that element is known. Silicon exists as a mixture of three isotopes. Determine it’s average atomic mass based on the following data. Isotope Mass (u) Abundance 28Si 27.9769265 92.23 % 29Si 28.9764947 4.67 % 30Si 29.9737702 3.10 %

Average atomic masses 92.23 (27.9769265 u) = 25.80 u 28Si 100 4.67 3.10 (29.9737702 u) = 0.929 u 28Si 29Si 30Si Average atomic mass for silicon = 28.08 u

The mole The number of atoms in 12.000 grams of 12C can be calculated. One atom 12C = 12.000 u = 12 x (1.661 x 10-24 g) = 1.993 x 10-23 g / atom # atoms = 12.000 g (1 atom / 1.993 x 10-23 g) = 6.021 x 1023 atoms The number of atoms of any element needed to equal its atomic mass in grams will always be 6.02 x 1023 atoms - the mole.

Moles and masses Atoms come in different sizes and masses. A mole of atoms of one type would have a different mass than a mole of atoms of another type. H - 1.008 grams / mol O - 16.00 grams / mol Mo - 95.94 grams / mol Pb - 207.2 grams / mol We rely on a straight forward system to relate mass and moles.

1 mole of any element = 6.02 x 1023 atoms The mole 1 mole of any element = 6.02 x 1023 atoms = gram atomic mass Atoms, ions and molecules are too small to directly measure in u. Using moles gives us a practical unit. We can then relate atoms, ions and molecules, using an easy to measure unit - the gram.

Masses of atoms and molecules Atomic mass The average, relative mass of an atom in an element. Can be expressed in relative amtomic mass units (u) or grams / mole. Molecular or formula mass The total mass for all atoms in a compound.

Masses of atoms and molecules H2O - water 2 hydrogen 2 x 1.008 u 1 oxygen 1 x 16.00 u mass of molecule 18.02 u 18.02 g / mol Rounded off based on significant figures

The mole If we had one mole of water and one mole of hydrogen, we would have the same number of molecules of each. 1 mol H2O = 6.022 x 1023 molecules 1 mol H2 = 6.022 x 1023 molecules We can’t weigh out moles-we use grams. We would need to weigh out a different number of grams to have the same number of molecules

Converting units Factor label method Regardless of conversion, keeping track of units makes thing come out right Must use conversion factors - The relationship between two units Canceling out units is a way of checking that your calculation is set up right!

Molecular mass vs. formula mass Formula mass - Add the masses of all the atoms in formula; for molecular and ionic compounds. Molecular mass - Calculated the same as formula mass; only valid for molecules. Both have units of either u or grams / mole. Molar mass is the generic term for the mass of one mole of anything.

Formula mass, FM The sum of the atomic masses of all elements in a compound based on the chemical formula. You must use the atomic masses of the elements listed in the periodic table. CO2 1 atom of C and 2 atoms of O 1 atom C x 12.011 u = 12.011 u 2 atoms O x 15.9994 u = 31.9988 u Formula mass = 44.010 u or g / mol

Another example CH3CH2OH - ethyl alcohol 6 hydrogen 6 x 1.008 u 2 carbon 2 x 12.01 u 6 hydrogen 6 x 1.008 u 1 oxygen 1 x 16.00 u mass of molecule 46.07 u 46.07 g /mol

Molar masses Once you know the mass of an atom, ion, or molecule, just remember: Mass of one unit - use amu Mass of one mole of units - use grams / mole The numbers DON’T change -- just the units.

Example - (NH4)2SO4 How many atoms are in 20.0 grams of ammonium sulfate? Formula weight = 132.14 grams/ 1 mol Atoms in formula = 15 atoms / 1 formula unit X moles = 20.0 g x = 0.151 mol 1 mol 132.14 g atoms = 0.151 mol x x 6.02 x1023 units 1 mol 15 atoms 1 unit atoms = 1.36 x1024

Formula weight = 132.14 g / 1 mol (NH4)2SO4 Example - (NH4)2SO4 Other information can be derived from the chemical formula of a compound. How many moles of ammonium ions are in 20.0 grams of ammonium sulfate? Formula weight = 132.14 g / 1 mol (NH4)2SO4 2 moles NH4 / 1 mol (NH4)2SO4 x moles = 20.0 g x = 0.151 mol (NH4)2SO4 1 mol 132.14 g x moles NH4 = 0.151 mol (NH4)2SO4 x 2 moles NH4 1 mol (NH4)2SO4 moles NH4 = 0.302

Formula weight = 132.14 g / 1 mol (NH4)2SO4 Example - (NH4)2SO4 How many grams of sulfate ions are in 20.0 grams of ammonium sulfate? Formula weight = 132.14 g / 1 mol (NH4)2SO4 96.06 grams SO4 / 1 mol (NH4)2SO4 x moles = 20.0 g x = 0.151 mol (NH4)2SO4 1 mol 132.14 g x grams SO4 = 0.151 mol (NH4)2SO4 x 96.06 g SO4 1 mol (NH4)2SO4 grams SO4 = 14.5

Masses of atoms and molecules Law of Definite Composition - compounds always have a definite proportion of the elements that make it up. These proportions can be expressed as ratios of atoms, equivalent mass values, percentage by mass or volumes of gaseous elements. Ex. Water always contains 2 H atoms for every O atom, which is 2 g H for every 16 g O or 11.1% H and 88.9% O by mass.

Percent Composition by Mass Percent composition can also be determined from experimental data. Example: When 2.47 g KClO3 is heated strongly, 0.96 g of O2 gas is driven off. What is the % by mass of oxygen in KClO3? % 0 = 0.96 g O x 100 = 38.9 % O 2.47 g KClO3 Based upon the formula mass: % O = 48.00 u O x 100 = 39.17 % O 122.55 u KClO3

Gay-Lussac’s Law Law of of Combining Volumes. At constant temperature and pressure, the volumes of gases involved in a chemical reaction are in the ratios of small whole numbers. Studies by Joseph Gay-Lussac led to a better understanding of molecules and their reactions.

Gay-Lussac’s Law Example. Reaction of hydrogen and oxygen gases. Two ‘volumes’ of hydrogen will combine with one ‘volume’ of oxygen to produce two volumes of water. We now know that the equation is: 2 H2 (g) + O2 (g) 2 H2O (g) + H2 O2 H2O

Contain same number of moles of molecules Avogadro’s law Equal volumes of gas at the same temperature and pressure contain equal numbers of molecules (or moles of molecules). Contain same number of moles of molecules

Standard conditions (STP) Remember the following standard conditions. Standard temperature = 273.15 Kelvin (the normal freezing point of water, 0ºC) Standard pressure = 1 atmosphere (the normal air pressure at sea level, 14.7 psi) At these conditions: One mole of any gas has a volume of 22.4 liters at STP.

Applying Law of Definite Composition In an expermiment, 10.0 grams of water is decomposed by electrolysis. Problem: How many liters of O2 gas will be formed at STP? How many grams of H2 gas will be formed? X L O2 =10.0g H2O x x x 1 mol H2O 18.0 g H2O 0.5 mol O2 1 mol H2O 22.4 L O2 1 mol O2 liters O2 = 6.22 Note that 12.4 L H2 will also be formed. X g H2 =10.0g H2O x 2.02 g H2 18.02 g H2O = 1.12 g H2

Empirical formula This type of formula shows the ratios of the number of atoms of each kind in a compound. For organic compounds, the empirical formula can be determined by combustion analysis. Elemental analyzer An instrument in which an organic compound is quantitatively converted to carbon dioxide and water -- both of which are then measured.

Elemental analyzer furnace CO2 trap H2O O2 sample A sample is ‘burned,’ completely converting it to CO2 and H2O. Each is collected and measured as a weight gain. By adding other traps elements like oxygen, nitrogen, sulfur and halogens can also be determined.

Elemental analysis Example: A compound known to contain only carbon, hydrogen and nitrogen is examined by elemental analysis. The following information is obtained. Original sample mass = 0.1156 g Mass of CO2 collected= 0.1638 g Mass of H2O collected= 0.1676 g Determine the % of each element in the compound.

Elemental analysis Mass of carbon 12.01 g C = 0.04470 g C 44.01 g CO2 Mass of hydrogen Mass of nitrogen 0.1638 g CO2 12.01 g C 44.01 g CO2 = 0.04470 g C 0.1675 g H2O 2.016 g H 18.01 g H2O = 0.01875 g H 0.1156 g sample - 0.04470 g C - 0.01875 g H = 0.05215 g N

Elemental analysis Since we know the total mass of the original sample, we can calculate the % of each element. % C = x 100% = 38.67 % % H = x 100% = 16.22 % % N = x 100% = 45.11 % 0.04470 g 0.1156 g 0.01875 g 0.05215 g

Empirical formula Empirical formula The simplest formula that shows the ratios of the number of atoms of each element in a compound. Example - the empirical formula for hydrogen peroxide (H2O2) is HO. We can use either our mass or our percent composition information from the earlier example to determine an empirical formula.

Empirical formula 1 mol C = 0.003722 mol C 12.01 g C 1 mol H 0.01875 g H 1 mol H 1.008 g H = 0.0186 mol H 0.04470 g C 1 mol C 12.01 g C = 0.003722 mol C 0.05215 g N 1 mol N 14.01 g N = 0.003722 mol N

Empirical formula The empirical formula is then found by looking for the smallest whole number mole ratio. C 0.003722 / 0.003722 = 1.000 H 0.0186 / 0.003722 = 4.998 N 0.003722 / 0.003722 = 1.000 The empirical formula is CH5N

Empirical formula % C = 38.67 % % H = 16.22 % % N = 45.11 % From experimental analysis, we found that a compound had a composition of: If we assume that we have a 100.0 gram sample, then we can divide each percentage by the elements atomic mass and determine the relative number of moles of each. % C = 38.67 % % H = 16.22 % % N = 45.11 %

Empirical formula 16.22 g H 1 mol H 1.008 g H = 16.09 mol H 38.67 g C 1 mol C 12.01 g C = 3.220 mol C 45.11 g N 1 mol N 14.01 g N = 3.220 mol N

Empirical formula The empirical formula is found by looking for the smallest whole number ratio. C 3.220 / 3.220 = 1.000 H 16.09 / 3.220 = 4.997 N 3.220 / 3.220 = 1.000 The empirical formula is determined to be the same, CH5N, whether using actual masses of the elements present in the sample or by using their % composition by mass.

Molecular formula Molecular formula - shows the actual number of each type of atom in a molecule. They are multiples of the empirical formula. If you know the molecular mass, then the molecular formula can be found. For our earlier example, what would be the molecular formula if you knew that the molecular mass was 62.12?

Molecular formula Empirical formula CH5N Empirical formula mass 31.06 u Molecular mass 62.12 Ratio: 62.12 / 31.06 = 2 The molecular formula is C2H10N2 Note: This does not tell you how the atoms are arranged in the compound!

Hydrated Compounds The formula for hydrated compounds are solved in a similar fashion as empirical formulas. Example: When a 5.000 g sample of hydrated barium chloride is heated to dryness, 0.738 g H2O is lost. 5.000g hydrate - 0.738g H2O = 4.262g BaCl2

Hydrated Compounds BaCl2 0.0205 / 0.0205 = 1.00 4.262g BaCl2 (1 mol BaCl2) = 0.0205 mol BaCl2 (208.2 g BaCl2) 0.738 g H2O ( 1 mol H2O ) = 0.0410 mol H2O (18.0 g H2O ) BaCl2 0.0205 / 0.0205 = 1.00 H2O 0.0410 / 0.0205 = 2.00 The compound’s formula is BaCl22H2O

Molarity M = moles solute mol liters of solution L = Molarity Recognizes that compounds have different formula weights. A 1 M solution of sucrose contains the same number of molecules as 1 M ethanol. [ ] - special symbol which means molar ( mol/L )

Molarity M NaOH = 10 mol NaOH / 2.0 L = 5.0 M Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH. M NaOH = 10 mol NaOH / 2.0 L = 5.0 M

Solution preparation Solutions are typically prepared by: Dissolving the proper amount of solute and diluting to volume. Dilution of a concentrated solution. Lets look at an example of the calculations required to prepare known molar solutions using both approaches.

Making a solution You are assigned the task of preparing 250.0 mL of a 1.00 M solution of sodium hydroxide. What do you do? First, you need to know how many moles of NaOH are in 250.0 mL of a 1.00 M solution. mol = M x V (in liters) = 1.00 M x 0.2500 liters = 0.250 moles NaOH

Making a solution Next, we need to know how many grams of NaOH to weigh out. g NaOH = mol x molar massNaOH = 0.250 mol x 40.0 g/mol = 10.0 grams NaOH

Making a solution Finally, you’re ready to make the solution. Weigh out exactly 10.0 grams of dry, pure NaOH and transfer it to a volumetric flask, (or some other containing where the exact volume can be accurately measured.) Fill the flask about 1/2 of the way with distilled water and gently swirl until the solid dissolves. Now, dilute exactly to the mark, cap and mix.

Dilution Once you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutions M1V1 = M2V2 1 = initial 2 = final Any volume or concentration unit can be used as long as you use the same units on both sides of the equation.

Dilution How many ml of concentrated 12.0 M HCl must be diluted to produce 250.0 mL of 1.00 M HCl? M1V1 = M2V2 M1 = 12.0 M M2 = 1.00 V1 = ??? ml V2 = 250.0 mL V1 = M2V2 / M1 M2 = (1.00 M) (250.0 mL) = 20.8 mL (12.0 M)

Diluting an Acid When diluting concentrated acids, ALWAYS add the acid to water to help dissipate the heat released. Fill the volumetric flask (or some other containing where the exact volume can be accurately measured.) about 1/2 of the way with distilled water. Measure out exactly 20.8 mL of 12.0 M HCl and transfer it to a volumetric flask. Gently swirl to mix. Now, dilute exactly to the mark, cap and mix.

Other Methods of Expressing Concentration When making different solutions with a specific molarity, the number of milliliters of solvent needed to prepare 1 liter of solution will vary. Sometimes it is necessary to know the exact proportions of solute to solvent that are in a particular solution. Various methods have been devised to express these proportions.

Molality moles solute mol Molality (m) = = kilograms of solvent kg Recognizes that the ratio between moles of solute and kg of solvent can vary. A 1 m solution of sucrose contains the same number of molecules as 1 m ethanol. The freezing point of water is lowered by 1.86ºC/ m and the boiling point is raised by 0.51ºC/ m.

Density grams of solution g Density (D) = = milliliters of solution mL Focuses on the total solution and does not emphasize either the solute or solvent. g solution = g solute + g solvent Units may be expressed as other mass per volume ratios.

Percent Composition value of the part Percent Composition = x 100 Value of the whole Percent Composition = x 100 % by Mass = g solute / g solution x 100 % by Volume = mL solute / mL solution x100 % by Mass per Volume = g solute/mL solution x 100 Must specify which type of % composition.

Mole Fraction moles of solute or solvent Mole fraction = total moles of solute & solvent Mole fraction = Often used to compare ratio between moles of gases in a mixture. The mole ratio of gases in a mixture is equal to their pressure ratio and their volume ratio.

Parts per Million or Billion # grams of solute 1,000,000 g solution Parts per million (ppm) = Parts per Million or Billion Used to express concentrations for very dilute solutions. For aqueous solutions, the mixture is mostly water. Therefore, the density of the solution = 1 g/mL, and 1 ppm = 1 g/1000 L.

Stoich Baby Given the following equation ___N2 + ___H2  ___NH3 1 3 2 Given one mole of nitrogen gas, how many moles of ammonia would form? Assuming gases at STP, given one mole of nitrogen, how many liters of ammonia would form? Given 2 liters of nitrogen and 5 liters of hydrogen, how may liters of ammonia are formed? What is left over?

Stoichiometry Stoichiometry The study of quantitative relationships between substances undergoing chemical changes. Law of Conservation of Matter In chemical reactions, the quantity of matter does not change. The total mass of the products must equal that of the reactants.

Chemical equations Chemist’s shorthand to describe a reaction. It shows: All reactants and products The state of all substances Any conditions used in the reaction CaCO3 (s) CaO (s) + CO2 (g) Reactant Products  A balanced equation shows the relationship between the quantities of all reactants and products.

Balancing chemical equations Each side of a chemical equation must have the same number of each type of atom. CaCO3 (s) CaO (s) + CO2 (g) Reactants Products 1 Ca 1 Ca 1 C 1 C 3 O 3 O

Balancing chemical equations Step 1 Count the number of atoms of each element on each side of the equation. Step 2 Determine which atom numbers are not balanced. Step 3 Balance one atom at a time by using coefficients in front of one or more substances. Step 4 Repeat steps 1-3 until everything is balanced.

Example. Decomposition of urea (NH2)2CO + H2O ______> NH3 + CO2 2 N 1 N < not balanced 6 H 3 H < not balanced 1 C 1 C 2 O 2 O We need to double NH3 on the right. (NH2)2CO + H2O ______> 2NH3 + CO2

Mass relationships in chemical reactions Stoichiometry - The calculation of quantities of reactants and products in a chemical reaction. 2 H2 + O2 -----> 2 H2O You need a balanced equation and you WILL work with moles.

Stoichiometry, General steps. 1 Balance the chemical equation. 2 Calculate formula masses. 3 Convert masses to moles. 4 Use chemical equation to get the needed answer. Convert back to mass if needed. 5

Mole calculations The balanced equation shows the reacting ratio between reactants and products. 2C2H6 + 7O2 4CO2 + 6H2O For each chemical, you can determine the moles of each reactant consumed moles of each product made If you know the formula mass, mass quantities can be used.

Mole-gram conversion How many moles are in 14 grams of N2 ? Formula mass = 2 N x 14.01 g/mol = 28.02 g /mol moles N2 = 14 g x 1 mol /28.02 g = 0.50 moles

Mass calculations We don’t directly weigh out molar quantities. We can use measured masses like kilograms, grams or milligrams. The formula masses and the chemical equations allow us to use either mass or molar quantities.

Mass calculations How many grams of hydrogen will be produced if 10.0 grams of calcium is added to an excess of hydrochloric acid? 2HCl + Ca ______> CaCl2 + H2 Note: We produce one H2 for each calcium. There is an excess of HCl so we have all we need.

Mass calculations = 10.0 g x = 0.25 mol Ca 2HCl + Ca ____> CaCl2 + H2 First - Determine the number of moles of calcium available for the reaction. Moles Ca = grams Ca / FM Ca = 10.0 g x = 0.25 mol Ca 1 mol 40.08 g

Mass calculations 2HCl + Ca _____> CaCl2 + H2 10 g Ca = 0.25 mol Ca According to the chemical equation, we get one mole of H2 for each mole of Ca. So we will make 0.25 moles of H2. grams H2 produced = moles x FW H2 = 0.25 mol x 2.016 g/mol = 0.504 grams

Mass calculations OK, so how many grams of CaCl2 were made? 2HCl + Ca _____> CaCl2 + H2 10 g Ca = 0.25 mol Ca We would also make 0.25 moles of CaCl2. g CaCl2 = 0.25 mol x FM CaCl2 = 0.25 mol x 110.98 g / mol CaCl2 = 27.75 g CaCl2

Limiting reactant In the last example, we had HCl in excess. Reaction stopped when we ran out of Ca. Ca is considered the limiting reactant. Limiting reactant - the material that is in the shortest supply based on a balanced chemical equation.

Limiting reactant example

Example For the following reaction, which is limiting if you have 5.0 g of hydrogen and 10 g oxygen? Balanced Chemical Reaction 2H2 + O2 ________> 2H2O You need 2 moles of H2 for each mole of O2. Moles of H2 5 g = 2.5 mol Moles of O2 10g = 0.31 mol 1 mol 2.0 g 1 mol 32.0 g

Example Balanced Chemical Reaction 2H2 + O2 2H2O You need 2 moles of H2 for each mol of O2 You have 2.5 moles of H2 and 0.31 mol of O2 Need a ratio of 2:1 but we have a ratio of 2.5 : 0.31 or 8.3 : 1. Hydrogen is in excess and oxygen is the limiting reactant.

Theoretical, actual, and percent yields Theoretical yield The amount of product that should be formed according to the chemical reaction and stoichiometry. Actual yield The amount of product actually formed. Percent yield Ratio of actual to theoretical yield, as a %. Quantitative reaction When the percent yield equals 100%.

Yield Less product is often produced than expected. Possible reasons A reactant may be impure. Some product is lost mechanically since the product must be handled to be measured. The reactants may undergo unexpected reactions - side reactions. No reaction truly has a 100% yield due to the limitations of equilibrium.

Percent yield The amount of product actually formed divided by the amount of product calculated to be formed, times 100. % yield = x 100 In order to determine % yield, you must be able to recover and measure all of the product in a pure form. Actual yield Theoretical yield

Stoichiometry Step 1 Step 2 Step 3 Step 4 Step 5 Identify species present in solution and determine the reaction that occurs Step 2 Write the balanced net ionic equation Step 3 Calculate the moles of reaction solution = molarity x volume heterogeneous = grams  molar mass Step 4 Consider the limiting reactant Step 5 Answer the question using stoich!