Presentation on theme: "AP Chemistry Introductory Material Chemical Foundations Chapter 1 Scientific Method Observations Hypotheses Predictions Theory Or Model Predictions Experiment."— Presentation transcript:
AP Chemistry Introductory Material
Chemical Foundations Chapter 1 Scientific Method Observations Hypotheses Predictions 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 –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 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 corrected 6.4
Dimensional Analysis Do I really have to? –No, but it will cost you extra work explaining yourself –Units written out in Dim Analysis are self explanatory Its way easier! Way, way easier!! Just Do it!
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. mi 2 and an average depth of 20.1 ft. (5280. ft in a mile, 12 in in a foot, 2.54 cm in an inch, 10 6 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?
Its the Law Explain the following laws: –Conservation of Mass –Definite proportion –Multiple proportion Name four parts of Daltons 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 CRTs –Used CRT to determine charge to mass ratio –Discovered electron Rutherfords Gold Foil –Used alpha particles and gold foil –Discovered a dense, positive nucleus Millakans Oil Droplet –Discovered the charge of an electron –Calculated the mass of an electron with JJs reults
Modern Theory Subatomic particles are? –Electron, neutron and proton Nucleus is composed of ? –Neutron and proton Electrons are in clouds –What does that mean?
Symbol F 19 9 How many protons? How many neutrons? How many electrons?
Periodic Table Describe the following: Period Metals Non-metals Semi-metals Alkali Metals Alkali Earth Metals Transition Metals Halogens Noble gases
Modern periodic table H Li Na Cs Rb K TlHgAuHfLuBa Fr PtIrOsReWTa He RnAtPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb LrRa ZnCuTiScNiCoFeMnCrV InXeITeSbSn GaKrBrSeAsGe AlArClSPSi BNeFONC 1 213 14 15 16 17 18 I A II A III A IV A V A VI A VIIA 0 3 4 5 6 7 8 9 10 11 12 III B IVB V B VIB VIIB VIII B IB IIB 12345671234567 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 Li Na Cs Rb K TlHgAuHfLuBa Fr PtIrOsReWTaPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb LrRa ZnCuTiScNiCoFeMnCrV InSn Ga Al Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Er Fm Tm Md Dy Cf Ho Es AtRn B He NeFONC TeXe I Sb As Si KrBrSe ArClSP Ge La Ac * + * + H
Nonmetals H Li Na Cs Rb K TlHgAuHfLyBa Fr PtIrOsReWTaPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb LrRa ZnCuTiScNiCoFeMnCrV InSbSn GaGe Al Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Er Fm Tm Md Dy Cf Ho Es At Te As Si B He Rn XeI KrBrSe ArClS NeFO P NC La Ac * + * +
Semimetals or Metalloids He Rn Xe I KrBrSe ArClS NeFO P NC H Li Na Cs Rb K TlHgAuHfLuBa Fr PtIrOsReWTaPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb LrRa ZnCuTiScNiCoFeMnCrV InSbSn GaGe Al 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 At Te As Si B * + * +
Bonds and Stuff Ion Cation and anion Ionic Bond Covalent Bond Molecule Formula Unit Chemical Formula Structural Formula Explain the following
Important: note that there are no easily identified NaCl molecules in the ionic lattice. Therefore, we cannot use molecular formulas to describe ionic substances. Ions and Ionic Compounds
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 Ions are charged particles formed by the transfer of electrons between elements or combinations of elements. Cation Cation - a positively charged ion. Mg Mg 2+ + 2e - Anion Anion - a negatively charged ion. F 2 + 2e - 2F -
Writing Formulas Al 3+ O 2- All compounds are electrically neutral. The sum of the positive and negative charges must add up to zero. Al 2 O 3 Use subscripts to indicate how many of each ion is used. 32
Naming inorganic compounds When an element forms only onecompound with a given anion. name the cation -ide name the anion using the ending (-ide) for monatomic ions NaClsodium chloride MgBr 2 magnesium bromide Al 2 O 3 aluminum oxide K 3 Npotassium 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. Cu 1+ + O 2- = Cu 2 O copper(I) oxide copper(I) oxide Cu 2+ + O 2- = 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! FeBr 3 –The three bromides are each 1- so iron must be 3+ for the compound to have zero net charge. Iron (III) bromideIron (III) bromide
Examples FeCl 2 FeCl 3 SnS SnS 2 AgCl ZnS Note: Note: Some transition metals have only one oxidation state, so Roman numbers are omitted. iron (II) chloride iron (III) chloride tin (II) sulfide tin (IV) sulfide silver chloride zinc sulfide
Metals with multiple charges Transition metals.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.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
Tl +3 +1 Hg +2 +1 Au +3 +1 Hf +4 Lu +3 Li +1 Na +1 Cs +1 Rb +1 K +1 Fr +1 Pt +4 +2 Ir +4 +3 Os +8 +6 Re +7 +6 +4 W +6 +4 Ta +5 H +1 He Rn At Po +2 Bi +5 +3 Pb +4 +2 Cd +2 Ag +1 Zr +4 Y +3 Pd +4 +2 Rh +4 +3 +2 Ru +8 +6 +4 +3 Tc +7 +6 +4 Mo +6 +4 +3 Nb +5 +4 Lr +3 Ba +2 Be +2 Mg +2 Sr +2 Ca +2 Ra +2 Zn +2 Cu +2 +1 Ti +4 +3 +2 Sc 3+ Ni +2 Co +3 +2 Fe +3 +2 Mn +7 +6 +4 +3 +2 Cr +6 +3 +2 V +5 +4 +3 +2 In +3 Xe +6 +4 +2 I +7 +5 +1 Te +6 +4 -2 Sb +5 +3 -3 Sn +4 +2 Ga +3 Kr +4 +2 Br +5 +1 Se 6+ 4+ 2- As 5+ 3+ 3- Ge +4 -4 Al +3 Ar Cl +7 +5 +3 +1 S +6 +4 +2 -2 P +5 +3 -3 Si +4 -4 B +3 Ne F O -2 N +5 +4 +3 +2 +1 -3 C +4 -2 -4 Common oxidation numbers
Polyatomic ions A special class of ions where a group of atoms tend to stay together. NH 4 + ammonium NO 3 - nitrate SO 4 2- sulfate OH - hydroxide O 2 2- peroxide Your book contains a more complete list. NH 4 + ammonium NO 3 - nitrate SO 4 2- sulfate OH - hydroxide O 2 2- 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 NH 4 Cl sodium hydroxide NaOH potassium permanganate KMnO 4 ammonium sulfate (NH 4 ) 2 SO 4
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 ClO 4 1- …. -ate ClO 3 1- …. -ite ClO 2 1- hypo- …. -ite ClO 1- Naming Inorganic Compounds
Oxidation number and nomenclature Polyatomic anions containing oxygenPolyatomic 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 Increased #oxygen and Oxidation number
Oxidation number and nomenclature ExamplesExamples Cl oxidation numberFormulaName +7NaClO 4 sodium perchlorate +5NaClO 3 sodium chlorate +3NaClO 2 sodium chlorite +1NaClOsodium hypochlorite -1NaClsodium 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.
H Li Na Cs Rb K TlHgAuHfLuBa Fr PtIrOsReWTa He RnAtPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb LrRa ZnCuTiScNiCoFeMnCrV InXeITeSbSn GaKrBrSeAsGe AlArClSPSi BNeFONC 1 213 14 15 16 17 18 I A II A III A IV A VA VI A VIIA 0 3 4 5 6 7 8 9 10 11 12 III B IVB V B VIB VIIB VIII B IB IIB 12345671234567 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 has 3 Oxygens -ate has 4 Oxygens Slivkas Square -ate & -ite charges usually = -ide charge
Polyatomic Ions ate 4 Oxygens.. Inside Slivkas Square ex: SO 4 2- sulfate 3 Oxygens.. Borders the outside of the square ex: NO 3 1- nitrate
Polyatomic Ions ite 1 less Oxygen compared to the -ate ex: ClO 2 1- chlorite SO 3 2- sulfite
Polyatomic Ions per root name ate has 1 more O than the ate ex: IO 4 1- periodate hypo root name ite has 2 less O than the ate ex: ClO 1- hypochlorite
Polyatomic Ions Per-ate 1 more O - ate - ite 1 less O hypo-ite 2 less O (also notice oxidation # of nonmetal changes)
H Li Na Cs Rb K TlHgAuHfLuBa Fr PtIrOsReWTa He RnAtPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb LrRa ZnCuTiScNiCoFeMnCrV InXeITeSbSn GaKrBrSeAsGe AlArClSPSi BNeFONC 1 2 18 I A II A VIIIA 3 4 5 6 7 8 9 10 11 12 III B IVB V B VIB VIIB VIII B IB IIB 12345671234567 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 Group B Elements follow Group A patterns CrO 4 2- chromate MnO 4 1- permanganate III A IV A VA VI A VIIA 13 14 15 16 17
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 = Fe 2+ ferric = Fe 3+
Latin Root Words cuprous = Cu 1+ cupric = Cu 2+ stannous = Sn 2+ stannic = Sn 4+ plumbous = Pb 2+ plumbic = Pb 4+ aurous = Au 1+ auric = Au 3+ Note that there is no pattern between the -ous and -ic suffixes and the actual charge of the ions.
Naming Covalent Compounds sulfite ionsulfur (VI) oxide sulfite ion vs. sulfur (VI) oxide What is the difference between SO 3 2- SO 3 SO 3 2- and SO 3 polyatomic ionneutral compound polyatomic ion vs. neutral compound
Naming Covalent Compounds SCl 4 sulfur (IV) chloride SCl sulfur (VI) chloride CO carbon (II) oxide CO 2 carbon (IV) oxide 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.
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: NH 3. Greek prefixes are used to indicate the number of atoms in the molecule(subscripts). PCl 5 is phosphorus pentachloride Naming Inorganic Compounds
Names and Formulas of Binary Molecular Compounds Naming Inorganic 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. S S F F F F F F F F F F
Other Methods of Naming molecules composed of two nonmetalsFor 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 tetra = 4 penta = 5 hexa = 6 hepta = 7 octa = 8 deca = 10 hepta = 7 octa = 8 deca = 10
Other Methods of Naming molecules The rule may be modified to improve how a name sounds. Example Example - use monoxide not monooxide. N 2 O 5 CO 2 CO CCl 4 dinitrogen pentoxide carbon dioxide carbon monoxide carbon tetrachloride name elements in the formula. use prefixes to indicate how many atoms there are of each type.
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 H 2 S is hydrosulfuric acid -ate becomes -ic acid H 3 PO 4 is phosphoric acid -ite becomes -ous acid HNO 2 is nitrous acid Other Naming -Acids
Naming Inorganic Acids
Salt Name Formula Acid Name Formula Acid Name Formula Sodium acetate NaC 2 H 3 O 2 Acetic acid HC 2 H 3 O 2 Sodium chloride NaCl Hydrochloric acid HCl Sodium hyponitrite NaNO Hyponitrous acid HNO Sodium phosphite Na 3 PO 3 Phosphorous acid H 3 PO 3 Sodium sulfate Na 2 SO 4 Sulfuric acid H 2 SO 4
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; H 2 S is hydrosulfuric acid -ate becomes -ic acid; H 3 PO 4 is phosphoric acid -ite becomes -ous acid. HNO 2 is nitrous acid Naming Inorganic Compounds
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: Other Naming -Double & Triple Salts CO 3 2- carbonateHCO 3 - hydrogen carbonate CO 3 2- is named carbonate, but HCO 3 - is hydrogen carbonate (or bicarbonate) H 2 PO 4 - dihydrogen phosphate H 2 PO 4 - dihydrogen phosphate anion. Note that these are not named as acids, since another cation is still needed to balance the charge.
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. Other Naming -Double & Triple Salts NaHCO 3 sodium hydrogen carbonate (or sodium bicarbonate) AlK(SO 4 ) 2 aluminum potassium sulfate
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 compounds name. Other Naming - Hydrates copper (II) sulfate pentahydrate CuSO 4 5H 2 O pentahydrate 5H 2 O
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. Other Naming - Historical Names For example, H 2 O is called water, not dihydrogen monoxide Check out http://www.dhmo.orghttp://www.dhmo.org
A quick review of nomenclature Is a metal present as the first element? Can the metal have more than one oxidation state? Use Roman numerals or may use prefixes (mono, di, tri...) Roman numerals are not needed. Use Roman numerals or may use Latin name with -ous/-ic suffixes No Yes Is hydrogen first element? Name as an acid Yes Is a nonmetal the first element? No Yes -ides become hydro- -ic acids -ates become -ic acids -ites become -ous acids
A quick review of nomenclature Is it a binary compound? Use the -ide suffix for the negative ion Name the common polyatomic ion Yes Look up the name or formula Use per- -ate 1 more O -ite 1 less O hypo- -ite 2 less O Yes Does it contain one of the 8 common ions? Yes No Does it have more or less O atoms than one of the -ate ions? No
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 - dont 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, dont 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: 1 H = 99% 2 H = 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 IsotopesIsotopes Atoms of the same element but having different masses. » Each isotope has a different number of neutrons. Isotopes of hydrogenHHH Isotopes of carbon CCC 1111 2121 3131 12 6 13 6 14 6
Isotopes Most elements occur in nature as a mixture of isotopes. –ElementNumber 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 1/12 th of the mass of a 12 C 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 7 Li = 7.016 004 u 14 N = 14.003 074 01 u 29 Si = 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 its average atomic mass based on the following data. Isotope Mass (u) Abundance 28 Si27.976926592.23 % 29 Si28.9764947 4.67 % 30 Si29.9737702 3.10 %
Average atomic masses 92.23 100 (27.9769265 u) = 25.80 u 4.67 100 (28.9764947 u) = 1.35 u 3.10 100 (29.9737702 u) = 0.929 u 28 Si 29 Si 30 Si Average atomic mass for silicon = 28.08 u
The mole The number of atoms in 12.000 grams of 12 C can be calculated. One atom 12 C = 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 10 23 atoms themoleThe number of atoms of any element needed to equal its atomic mass in grams will always be 6.02 x 10 23 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.
The mole 1 mole of any element = 6.02 x 10 23 atoms = gram atomic mass uAtoms, ions and molecules are too small to directly measure in u. Using moles gives us a practical unit. the gramWe can then relate atoms, ions and molecules, using an easy to measure unit - the gram.
Masses of atoms and molecules Atomic massAtomic 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 massMolecular or formula mass The total mass for all atoms in a compound.
Masses of atoms and molecules H 2 O H 2 O - water 2 hydrogen 2 x1.008 u 1 oxygen1 x 16.00 u mass of molecule 18.02 u 18.02 g / mol Rounded off based on significant figures 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 H 2 O = 6.022 x 10 23 molecules 1 mol H 2 = 6.022 x 10 23 molecules We cant 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 methodFactor 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 massFormula mass - Add the masses of all the atoms in formula; for molecular and ionic compounds. Molecular massMolecular mass - Calculated the same as formula mass; only valid for molecules. Both have units of either u or grams / mole. Molar massMolar 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. CO 2 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 Formula mass =44.010 u » or g / mol
Another example CH 3 CH 2 OH CH 3 CH 2 OH - ethyl alcohol 2 carbon2 x12.01 u 6 hydrogen6 x1.008 u 1 oxygen1 x16.00 u mass of molecule46.07 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 DONTThe numbers DONT change -- just the units.
Example - (NH 4 ) 2 SO 4 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 15 atoms 1 unit 6.02 x10 23 units 1 mol atoms = 1.36 x10 24
Example - (NH 4 ) 2 SO 4 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? x moles = 20.0 g x = 0.151 mol (NH 4 ) 2 SO 4 1 mol 132.14 g x moles NH 4 = 0.151 mol (NH 4 ) 2 SO 4 x 2 moles NH 4 1 mol (NH 4 ) 2 SO 4 moles NH 4 = 0.302 Formula weight = 132.14 g / 1 mol (NH 4 ) 2 SO 4 2 moles NH 4 / 1 mol (NH 4 ) 2 SO 4
Example - (NH 4 ) 2 SO 4 How many grams of sulfate ions are in 20.0 grams of ammonium sulfate? x moles = 20.0 g x = 0.151 mol (NH 4 ) 2 SO 4 1 mol 132.14 g x grams SO 4 = 0.151 mol (NH 4 ) 2 SO 4 x 96.06 g SO 4 1 mol (NH 4 ) 2 SO 4 grams SO 4 = 14.5 Formula weight = 132.14 g / 1 mol (NH 4 ) 2 SO 4 96.06 grams SO 4 / 1 mol (NH 4 ) 2 SO 4
Masses of atoms and molecules Law of Definite Composition 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 –% 0 = 0.96 g O x 100 = 38.9 % O 2.47 g KClO 2.47 g KClO 3 –Based upon the formula mass: –% O = 48.00 u O x 100 = 39.17 % O 122.55 u KClO 122.55 u KClO 3 Percent composition can also be determined from experimental data. Example: When 2.47 g KClO 3 is heated strongly, 0.96 g of O 2 gas is driven off. What is the % by mass of oxygen in KClO 3 ?
Gay-Lussacs Law Law of of Combining Volumes.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-Lussacs Law Example.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 H 2 (g) + O 2 (g) 2 H 2 O (g) +H2H2 H2H2 O2O2 H2OH2OH2OH2O
Avogadros 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: 22.4 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 O 2 gas will be formed at STP? How many grams of H 2 gas will be formed? X L O 2 =10.0g H 2 O x x x 1 mol H 2 O 18.0 g H 2 O liters O 2 = 6.22 Note that 12.4 L H 2 will also be formed. 0.5 mol O 2 1 mol H 2 O 22.4 L O 2 1 mol O 2 X g H 2 =10.0g H 2 O x 2.02 g H 2 18.02 g H 2 O = 1.12 g H 2
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 analyzerElemental 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 CO 2 trap H 2 O trap O2O2 sample A sample is burned, completely converting it to CO 2 and H 2 O. 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 ExampleExample: 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 CO 2 collected= 0.1638 g Mass of H 2 O collected= 0.1676 g Determine the % of each element in the compound.
Elemental analysis Mass of carbonMass of carbon Mass of hydrogenMass of hydrogen Mass of nitrogenMass of nitrogen 0.1638 g CO 2 12.01 g C 44.01 g CO 2 = 0.04470 g C 0.1675 g H 2 O 2.016 g H 18.01 g H 2 O = 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 0.1156 g 0.05215 g 0.1156 g
Empirical formula Empirical formulaEmpirical formula The simplest formula that shows the ratios of the number of atoms of each element in a compound. ExampleExample - the empirical formula for hydrogen peroxide (H 2 O 2 ) is HO. We can use either our mass or our percent composition information from the earlier example to determine an empirical formula.
Empirical formula 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. C0.003722 / 0.003722 = 1.000 H0.0186 / 0.003722 = 4.998 N 0.003722 / 0.003722 = 1.000 The empirical formula is CH 5 N The empirical formula is CH 5 N
Empirical formula 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 % % 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. C3.220 / 3.220= 1.000 H16.09 / 3.220= 4.997 N3.220 / 3.220= 1.000 The empirical formula is determined to be the same, CH 5 N, whether using actual masses of the elements present in the sample or by using their % composition by mass.The empirical formula is determined to be the same, CH 5 N, whether using actual masses of the elements present in the sample or by using their % composition by mass.
Molecular formula Molecular formulaMolecular 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 formulaCH 5 N Empirical formula mass31.06 u Molecular mass62.12 Ratio:62.12 / 31.06= 2 The molecular formula is C 2 H 10 N 2 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 H 2 O is lost. 5.000g hydrate - 0.738g H 2 O = 4.262g BaCl 2
Hydrated Compounds 4.262g BaCl 2 (1 mol BaCl 2 ) = 0.0205 mol BaCl 2 (208.2 g BaCl 2 ) 0.738 g H 2 O ( 1 mol H 2 O ) = 0.0410 mol H 2 O (18.0 g H 2 O ) BaCl 2 0.0205 / 0.0205 = 1.00 H 2 O 0.0410 / 0.0205 = 2.00 The compounds formula is BaCl 2 2H 2 O
Molarity M = moles solutemol 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 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 mass NaOH = 0.250 mol x 40.0 g/mol = 10.0 grams NaOH
Making a solution Finally, youre 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 M 1 V 1 = M 2 V 2 –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? M 1 V 1 = M 2 V 2 M 1 = 12.0 MM 2 = 1.00 V 1 = ??? mlV 2 = 250.0 mL V 1 = M 2 V 2 / M 1 M 2 = (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 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. moles solute mol kilograms of solvent kg Molality ( m ) = =
Density 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. grams of solution g milliliters of solution mL Density ( D ) = =
Percent Composition % 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. value of the part Value of the whole Percent Composition = x 100
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. moles of solute or solvent total moles of solute & solvent Mole fraction =
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. # grams of solute 1,000,000 g solution Parts per million (ppm) =
Stoich Baby Given the following equation ___N 2 + ___H 2 ___NH 3 Assuming gases at STP, given one mole of nitrogen, how many liters of ammonia would form? Given one mole of nitrogen gas, how many moles of ammonia would form? 1 3 2 Given 2 liters of nitrogen and 5 liters of hydrogen, how may liters of ammonia are formed? What is left over?
Stoichiometry StoichiometryStoichiometry The study of quantitative relationships between substances undergoing chemical changes. Law of Conservation of MatterLaw 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 Chemists shorthand to describe a reaction. It shows: All reactants and products The state of all substances Any conditions used in the reaction » CaCO 3 (s) CaO (s) + CO 2 (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. –CaCO 3 (s) CaO (s) + CO 2 (g) –Reactants Products »1 Ca 1 Ca »1 C 1 C »3 O 3 O
Balancing chemical equations Step 1 Step 1Count the number of atoms of each element on each side of the equation. Step 2 Step 2 Determine which atom numbers are not balanced. Step 3 Step 3 Balance one atom at a time by using coefficients in front of one or more substances. Step 4 Step 4Repeat steps 1-3 until everything is balanced.
Example. Decomposition of urea –(NH 2 ) 2 CO + H 2 O ______ > NH 3 + CO 2 < not balanced 2 N1 N < not balanced < not balanced6 H3 H < not balanced 1 C1 C 2 O2 O We need to double NH 3 on the right. 2 –(NH 2 ) 2 CO + H 2 O ______ > 2NH 3 + CO 2
2 H 2 + O 2 -----> 2 H 2 O You need a balanced equation and you WILL work with moles. You need a balanced equation and you WILL work with moles. Mass relationships in chemical reactions StoichiometryStoichiometry - The calculation of quantities of reactants and products in a chemical reaction.
Stoichiometry, General steps. 1 1 Balance the chemical equation. 3 3 Convert masses to moles. 2 2 Calculate formula masses. 4 4 Use chemical equation to get the needed answer. Convert back to mass if needed. 5 5
Mole calculations The balanced equation shows the reacting ratio between reactants and products. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O –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 N 2 ? Formula mass –=2 N x 14.01 g/mol –= 28.02 g /mol moles N 2 –= 14 g x 1 mol /28.02 g –= 0.50 moles
Mass calculations We dont 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. We dont 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 ______ > CaCl 2 + H 2 –Note: We produce one H 2 for each calcium. There is an excess of HCl so we have all we need.
Mass calculations 2HCl + Ca ____ > CaCl 2 + H 2 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 _____ > CaCl 2 + H 2 » 10 g Ca = 0.25 mol Ca According to the chemical equation, we get one mole of H 2 for each mole of Ca. So we will make 0.25 moles of H 2. grams H 2 produced = moles x FW H 2 » = 0.25 mol x 2.016 g/mol » = 0.504 grams
Mass calculations OK, so how many grams of CaCl 2 were made? »2HCl + Ca _____ > CaCl 2 + H 2 » 10 g Ca = 0.25 mol Ca –We would also make 0.25 moles of CaCl 2. –g CaCl 2 = 0.25 mol x FM CaCl 2 »= 0.25 mol x 110.98 g / mol CaCl 2 »= 27.75 g CaCl 2
Limiting reactant In the last example, we had HCl in excess. Reaction stopped when we ran out of Ca. limiting reactant.Ca is considered the limiting reactant. Limiting reactantLimiting 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 ReactionBalanced Chemical Reaction 2H 2 + O 2 ________ > 2H 2 O You need 2 moles of H 2 for each mole of O 2. Moles of H 2 5 g = 2.5 mol Moles of O 2 10g = 0.31 mol 1 mol 32.0 g 1 mol 2.0 g
Example Balanced Chemical ReactionBalanced Chemical Reaction 2H 2 + O 2 2H 2 O You need 2 moles of H 2 for each mol of O 2 You have 2.5 moles of H 2 and 0.31 mol of O 2 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 yieldTheoretical yield The amount of product that should be formed according to the chemical reaction and stoichiometry. Actual yieldActual yield The amount of product actually formed. Percent yieldPercent yield Ratio of actual to theoretical yield, as a %. Quantitative reactionQuantitative reaction When the percent yield equals 100%.
Yield Less product is often produced than expected. Possible reasonsPossible 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 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!