Significant Figures When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer.

Significant Figures When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer. There are 2 different types of numbers – Exact – Measured Exact numbers are infinitely important Measured number = they are measured with a measuring device so these numbers have ERROR. When you use your calculator your answer can only be as accurate as your worst measurement 2

3 Exact Numbers An exact number is obtained when you count objects or use a defined relationship. Counting objects are always exact 2 soccer balls 4 pizzas Exact relationships, predefined values, not measured 1 foot = 12 inches 1 meter = 100 cm For instance is 1 foot = 12.000000000001 inches? No 1 ft is EXACTLY 12 inches.

Measurement and Significant Figures Every experimental measurement has a degree of uncertainty. The volume, V, at right is certain in the 10’s place, 10mL<V<20mL The 1’s digit is also certain, 17mL<V<18mL A best guess is needed for the tenths place. 4

5 What is the Length? We can see the markings between 1.6-1.7cm We can’t see the markings between the.6-.7 We must guess between.6 &.7 We record 1.65 cm as our measurement The last digit an 5 was our guess...stop there

Measured Numbers Do you see why Measured Numbers have error…you have to make that Guess! All but one of the significant figures are known with certainty. The last significant figure is only the best possible estimate. To indicate the precision of a measurement, the value recorded should use all the digits known with certainty. 6

Note the 4 rules When reading a measured value, all nonzero digits should be counted as significant. There is a set of rules for determining if a zero in a measurement is significant or not. ► RULE 1. Zeros in the middle of a number are like any other digit; they are always significant. Thus, 94.072 g has five significant figures. ► RULE 2. Zeros at the beginning of a number are not significant; they act only to locate the decimal point. Thus, 0.0834 cm has three significant figures, and 0.029 07 mL has four. 7

RULE 3. Zeros at the end of a number and after the decimal point are significant. It is assumed that these zeros would not be shown unless they were significant. 138.200 m has six significant figures. If the value were known to only four significant figures, we would write 138.2 m. RULE 4. Zeros at the end of a number and before an implied decimal point may or may not be significant. We cannot tell whether they are part of the measurement or whether they act only to locate the unwritten but implied decimal point. 8

RULE 1. In carrying out a multiplication or division, the answer cannot have more significant figures than either of the original numbers. 9

RULE 2. In carrying out an addition or subtraction, the answer cannot have more digits after the decimal point than either of the original numbers. 10

White Board: Sig Fig Practice Calculation Calculator Says: Answer 3.24 m x 7.0 m 22.68 m 2 23 m 2 100.0 g ÷23.7 cm 3 4.219409283 g/cm 3 4.22 g/cm 3 713.1 L – 3.872 L 709.228 L709.2 L 0.02 cm + 2.371 cm 2.391 cm2.39 cm 710 m ÷3.0 s236.6666667 m/s240 m/s 2.030 mL – 1.870 mL0.16 mL 0.160 mL 1818.2 m x 3.23 o C 5872.786 m o C5870 m o C 1818.2 g + 3.37 g1821.57 g1821.6 g

Cambridge Standards 1.1 Relative masses of atoms and molecules 1.2 The mole and the Avogadro constant 1.3 The determination of relative atomic masses 1.4 The calculation of empirical and molecular formulae 1.5 Reacting masses and volumes (of solutions and gases)

Define and use the terms: – Relative atomic mass – Isotopic mass – Formula mass – Empirical formula – Molecular formula – Mole Analyse and use mass spectra to calculate the relative atomic mass of an element Calculate empirical and molecular formula using combustion data or composition by mass Write and construct balanced equations Perform calculations, including use of the mole concept involving – Reacting mass (from formula and equations) – Volumes of gases – Volumes and concentrations of solutions Deduce stoichiometric relationships from calculations involving reacting masses, volumes of gases and volumes and concentrations of solutions.

Introduction – Why is chemistry so important?

Relative Atomic Mass Relative Isotopic Mass Relative Molecular Mass Relative Formula Mass Masses (wrt C-12)

Relative atomic mass, A r The mass of an atom is too small to measure. We can only measure the mass of an atom by comparing its mass to the mass of another atom.

Relative atomic mass, A r What is relative atomic mass??? The mass of an atom when compared to a standard atom is known as its relative atomic mass (A r ). It has no unit.

Relative atomic mass, A r Since hydrogen atom is the lightest atom, chemists first started comparing masses of other atoms with the mass of one hydrogen atom. We can consider the relative atomic mass of hydrogen to be 1. We can compare the mass of other atom to the mass of the hydrogen atom. An atom of carbon-12 is 12 times heavier than one atom of hydrogen. The relative atomic mass, A r of one carbon-12 atom is 12.

Relative atomic mass, A r

This method, however is not always convenient because Hydrogen is a gas. Therefore it is difficult to measure its mass. It has a number of isotopes with different masses. Thus it is difficult to measure exactly. Oxygen was then used as a standard to compare the masses of atoms, however, this also posed some problems.

At present carbon-12 is used as the standard of comparison because It is solid and can be easily handled, Its mass can be more easily measured with a mass spectrometer. Carbon-12 is easily found as its compounds are abundant. Carbon-12 isotopes is assigned a mass of exactly 12 units. Relative atomic mass, A r

The relative atomic mass of an element is the average mass of one atom of the element when compared with of the mass of a carbon -12.

Relative atomic mass, A r Since relative atomic mass compares the masses of atoms, it does not have any units. The relative atomic mass is also related to the number of protons and neutrons in the atom.

Relative atomic mass An element can have several naturally occurring isotopes. These isotopes of a element behave in the same way. In calculating the relative atomic mass of an element with isotopes, the relative mass and proportion or percentage of each is taken into account.

Calculating relative atomic mass IsotopeRelative isotopic mass Relative abundance (%) Cl34.96975.80 Cl36.96624.20 A r = (relative isotopic mass X % abundance) + relative isotopic mass X % abundance) 100 A r (Cl) = (34.969 X 75.8) + ( 36.966 X 24.2) 100 A r (Cl) = 2650.65 + 894.58 100 A r (Cl) = 35.45 amu (atomic mass unit)

Relative isotopic mass The relative isotopic mass (I r ) of an isotope is the mass of an atom of that isotope relative to the mass of an atom of 12 C taken as 12 units exactly

Relative Isotopic Mass – What is an isotope?

Relative Isotopic Mass – The mass of a particular isotope of an element on a scale where an atom of carbon-12 has a mass exactly 12 units. Nucleon number

The symbols, relative mass and the charge of subatomic particles Subatomic particle SymbolRelative mass Charge Protonp1+1 Electrone Neutronn10

Examples of isotopes of some elements ElementProton number Nucleon number Number of protons Number of neutrons Percentage abundance Hydrogen, Deuterium, Tritium, 111111 123123 111111 012012 99.985% 0.015% Man-made isotope Carbon-12, Carbon-13, Carbon-14, 666666 12 13 14 666666 678678 98.1% 1.1% 0.8 % Chlorine-35, Chlorine-37, 17 35 37 17 18 20 75.5% 24.5% Oxygen-16, Oxygen-17, Oxygen-18, 888888 16 17 18 888888 8 9 10 99.757% 0.038% 0.205%

Example How many times heavier is a krypton atom compared to a helium atom? Solution: mass of a krypton atom = A r of krypton mass of a helium atom A r of helium = 84 4 = 21 times

Relative Molecular Mass, M r The idea of relative atomic mass can be extended to molecules. The Relative molecular mass of a molecule is the average mass of the molecule when compared with of the mass of one atom of carbon-12. Example: Relative Molecular Mass of CO 2 is 44. This means that the molecule of carbon dioxide is 44 times heavier than of one atom of carbon-12

Relative Molecular Mass, M r Since a molecule is made up of atoms, the relative molecular mass of a molecule can be calculated by adding up the relative atomic masses of all the atom present in the molecule. The relative molecular mass of a compound is the mass of one molecule of that substance relative to the mass of a 12 C This is calculated by taking the sum of the relative atomic masses of the elements in the molecular formula (i.e. covalent compounds) It is called relative formula mass for ionic compounds

Calculating relative molecular mass Normally to find the relative atomic mass you just look in the periodic table!! Oxygen (O 2 ) M r = 2 X A r (0) = 2 X 16.0 = 32.0 amu Carbon Dioxide (CO 2 )M r = A r (C) + 2 X A r (0) = 12.0 + (2 X 16.0) = 44.0 amu

Calculation of Relative Molecular Mass Step 1: Determine the molecular formula. Step 2: Find the relative atomic mass of each element in the molecule. Step 3: Add up all the relative atomic masses of the element or elements.

Relative Molecular Mass, M r MoleculeMolecular formula Relative molecular mass ChlorineCl 2 2 x 35.5 = 72 NitrogenN2N2 2 x 14 = 28 AmmoniaNH 3 14 + (3 x 1) = 17 EthanolC 2 H 5 OH(2 x 12) + (5 x 1) + 16 + 1 = 46 Carbon dioxide CO 2 12 + (2 x 16) = 44

Relative formula masses Some substances consists of ions and not molecules. For these ionic substances, the relative formula masses are used in place of relative molecular masses.

Relative formula masses Ionic substance Ionic formulaRelative formula mass Sodium chloride NaCl23 + 35.5 = 58.5 Hydrated magnesium sulphate MgSO 4. 7H 2 O24 + 32 + (4x 16) + 7(2 + 16) = 246 Calcium oxideCaO40 + 16 = 56

Determination of A r from mass spectra

Mass Spectrometry – Can be used to measure the mass of each isotope present in an element – Compares how much of each isotope is present

Mass Spectrometer

Purpose of Mass Spectrometry  Produces spectra of masses from the molecules in a sample of material, and fragments of the molecules.  Used to determine  the elemental composition of a sample  the masses of particles and of molecules  potential chemical structures of molecules by analyzing the fragments  the identity of unknown compounds by determining mass and matching to known spectra  the isotopic composition of elements in a molecule  Produces spectra of masses from the molecules in a sample of material, and fragments of the molecules.  Used to determine  the elemental composition of a sample  the masses of particles and of molecules  potential chemical structures of molecules by analyzing the fragments  the identity of unknown compounds by determining mass and matching to known spectra  the isotopic composition of elements in a molecule

Mass Spectrometer is an analytical technique that can be used to determine the chemical composition of a sample It is a machine that separates the individual isotopes in a sample From this you can calculate the relative atomic mass.

Stages The ionizer converts some of the sample into ions. Mass analyzers separate the ions according to their mass-to-charge ratio. The detector records either the charge induced or the current produced when an ion passes by or hits a surface

The substance to be analysed is injected into a high vacuum where the particles are ionized by colliding with beam of high speed electrons If the sample is not already a gas, then a liquid or solid substance must be vaporized The resulting (+) ions are accelerated down a tube and then through a powerful magnetic field. The charged or ionized particles are deflected by this powerful magnetic field How much they are deflected depends on the particle mass and the speed of the particle and the strength of a magnetic field i.e. lighter particles of lower mass (and momentum) are deflected more than heavier particles of bigger mass

Mass Spectrometry

https://youtu.be/mBT73Pesiog

Mass spectrum – graph of a mass spectrometer. Number of peaks = number of isotopes Position of peaks = relative isotopic mass Height of peaks = relative abundance of the isotope, in comparison to the other isotopes of that element.

A typical mass spectrometer output Molybdenum

A. Mass Spectrometry Magnesium If you know what isotopes are present and in what percent, how can you find the relative atomic mass using this information?

Mass Spectrum of CO 2 Molecular ion peak [CO 2 ] + = 44 [CO] + = 28 [O] + = 16 [C] + = 12 Fragment Peaks

The mass spectrum for boron The tallest peak is given an arbitrary height of 100. Hence, the two isotopes (with their relative abundances) are: boron-1023 boron-11100 Suppose you had 123 typical atoms of boron. 23 of these would be 10 B and 100 would be 11 B. The total mass of these would be: (23 x 10) + (100 x 11) = 1330 The average mass of these 123 atoms would be 1330 / 123 = 10.8 10.8 is the relative atomic mass of boron

The mass spectrum for zirconium 5 isotopes with relative isotopic masses of 90, 91, 92, 94 and 96 on the 12 C scale. Relative percentage abundances) are: zirconium-90 51.5 zirconium-91 11.2 zirconium-92 17.1 zirconium-94 17.4 zirconium-96 2.8 Suppose you had 100 typical atoms of zirconium The total mass of these 100 typical atoms would be (51.5 x 90) + (11.2 x 91) + (17.1 x 92) + (17.4 x 94) + (2.8 x 96) = 9131.8 The average mass of these 100 atoms would be 9131.8 / 100 = 91.3 91.3 is the relative atomic mass of zirconium.

The mass spectrum for chlorine Chlorine consists of molecules. In the ionisation chamber a molecular ion, Cl 2 + forms which undergoes fragmentation. Possible combinations of chlorine-35 and chlorine-37 atoms in a Cl 2 + ion are: 35 + 35 = 70 35 + 37 = 72 37 + 37 = 74

Mass Spectrum of Bromine, Br 2 Bromine has two isotopes: 50.69% 79 Br and 49.31% 81 Br 79 Br +81 Br + [ 79 Br 81 Br] + [ 79 Br 79 Br] + [ 81 Br 81 Br] + Molecular Ion Peaks Fragments

Practice: Methyl Bromide, CH 3 Br

Answers: Methyl Bromide, CH 3 Br [CH 3 81 Br] + [CH 2 81 Br] + [C 81 Br] + and [CH 2 79 Br] + [CH 81 Br] + and [CH 3 79 Br] + [CH 79 Br] + [CH 3 ] + [C 79 Br] + [ 81 Br] + [ 79 Br] +

Practice: Methylene Chloride (CH 2 Cl 2 ) Chlorine is 75.77% 35 Cl and 24.23% 37 Cl

Practice: Vinyl Chloride (CH 2 CHCl) Chlorine is 75.77% 35 Cl and 24.23% 37 Cl

Spectra of Larger Molecules Spectra of large molecules have many fragments, and the interpretation of their spectra is beyond the scope of this course. Codeine, C 18 H 21 NO 3

Toluene ( C 7 H 8 ) Fragmentation **Note that the molecular ion peak is NOT the dominant peak in the spectrum

Mass Spectrometry in Forensics  Mass spectrometry is used to confirm the identify of unknowns, such as illegal drugs  Unknowns are often not pure, and must be separated from a mixture  Gas chromatography is used to separate the components of the mixture  Mass spectrometry “fingerprints” the components, so that they can be matched to existing known spectra  Mass spectrometry is used to confirm the identify of unknowns, such as illegal drugs  Unknowns are often not pure, and must be separated from a mixture  Gas chromatography is used to separate the components of the mixture  Mass spectrometry “fingerprints” the components, so that they can be matched to existing known spectra Gas Chromatograph Mixture Mass Spectrometer

Where did the mole come from? The mole is a unit of measurement based on work done almost 200 years ago by Amadeo Avogadro as he studied gas behavior. His work led to the association of a number, 6.02 x 10 23, with the mole. 6.02 x 10 23, is also called Avogadro’s constant or number It allows particles to be "counted.“ The word "mole" is derived from "gram molecular weight"

Molar Mass Molar mass (M) is the mass of one mole of a substance (chemical element or chemical compound). The unit of molar masses is: grams per mole ( g or g/mol or g mol –1 ) mol

Atoms to Moles I have 4.673 x 10 11 atoms of Zinc. How many moles do I have? 4.673 x 10 11 atoms of Zinc 1 mol of Zinc 6.02 x 10 23 atoms of Zinc = 7.7 x10 -11 mols of zinc Always check the units have cancelled out This number is less than one therefore it is less than one mol as it should be!

challenge How many moles of seconds have we experienced (if the world is 15 billion years old) Show units and unit cancellation 1 yr = 365 days 1 mol = 6.0 x 10 23 sec 7.8578 x 10 -7 mol

Reacting masses When reacting chemicals together we need to know what mass of each reactant to use so that they react exactly without any waste. Need a balanced equation This shows us the mole ratio = stoichiometry

Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g)  In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water  In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water.  This equation could also be read as 2 moles of benzene reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water. MOLE RATIO Since the relationship between the actual number of molecules and the number of moles present is 6.02 x 10 23, a common factor between all species involved in the equation, a MOLE RATIO relationship can be discussed.

Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g) The MOLE RATIO for benzene and oxygen is 2 : 15. It can be written as: 2 moles C 6 H 6 or as 15 moles of O 2 15 moles O 2 2 moles of C 6 H 6 The MOLE RATIO for oxygen and carbon dioxide is 15 : 12. It can be written as: 12 moles CO 2 or as 15 moles of O 2 15 moles O 2 12 moles of CO 2 NOTE: The MOLE RATIO is used for converting moles of one substance into moles of another substance. Without the balanced equation there is no other relationship between two different compounds.

Using the mole ratio to relate the moles of one compound to the moles of another compound is the part of chemistry called STOICHIOMETRY !!!!! 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q. How many mole of hydrogen are necessary to react with 2 moles of oxygen in order to produce exactly 4 moles of water? A. 2 mol O 2 (2 moles H 2 / 1 mole O 2 ) = 4 mole H 2

STOICHIOMETRY The Stoichiometry Flow Chart Use Molar mass (A) Use mole ratio from equation Use Molar mass (B)

STOICHIOMETRY 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) How many moles of hydrogen are necessary to react with 15.0 g of oxygen? 15.0g O 2 1 mole O 2 2 mole H 2 = 0.938 moles H 2 32.0 g 1 mole O 2 How many grams of hydrogen are necessary to react with 15.0 g of oxygen ? 15.0g O 2 1 mole O 2 2 mole H 2 2.016 g H 2 = 1.89 g H 2 32.0 g 1 mole O 2 1 mole H 2

STOICHIOMETRY 2 H 2 (g) + O 2 (g)  2 H 2 O (g) How many grams of water are produced from 15.0 g of oxygen? 15.0g O 2 1 mole O 2 2 mole H 2 O 18.0 g H 2 O = 16.9 g H 2 O 32.0 g 1 mole O 2 1 mole H 2 O How much hydrogen and oxygen is needed to produce 25.0 grams of water? 25.0g H 2 O 1 mole H 2 O 2 mole H 2 2.016 g H 2 = 2.80 g H 2 18.0 g 2 mole H 2 O 1 mole H 2 25.0g H 2 O 1 mole H 2 O 1 mole O 2 32.0 g O 2 = 22.2 g O 2 18.0 g 2 mole H 2 O 1 mole O 2 Notice that the Law of Conservation of Mass still applies.

How many grams of solid are formed when 10.0 g of lead reacts with excess phosphoric acid? 1.Write the chemical equation: Pb + H 3 PO 4  ? You recognize that this is a single displacement (replacement) reaction. So Pb (a metal) will displace (replace) H (the cation). Pb + H 3 PO 4  Pb 3 (PO 4 ) 2 + H 2 2. Balance the equation: 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 + 3 H 2 3. Make a list under the appropriate substance 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 (s) + 3 H 2 (g) 10.0g m=? Start with what is given: 10.0gPb 1 mole Pb 1 mole Pb 3 (PO 4 ) 2 811 g Pb 3 (PO 4 ) 2 = 207 g Pb 3 mole Pb 1 mole Pb 3 (PO 4 ) 2 13.1 g Pb 3 (PO 4 ) 2

Percentage composition Percentage composition of a compound is a relative measure of the mass of each different element present in the compound. We can use the formula of a compound and relative atomic masses to calculate the percentage by mass of a particular element in a compound. E.g. What is the percentage of Zn in ZnCl 2 or the percentage of Al in Al 2 O 3

Calculate the percentage of Al in Al 2 O 3 % by mass of = mass of element in 1 mole of compound x 100 the element mass of 1 mole of the compound Step 1: Find the molar mass of the compound M r (Al 2 O 3 ) = (2 x 27) + (3 x16) = 102 g/mol Step 2: Find the molar mass of the element in the compound mass of Al in Al 2 O 3 = (2 x 27) = 54g/mol Step 3 : Determine the % of the element in the compound % Al in Al 2 O 3 = 54g/mol x 100 102g/mol = 52.9%

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

Formulas  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

Empirical formula The empirical formula is a simple expression of the relative numbers of each type of atom in it. A person has two hands and ten fingers, or H 2 F 10. The empirical formula for that would be HF 5 Benzene, C 6 H 6. The empirical formula is CH. This empirical formula tells us that the ratio of C to H is 1 to 1; there is one H atom for every C atom.

81 Empirical Formula The empirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxideH 2 O 2 OH BenzeneC 6 H 6 CH EthyleneC 2 H 4 CH 2 PropaneC 3 H 8 C 3 H 8

Calculating an Empirical Formula from Data Obtain the mass (g) of each element in the compound. If percent (%) is given, change the percent sign (%) to grams (g) Divide the mass (g) by the molar mass of the element to get the number of moles Divide the number of moles of each element by the smallest number of moles Determine the formula from the numbers in the mole ratios “Percent to mass, Mass to mole, Divide by smallest Times ‘till whole”

Calculation of Molecular Formulas x(empirical formula) = molecular formula determine empirical formula (if not given) calculate empirical formula mass from empirical formula determine x : x= molecular formula mass (given in problem) empirical formula mass multiply empirical formula by x to obtain molecular formula

84 Percentage Composition Glucose has the molecular formula C 6 H 12 O 6. What is its empirical formula, and what is the percentage composition of glucose? Empirical Formula = smallest whole number ratio CH 2 O

85 Percentage Composition CH 2 O Total mass = 12.01 + 2.02 + 16.00 = 30.03 %C = 12.01/30.03 x 100% = 39.99% %H = 2.02/30.03 x 100% = 6.73% %O = 16.00/30.03 x 100% = 53.28%

86 Percentage Composition Saccharin has the molecular formula C 7 H 5 NO 3 S. What is its empirical formula, and what is the percentage composition of saccharin? Empirical Formula is same as molecular formula MW = 183.19 g/mole %C = (7 x 12.011)/183.19 x 100% = 45.89% etc.

87 Empirical Formula A compound’s empirical formula can be determined from its percent composition. A compound’s molecular formula is determined from the molar mass and empirical formula.

88 Empirical Formula A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula for the compound? Assume 100 g of sample, then 82.67 g are C and 17.33 g are H Convert masses to moles: 82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Find relative # of moles (divide by smallest number)

89 Empirical Formula Convert moles to ratios: 6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens C 2 H 5 Empirical Formula is: C 2 H 5 Formula weight is: 29.06 g/mole If the molecular weight is known to be 58.12 g/mole Then the molecular formula must be: C 4 H 10

How to calculate empirical formulas

A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. CarbonOxygen Calculate the amount in moles 27.3 = 2.27 12.0 72.7 = 4.54 16.0 Divide the number of moles by smallest 2.27 = 1 2.27 4.54 = 2 2.27 Obtain the simplest whole number mole ratio 12

9.0g of a compound of only C,H,O contains 4.8 g of O and 3.6g of C CHO m(g)3.69 – (4.8+3.6) = 0.6 4.8 n(mol)3.6 =0.3 12.0 0.6 = 0.6 1.0 4.8 = 0.3 16.0 Divide by smallest mol 0.3 = 1 0.3 0.6 = 2 0.3 0.3 = 1 0.3 Ratio121

Molecular Formulas A molecular formula gives the actual number of atoms in one molecule of the compound The empirical and molecular formula can be the same The molecular formula is always a whole number multiple o the empirical formula

A compound has the empirical formula of CH. The molar mass of the compound is 78 g/mol. What is the molecular formula of the compound? The molar mass of CH is 13g/mol Compound molar mass is 78g/mol No. of CH unit in a molecule = 78g/mol 13g/mol = 6 CH x 6 = C 6 H 6 = molecular formula

A sample of hydrocarbon was found to contain 7.2g of C and 1.5g of hydrogen. The molar mass of the compound is 58g/mol. What is the molecular formula? 1. Determine empirical formula CH 7.21.5 7.2 = 0.6 12.0 1.5 = 1.5 1 0.6 = 1 0.6 1.5 = 2.5 0.6 25

2. Determine the molecular formula Empirical formula is C 2 H 5 The molar mass of C 2 H 5 is 29g/mol Compound molar mass is 58g/mol No. of C 2 H 5 unit in a molecule = 58g/mol 29g/mol = 2 C 2 H 5 x 2 = C 4 H 10 = molecular formula

98 Empirical Formula Combustion analysis is one of the most common methods for determining empirical formulas. A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram. It is particularly useful for analysis of hydrocarbons.

99 Combustion Analysis Combustion Analysis: the technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion. X + O 2 → CO 2 + H 2 O 0.250 g of compound X produces: 0.686 g CO 2 and 0.562 g H 2 O

100 Combustion Analysis X + O 2 → CO 2 + H 2 O 1. Find the mass of C & H that must have been present in X (multiply masses of products by percent composition of the products). C: 0.686 g x 12.01 g/44.01 g = 0.187 g C H: 0.562 g x (2 x 1.008 g)/18.02 g = 0.063 g H

101 Combustion Analysis X + O 2 → CO 2 + H 2 O 0.187 g C + 0.063 g H = 0.250 g total So compound X must contain only C and H!! 2. Find the number of moles of C and H C: 0.187 g x mole/12.01 g = 0.0156 moles C H: 0.063 g x mole/1.008 g = 0.063 moles H

102 Combustion Analysis X + O 2 → CO 2 + H 2 O 3. Find the RELATIVE number of moles of C and H in whole number units (divide by smallest number of moles) C: 0.0156/0.0156 = 1 H: 0.063/0.0156 = 4 If these numbers are fractions, multiply each by the same whole number.

103 Combustion Analysis X + O 2 → CO 2 + H 2 O 4. Write the Empirical Formula (use the relative numbers as subscripts) CH 4

104 Combustion Analysis Summary 1.Find the mass of C and H in the sample. 2.Find the actual number of moles of C and H in the sample. 3.Find the relative number of moles of C and H in whole numbers. 4.Write the empirical formula for the unknown compound.

105 Combustion Analysis NOTE! In step # 1 always check to see if the total mass of C and H adds up to the total mass of X combusted. If the combined mass of C and H is less than the mass of X, then the remainder is an unknown element (unless instructed otherwise). If a third element is known, calculate the mass of that element by subtraction (at the end of step 1), and include the element in the remaining steps.

106 Combustion Analysis Combustion Analysis provides the Empirical Formula. If a second technique provides the molecular weight, then the molecular formula may be deduced. 1.Calculate the empirical formula weight. 2.Find the number of “formula units” by dividing the known molecular weight by the formula weight. 3.Multiply the number of atoms in the empirical formula by the number of formula units.

107 Combustion Analysis The molecular weight of glucose is 180 g/mole and its empirical formula is CH 2 O. Deduce the molecular formula. 1.Formula weight for CH 2 O is 30.03 g/mole 2.# of formula units = 180/30.03 = 6 3.Molecular formula = C 6 H 12 O 6

Using state symbols It is useful to specify the physical states of the reactants and products in a chemical reaction. It is especially important in problems involving chemical equilibrium and rates of reactions

Symbols used in Chemical Equations symboldefinition +written between multiple reactants or products Δwritten above the arrow to indicate heat added to a reaction (s)indicates a substance in the solid state (l)indicates a substance in the liquid state (g)indicates a substance in the gaseous state (aq)indicates a substance dissolved in water

Balancing ionic equations When ionic compounds dissolve in water, the ions separate from each other. Ionic compounds include salts, acids, bases Sometimes only a few of the ions produced in an aqueous solution will participate in a reaction The ions that play no part in the reaction are called spectator ions

WARNING! Don’t mess with the insides of polyatomic ions – put parenthesis around them – treat the WHOLE polyatomic ion as though it were an element! Don’t ever play around with subscripts (those little numbers that tell you how many atoms are in a molecule) C 6 H 22 O 11

113 Writing Chemical Equations The law of conservation of mass is based on the concept that atoms are not created or destroyed, but just rearranged in a chemical reaction. It is because of this law that we balance chemical equations so that the number of atoms of each element on the reactant side is equal to that on the product side. Example:Propane gas reacts with oxygen gas to form carbon dioxide gas and water vapor. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Reactants: Propane and oxygen Products: Carbon dioxide and water Coefficients: 1, 5, 3, 4 Subscripts: 3, 8, 2, 2, 2

114 Try Writing Balanced Equations 1.Solid aluminum reacts with hydrochloric acid to form aqueous aluminum chloride and hydrogen gas. a.Al (s) + HCl (aq)  AlCl 3 (aq) + H 2 (g) b.Tally up the number of each atom on the reactant and product side of the equation c.Change the coefficients to balance the reactant side with the product side i.Do not start with an element that is already balanced ii.Do not start with the most difficult element iii.When a polyatomic ion appears in the reactant and product, keep that polyatomic ion together and balance it as you would an individual element

2 HgO(s) -> 2 Hg(l) + O 2 (g) Solid mercury (II) oxide decomposes upon heating to form liquid mercury and oxygen gas.

Writing an ionic equation An ionic equation is simpler than a full chemical equation. It shows only the ions or other particles that are reacting Spectator ions are omitted

Writing an ionic equation 1.Write a full chemical equation 2.Rewrite ionic compounds to include ionic charges 3.Cancel out spectator ions 4.Write ionic equation

Zinc reacts with aqueous copper (II) sulfate produces aqueous zinc sulfate plus copper Write a full chemical equation Zn (s) + CuSO 4 (aq) -> ZnSO 4 (aq) + Cu (s) Rewrite ionic compounds to include ionic charges Zn (s) + Cu 2+ SO 4 2- (aq) -> Zn 2+ SO 4 2- (aq) + Cu (s) Cancel out spectator ions Zn (s) + Cu 2+ SO 4 2- (aq) -> Zn 2+ SO 4 2- (aq) + Cu (s) Write ionic equation Zn (s) + Cu 2+ (aq) -> Zn 2+ (aq) + Cu (s)

Write a full chemical equation Zn(s) + CuSO 4 (aq) -> ZnSO 4 (aq) + Cu(s) Rewrite ionic compounds to include ionic charges Zn(s) + Cu 2+ SO 4 2- (aq) -> Zn 2+ SO 4 2- (aq) + Cu(s) Cancel out spectator ions Zn(s) + Cu 2+ SO 4 2- (aq) -> Zn 2+ SO 4 2- (aq) + Cu(s) Write net ionic equation Zn(s) + Cu 2+ (aq) -> Zn 2+ (aq) + Cu(s) Note: spectator ions In the net ionic equation there are no sulfate ions- they are spectator ions In the net ionic equation both the charges and the atoms are balanced In the net ionic equation both the charges and the atoms are balanced

Chemists prefer to write ionic equations (net ionic equation) for precipitation reactions. A precipitation reaction is a reaction where two aqueous solutions react to form a solid. An aqueous solution of iron (III) sulfate reacts with an aqueous solution of sodium hydroxide. A precipitate of iron (III) hydroxide is formed, together with an aqueous solution of sodium sulfate. FeSO 4 (aq) + 2NaOH (aq) -> Fe(OH) 2 (s) + Na 2 SO 4 (aq) Fe 2+ (aq) + 2OH - (aq) -> Fe(OH) 2 (s)

121 1)Combination (Synthesis) Reaction: Reactants combine to give one product. A + B  AB 4Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) 2Al (s) + 3Br 2 (l)  2AlBr 3 (s) 2)Decomposition Reaction: The reactant is broken down into two or more smaller substances. AB  A + B 2NaN 3 (s)  2Na (s) + 3N 2 (g) CaCO 3 (s)  CaO (s) + CO 2 (g)

122 3)Single-Displacement (Replacement) Reaction: One element replaces another element in a compound so long as the element being replaced is less reactive. Note that one element is oxidized while another is reduced. A + BC  AC + B Br 2 (l) + 2KI (aq)  I 2 (s) + 2KBr (aq); Br is more reactive than I; I is oxidized while Br is reduced I 2 (s) + KBr (aq)  NO REACTION!!!; I is NOT more reactive than Br 4)Double-Displacement (Replacement) Reaction: The cations and anions of two compounds are exchanged to produce two different compounds. Reaction will only occur if a product that is insoluble or molecular is produced. AB + CD  AD + CB 2AgNO 3 (aq) + BaCl 2 (aq)  2AgCl (s) + Ba(NO 3 ) 2 (aq) CaCl 2 (aq) + KNO 3 (aq)  NO REACTION!!!

123 Types of Chemical Reactions Continued 5)Combustion Reaction: The reaction of a compound with molecular oxygen to form products in which all elements are combined with oxygen. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) 6) Neutralization (a special type of DD reaction): The products are a salt and water. HA(aq) + BOH(aq)  BA (aq or s) + H 2 O (l) HCl(aq) + NaOH(aq)  NaCl (aq) + H 2 O (l) H 2 SO 4 (aq) + Al(OH) 3 (aq)  Al 2 (SO 4 ) 3 (aq) + H 2 O (l)

Calculating the concentration of a solution A titration is often used to find the exact concentration of a solution Remember that a the concentration of a solution is molarity (M) Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution:

Concentrations of Solutions –Water must be tested continually to ensure that the concentrations of contaminants do not exceed established limits. These contaminants include metals, pesticides, bacteria, and even the by- products of water treatment. –(molarity; % v/v; %m/m; ppm) 16.2

Molarity The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. –A dilute solution is one that contains a small amount of solute. –A concentrated solution contains a large amount of solute. 16.2

Molarity Molarity (M) is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, divide the moles of solute by the volume of the solution. 16.2

Molarity To calculate molarity: Calculate the number of moles of solute present. Calculate the number of liters of solution present. Divide the number of moles of solute by the number of liters of solution.

Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution. Other common rearrangements:

MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water. First calculate the moles of solute: 1.5 g NaCl x 1 mole NaCl = 0.0257 moles of NaCl 58.45 g NaCl Next convert mL to L: 0.500 L of solution Last, plug the appropriate values into the correct variables in the equation: M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L

16.2 Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90g NaCl in exactly 100 mL of solution. What is the molarity of the solution?

MOLARITY M = n / V = mol / L How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution? First calculate the moles of solute needed: M = n / V, now rearrange to solve for n: n = MV n = (1.25 mol / L ) (0.2500 L) = 0.3125 moles of solute needed Next calculate the molar mass of LiOH: 23.95 g/mol Last, use diminsional analysis to solve for mass: 0.3125 moles 23.95 g LiOH 1 mol LiOH = 7.48 g of LiOH

What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL? …. Great question Ms. Strauss…. Thank you, glad you like it First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute therefore: 47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL density of solute Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass. 0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL Next convert mL to L: 0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M MOLARITY M = n / V = mol / L

MOLARITY & DILUTION M 1 V 1 = M 2 V 2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. Since the amount or moles of solute before dilution (n b ) and the moles of solute after the dilution (n a ) are the same: n b = n a And the moles for any solution can be calculated by n=MV A relationship can be established such that M b V b = n b = n a = M a V a Or simply : M b V b = M a V a

Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M 1 = 0.05 mol/LM 2 = ? V 1 = 25.0 mLV 2 = 50.0 + 25.0 = 75.0 mL M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V 2 75.0 mL MOLARITY & Dilution

Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl? M 1 = 6.00 mol/LM 2 = 0.150 V 1 = ? mLV 2 = 250.0 mL M 1 V 1 = M 2 V 2 M 2 V 2 = V 1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl M 1 6.00 mol/L You would need 6.25 mL of the 6.00 M HCl reagent which would be added to about 100 mL of DI water in a 250.0 mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250.0 mL. Mix well. MOLARITY & dilution

Preparation of Molar Solutions Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?  Step #1: Ask “How Much?” (What volume to prepare?) 1.500 L  Step #2: Ask “How Strong?” (What molarity?) 0.500 mol 1 L  Step #3: Ask “What does it weigh?” (Molar mass is?) 58.44 g 1 mol = 43.8 g

Molarity half filled with distilled water. Put some dH 2 O in before the solid!!! a) To make a 0.5 molar (0.5M) solution, first add 0.5 mol of solute to a 1L volumetric flask half filled with distilled water. Put some dH 2 O in before the solid!!! 16.2

Molarity carefully b) Swirl the flask carefully to dissolve the solute. 16.2 exactly c) Fill the flask with water exactly to the 1-L mark.

Label the volumetric Use a small amount of masking tape… NOT entire roll Needed information: – Chemical formula – Molarity – Your name – Date solution was made – Write small and neat Tape does NOT go on the neck of volumetric flask

Excellent job at labelling volumetric flask

Concentration of Solutions Dilution is the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution

16.2 Making Dilutions –What effect does dilution have on the total moles of solute in a solution? –Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change.

Making Dilutions The total number of moles of solute remains unchanged upon dilution, so you can write this equation. M 1 and V 1 are the molarity and volume of the initial solution, and M 2 and V 2 are the molarity and volume of the diluted solution. 16.2

Making Dilutions Making a Dilute Solution: Adding solvent to a concentrated solution lowers the concentration, but the total number of moles of solute present remains the same. 16.2

Making Dilutions Volume-Measuring Devices: Buret graduated cylinder volumetric flask volumetric pipet. Note: beaker is not on the list! 16.2

Making Dilutions To prepare 100 ml of 0.40M MgSO 4 from a stock solution of 2.0M MgSO 4 : A.measure 20 mL of the stock solution with a 20mL pipet. B.transfer the 20 mL to a 100mL volumetric flask C.carefully add distilled water to the mark to make 100 mL of solution. D.Precision is essential! 16.2

Finding the formula of washing soda lab Sources of error? 1.Average titre does not include the practice titration 2.Ave ml titre from #1 0.1 mol = mol HCl 1000 ml

3 calc # of moles of Na 2 CO 3 Use formula V acid x M acid x mol base = V base x M base x mol acid V acid – ave titre M acid- 0.1 M V base- 25 cm3 M base- this is what we are looking for Mol acid- get from balanced equation coefficient Mol base- get from balance equation coefficient

4 calc # of moles of Na 2 CO 3 Use the molarity you just calculated to figure out the moles For example: this is not your molarity 0.0251 mol 250 cm3 = moles of Na 2 CO 3 1000 ml

5. Calc formula of Na 2 CO 3 Mr = m/n Mr = 1.3 grams/ moles you just calculated Take this # which is grams/mol and substrate the FW of Na 2 CO 3. This gives you the grams/mol of H 2 O Divide by FW of H 2 O and you get the moles of H 2 O in the formula Na 2 CO 3 *xH 2 O

How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M f = 0.200V f = 0.06 L V i = ? L V i = MfVfMfVf MiMi = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

Concentration of Solutions In an experiment, a student needs 250.0 mL of a 0.100 M CuCl 2 solution. A stock solution of 2.00 M CuCl 2 is available. How much of the stock solution is needed? (2.00 M CuCl 2 )(L) = (0.100 M CuCl 2 )(0.2500 L) L c = 0.0125 L or 12.5 mL To make the solution: 1)Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask. 2) Carefully dilute to the calibration mark. M c × V c = M d × V d

Serial Dilutions Many procedures performed in modern biology and chemistry laboratories require sets of solutions that cover a range of concentrations. These include quantifying the number of bacteria in a sample using plate counts and development of standard curves for quantitative colorimetric, radiometric and enzymatic assays. Sets of solutions over a range of concentrations are prepared via serial dilution.

Typically, the dilution factor remains constant for each dilution resulting in an exponential decrease in concentration. 10-fold serial dilution results: 1M, 0.1M, 0.01M, 0.001M, an so on Serial dilutions are used to accurately create extremely diluted solutions as well as solutions for experiments that require a concentration curve with an exponential or logarithmic scale Widely used in biochemistry, pharmacology, microbiology, and physics

Serial Dilutions

In the serial dilution on the previous slide, 1 ml of stock solution is mixed with 9 ml of diluent, for a 1/10 dilution. Then 1 ml of the 1/10 dilution is mixed with another 9 ml of diluent. The second tube also has a 1/10 dilution, but the concentration of stock in the second tube is 1/10 x 1/10 for a 1/100 dilution.

Serial Dilutions Continuing with the serial dilution, in the third tube, you mix 1 ml of the 1/100 dilution from the second tube with 9 ml of diluent in the third tube. Again you have a 1/10 dilution in the third tube, but the concentration of stock in the third tube is 1/10 x 1/10 x 1/10 for a 1/1000 dilution. This dilution could be carried out over many subsequent tubes.

Serial dilutions are most often used in serological procedures, where technicians need to make dilutions of patient’s serum to determine the weakest concentration that still exhibits a reaction of some type. The concentration exhibiting a reaction is called a “titer”.

Serial Dilutions Example of determining a titer: A technician makes a serial dilution using patient serum: Tube #1 = 1/10 Tube #2 = 1/100 Tube #3 = 1/1000 Tube #4 = 1/10,000 Tube #5 = 1/100,000 Reactions occur in tubes 1 through 3, but NOT in tubes 4 or 5. The titer = 1000.

Serial Dilution Strategy (a) M d = Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions (1 to 4) are prepared by sequential diluting 10.00 mL of each solution to 250.00 mL. Determine (a) the concentrations of all four standard solutions and (b) the number of moles of HCl in each solution. Solution (a) M d1 = M d2 = M d3 = M d4 = 2.00 M × 10.00 mL 250.00 mL M c × mL c mL d ; (b) mol = M×L, 250.00 mL = 2.500×10 -1 L = 8.00×10 -2 M 8.00×10 -2 M × 10.00 mL 250.00 mL = 3.20×10 -3 M 3.20×10 -3 M × 10.00 mL 250.00 mL = 1.28×10 -4 M 1.28×10 -4 M × 10.00 mL 250.00 mL = 5.12×10 -6 M

Percent Solutions –What are two ways to express the percent concentration of a solution? –The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution. 16.2

Percent Solutions Isopropyl alcohol (2-propanol) is sold as a 91% solution. This solution consist of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution. 16.2 Concentration in Percent (Volume/Volume)

170 Solution Stoichiometry Solution Stoichiometry uses molarity as a conversion factor between volume and moles of a substance in a solution.

Stoichiometry The stoichiometry of an acid-base neutralization reaction is the same as that of any other reaction that occurs in solution (they are double displacement reactions, after all). For example, in the reaction of sodium hydroxide and hydrogen chloride, 1 mol of NaOH neutralizes 1 mol of HCl: NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l) Stoichiometry provides the basis for a procedure called titration, which is used to determine the concentrations of acidic and basic solutions. 171

Titration Titration is a method for determining the concentration of a solution by reacting a known volume of that solution with a solution of known concentration. If you wish to find the concentration of an acid solution, you would titrate the acid solution with a solution of a base of known concentration. You could also titrate a base of unknown concentration with an acid of known concentration. 172

In the titration of an acid by a base, the pH meter measures the pH of the acid solution in the beaker as a solution of a base with a known concentration is added from the buret. http://wps.prenhall.com/wps/media/objects/3 312/3392202/blb1703.html 173

Titration Procedure 1)A measured volume of an acidic or basic solution of unknown concentration is placed in a beaker. The electrodes of a pH meter are immersed in this solution, and the initial pH of the solution is read and recorded. 2)A buret is filled with the titrating solution of known concentration. This is called the standard solution, or titrant. 3)Measured volumes of the standard solution are added slowly and mixed into the solution in the beaker. The pH is read and recorded after each addition. This process continues until the reaction reaches the equivalence point, which is the point at which moles of H + ion from the acid equal moles of OH - ion from the base. 174

In the titration of a strong acid by a strong base, a steep rise in the pH of the acid solution indicates that all of the H + ions from the acid have been neutralized by the OH - ions of the base. The point at which the curve flexes is the equivalence point of the titration. Bromthymol blue is an indicator that changes color at this equivalence point. Notice that phenolphthalein and methyl red don’t match the exact equivalence point, but the slope is so steep that it doesn’t matter. 175

Strong-Strong Titration The previous slide shows how the pH of the solution changes during the titration of 50.0 mL of 0.100 M HCl, a strong acid with 0.100 M NaOH, a strong base. The inital pH of the 0.100 M HCl is 1.00. – As NaOH is added, the acid is neutralized and the solution’s pH increases gradually. – When nearly all of the H + ions from the acid have been used up, the pH increases dramatically with the addition of an exceedingly small volume of NaOH. – This abrupt change in pH occurs at the equivalence point of the titration. – Beyond the equivalence point, the addition of more NaOH again results in the gradual increase in pH. 176

The equivalence point here is not at a pH of 7. Phenolphthalein is an indicator that changes color at this equivalence point. Notice that the starting pH is different and the region of change is smaller. 177

Strong-Weak Titrations You might think that all titrations must have an equivalence point at pH 7 because that is the point at which concentrations of hydrogen ions and hydroxide ions are equal and the solution is neutral. This is not the case – some titrations have equivalence points at pH values less than 7, and some have equivalence points at pH values greater than 7. These differences occur because of reactions between the newly formed salts and water – salt hydrolysis. – Some salts are basic (weak acid, strong base) and some salts are acidic (strong acid, weak base). The previous slide shows that the equivalence point for the titration of HPr (a weak acid) with NaOH (a strong base) lies at pH 8.80. 178

Acid-Base Indicators Chemists often use a chemical dye rather than a pH meter to detect the equivalence point of an acid-base titration. Chemical dyes whose colors are affected by acidic and basic solutions are called acid-base indicators. Many natural substances act as indicators. – If you use lemon juice in your tea, you might have noticed that the brown color of tea gets lighter when lemon juice is added. – Tea contains compounds called polyphenols that have slightly ionizable hydrogen atoms and therefore are weak acids. – Adding acid in the form of lemon juice to a cup of tea lessens the degree of ionization, and the color of the un-ionized polyphenols becomes more apparent. Chemists have several choices in selecting indicators. – Bromthymol blue is a good choice for the titration of a strong acid with a strong base, and phenolphthalein changes color at the equivalence point of a titration of a weak acid with a strong base. 179

Phenolphthalein 181

Methyl Orange 182

Equivalence Point

Indicators and Titration End Point Many indicators used for titrations are weak acids. – Each has its own particular pH or pH ranges over which it changes color. The point at which the indicator used in a titration changes color is called the end point of the titration. – It is important to choose an indicator for a titration that will change color at the equivalence point of the titration. – Remember that the role of the indicator is to indicate to you, by means of a color change, that just enough of the titrating solution has been added to neutralize the unknown solution. Equivalence point ≠ End point! – BUT for strong-strong titrations, the pH change is so steep and so large, that the are approximately equal. 184

Titration with an Indicator 185

What’s the Point of a Titration Again? To find the unknown concentration of an acid or a base. So you perform the actual titration noting the volume you started with and how much volume of the titrant you added and then... Math! (Oh no! Not math! Anything but math!) 186

Titration Calculations The balanced equation of a titration reaction is the key calculating the unknown molarity. For example, sulfuric acid is titrated with sodium hydroxide according to this equation: H 2 SO 4 (aq) + 2 NaOH (aq)  Na 2 SO 4 (aq) + 2 H 2 O (l) 1)Calculate the moles of NaOH in the standard from the titration data: molarity of the base (M B ) and the volume of the base (V B ). In other words, M B V B = (mol/L)(L) = mol NaOH in standard 2) From the equation, you know that the mole ratio of NaOH to H 2 SO 4 is 2:1. Two moles of NaOH are required to neutralize 1 mol of H 2 SO 4. mol H 2 SO 4 titrated = mol NaOH in standard x (1 mol H 2 SO 4 / 2 mol NaOH) 3)M A represents the molarity of the acid and V A represents the volume of the acid in liters. M A = mol H 2 SO 4 titrated/V A 187

In Short Form... M A V A = M B V B (mol acid/mol base) This is the mole ratio Does this make sense? Let’s find out using the definition of molarity (mol/L) and dimensional analysis... M A V A = M B V B (mol acid/mol base) (mol acid/L acid)(L acid) = (mol base/L base)(L base) (mol acid/mol base) mol acid = mol base (mol acid/mol base) mol acid = mol acid 188

Titrations Analyte: the solution of unknown concentration but known volume. Titrant: the solution of known concentration. Process: Add titrant until all of the analyte has reacted, then record the volume of titrant added. 189 Analyte + Titrant → Products

190 Titrations Equivalence Point: the point at which exactly the right volume of titrant has been added to complete the reaction. Indicator: substance that changes color when an excess of titrant has been added (phenolphthalein, bromocresol green).

Titrations Titration Calculations: 1.Find the number of moles of titrant added to reach the endpoint. 2.Determine the moles of analyte that must have been present (use stoichiometric coefficients). 3.Determine the concentration of analyte that must have been present in the flask (use the volume of analyte). 4.Calculate the concentration of analyte in the original sample. Analyte: a substance whose chemical constituents are being identified and measured.

192 Titrations 14.84 mL of an HCl solution of unknown concentration is titrated with standard NaOH solution. At the equivalence point, 25.0 mL of the 0.675 M NaOH has been added. Calculate the concentration of the HCl solution. NaOH + HCl → NaCl + H 2 O Analyte (HCl) 14.84 mL ?M Titrant (NaOH) 25.0 mL 0.675 M (25.0mL)(0.675M) 14.84mL = 1.14 M HCl

193 An antacid tablet containing sodium bicarbonate (NaHCO 3 ) and weighing 4.00 g is dissolved in water. The solution is titrated to the equivalence point with 50.0 mL of 0.200 M HCl. Calculate the mass% of sodium bicarbonate in the tablet. NaHCO 3 + HCl → NaCl + H 2 O + CO 2 (0.05L HCl)(0.200M HCl) = 0.01 moles HCl Look at balanced equation, for every 1 mole HCl = 1 mol NaHCO 3. Therefore, 0.01 mole HCl = 0.01 mole NaHCO 3. 0.01 mole NaHCO 3 x 84 g/mol = 0.84 g NaHCO 3 in the tablet 0.84 g / 4g x 100% = 21%

Using molar gas volumes STP 1 mole of any gas = 22.4 L rtp 1 mole of any gas = 24.0 dm 3

Significant Figures When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer.

Similar presentations

Presentation on theme: "Significant Figures When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Significant Figures When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer.

Similar presentations

Presentation on theme: "Significant Figures When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer."— Presentation transcript:

Similar presentations

About project

Feedback