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Stoichiometry How much is consumed and produced in chemical reactions Labs #6 Formula of a Hydrate #7 Empirical Formulas #8 Limiting Reactant Chemical Equations Chapter 7

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Atomic Mass Unit Standard/unit needed to measure any mass –SI-standard is platinum-iridium cylinder –Unit is kilogram Masses of individual atoms cannot be measured with balance but relative masses of atoms of different elements can be measured –Atomic mass scale used Standard is 1 atom C-12 (assigned mass of exactly 12 amu) 1 u = x kg Unit is atomic mass unit (u)

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Masses of other atoms, molecules, ions, and subatomic particles measured by mass spectrometer and values are reported relative to mass of carbon-12 –Proton x kg = u –Neutron x kg = u –Electron x kg = u Use 1 for protons/neutrons while electrons mass is small enough ( u) to be neglected for most purposes

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Formula weights Sum of atomic weights of each atom in chemical formula FW of H 2 SO 4 = 2(AW H) + (AW S) + 4(AW O) –= 2(1.0 amu) amu + 4(16.0 amu) –= 98.1 amu Of atom = atomic weight of element Of molecule = molecular weight

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Modern Atomic Masses Most elements mixtures of isotopes-we need to know –Which isotopes are present naturally –Masses of each isotopes –Abundance of each isotope in element Atomic mass of element is weighted average of masses of naturally occurring isotopes forming element –Calculate weighted average by multiplying mass of each isotope by decimal equivalent of its abundance and then we add each of these products together

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For example, the element lithium has two isotopes that occur in nature: with 7.5 percent abundance, and with 92.5 percent abundance. The atomic mass of lithium-6 is amu, and that of lithium-7 is amu. The average mass of such a mixture of Li atoms is given by: –average atomic mass = (fraction of isotope X)(mass of isotope X) + (fraction of isotope Y)(mass of isotope Y) –= (0.075)( amu) + (0.925)( amu) = 0.45 amu amu = 6.94 amu Note that neither lithium-6 nor lithium-7 has an atomic mass of 6.94 amu. This is the average value for the mixture of the two Li isotopes.

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A sample of metal M is vaporized and injected into a mass spectrometer. The mass spectrum tells us that 60.10% of the metal is present as 69 M and 39.90% is present as 71 M. The mass value for 69 M and 71 M are amu and amu, respectively. –What is the average atomic mass of the element? amu –What is the element? Ga

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The element indium exists naturally as two isotopes. 113 In has a mass of amu, and 115 In has a mass of amu. The average atomic mass of indium is Calculate the percent relative abundance of the two isotopes of indium. – = (X) (Y) –X + Y = 1 so Y = 1 –X –Substitute and get 113 In = 4.2% and 115 In = 95.8%

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The Mole (mol) Mole used because atoms, ions, and simple molecules very small Must work with very large numbers of them to obtain quantities measurable in lab = #C atoms in exactly 12 grams of pure 12 C –SI unit defined in relation to mass of C-12 isotope –Amount of substance that contains as many elementary entities as there are atoms in kg of carbon-12 –In kg of carbon-12 there are x carbon-12 atoms (experimentally determined )

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Molar mass Molar mass of element is mass equal to its atomic mass expressed in grams Any element will contain same number of atoms as molar mass of any other element –28.09 g Si contains same # of atoms as g C # atoms in a molar mass, called Avogadro's number, is equal to x atoms Quantity of a substance that contains Avogadro's number of atoms or other entities is called mole –1 mole Na+ ions = x Na + ions = g Na + –1 mole O 2 molecules = x O 2 molecules = g O 2 –1 mole O 3 molecules = x O 3 molecules = g O 3

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How many grams does a sample containing 34 atoms of neon weigh? 34 atoms Ne amu Ne 1 g Ne = x g Ne 1 atom Ne x amu Ne A sample of elemental silver (Ag) has a mass of g. –How many moles of silver are in the sample? –How many atoms of silver are in the sample? g Ag 1 mol Ag = mol Ag g Ag mol Ag x atoms Ag = x atoms Ag 1 mol Ag

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The molar mass of a compound is its molecular mass expressed in grams 1 mol NO 2 = x NO 2 molecules = g NO 2 1 mol N 2 O 5 = x N 2 O 5 molecules = g N 2 O 5 Also note that 1 mole of NO 2 consists of 1 mole of N atoms, and 2 moles of O atoms 1 mol NO 2 contains 1 mol N atoms = g N 1 mol NO 2 contains 2 mol O atoms= g O 1 mol NO 2 = g NO 2 Just as molecular mass is sum of atomic masses, molar mass of compound is sum of molar masses of atoms in molecule

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Molecular Mass-Formula Mass Formula mass is an extension of atomic mass Multiply # atoms of each element in formula by atomic mass of element and add results together For example, the molecular masses of two nitrogen oxides, NO 2 and N 2 O 5, are as follows: –molecular mass of NO 2 = atomic mass of N + 2(atomic mass of O) = amu + 2(16.00 amu) = amu –molecular mass of N 2 O 5 = 2( atomic mass of N) + 5(atomic mass of O) = 2(14.01 amu) + 5(16.00 amu) = amu

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Homework: Read , pp Q pp. 22, 24, 28, 32, 46, 48 b/c, 52

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Percent Composition Relative contribution of mass of element to mass of formula in which it appears Mass composition-when mass of each element in a substance is specified, either as a % or in grams 3 steps: –Compute molecular mass of compound –Calculate how much of molecular mass comes from each element –Divide each elements mass contribution by total molecular mass and multiply by 100 to convert to %

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Sodium chloride (NaCl)-The molar mass, g, is the sum of the mass of 1 mole of Na, g, and the mass of 1 mole of Cl, g. –% Na by mass is g Na/58.44 g NaCl x 100 = 39.33% –% Cl by mass is g Cl/58.44 g NaCl x 100 = 60.66%

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Calculate the mass percent of each element in potassium ferricyanide, K 3 Fe(CN) 6. –35.62% K –16.96% Fe –21.88% C –25.53% N

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Formulas molecular formula = (empirical formula) n –[n = integer] ( actual ratio of atoms in compound ) molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH ( simplest whole- number ratio of atoms in compound )

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Empirical Formula Determination 1.Base calculation on 100 grams of compound (If % given, assume 100 g) 2.Determine moles of each element in 100 grams of compound (divide given mass by atomic mass) 3.Divide each value of moles by smallest of values 4.Multiply each number by integer to obtain all whole numbers

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1.Calculated from empirical formula when molar mass is known 2.Divide molar mass of compound by empirical molar mass 3.Multiply empirical formula by quotient obtained from division Molecular Formula Determination

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Determine empirical/molecular formulas for a deadly nerve gas that gives the following mass percent analysis: C=39.10%3.26 mol C/ H=7.67%7.61 mol H/ O=26.11%1.63 mol O/ P=16.82%0.543 mol P/ F=10.30%0.542 mol F/ Known molar mass = g. C 6 H 14 O 3 PF Empirical formula = g. So molecular formula also.

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When CH and CHO are combusted in an excess of O 2, the only products formed are CO 2 and H 2 O. By measuring mass of original sample and masses of products, can calculate empirical formula of compound Since all of C/H appear as CO 2 and H 2 O respectively, masses of these elements can be determined If original compound contains oxygen, its mass is determined by subtracting mass of compound from sum of masses of C/H Masses of element are converted to moles and empirical formula is determined.

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Combustion of 11.5 grams of ethanol produces 22.0 grams of CO 2 and grams of H 2 O. Determine the empirical formula of ethanol.

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A g sample of an unknown produced g of CO 2 and g of H 2 O. Determine the empirical formula of compound g CO 2 1 mol CO 2 1 mol C g C = g C g CO 2 1 mol CO 2 1 mol C g H 2 O 1 mol H 2 O 2 mol H 1.01 g H = g H g H 2 O 1 mol H 2 O 1 mol H – ( ) = g O /12.01 = mol C/ /1.01 = mol H/ /16.00 = mol O/ C 3.5 H 3 O = multiply everything by 2 C 7 H 6 O 2

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Homework: Read , pp Q pp , #54a, 56, 58 (youll love this one), 65, 68

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Chemical Equation- symbolic way of representing chemical reaction Reactants on the left side of equation. Products on the right side. Symbols Used in Writing Chemical Equations SymbolMeaning yields, produces reversible reaction, equilibrium (s)solid phase (l)liquid phase (g)gas phase (aq)aqueous solution +added to, and

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Types of chemical reactions Combination –A + B C Decomposition –C A + B Combustion –Hydrocarbon (may have -OH)+ O 2 CO 2 + H 2 O –Produce flame Exchange/metathesis reactions –Single replacement A + BC B + AC –Double replacement AB + CD AD + CB

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Chemical Equation C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O Equation balanced 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

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Balancing equations- using coefficients Number placed before chemical formula in equation changes amount of substance Multiplier for formula –2H 2 O means two molecules of water Two molecules of water consists of 4 hydrogen atoms and 2 oxygen atoms. –3H 2 O means three molecules of water, which stands for 6 H atoms and 3 O atoms Absence of coefficient is understood to mean one

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Calculating masses of reactants and products: Balance equation Convert known mass of reactant or product to moles of that substance Use balanced equation to set up appropriate mole ratios Use appropriate mole ratios to calculate number of moles of desired reactant or product Convert from moles back to grams if required

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Example If 1.5 moles of C 2 H 6 reacts, how many moles of H 2 O will be formed? 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 1.50 mol C 2 H 6 7 mol O g O 2 = 168 g O 2 2 mol C 2 H 6 1 mol O 2

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Example If 160. grams of O 2 reacts, how many grams of CO 2 will be formed? 160.g O 2 1 mol O 2 4 mol CO 2 1 mol CO 2 = 126 g CO g O 2 7 mol O g CO 2

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Example A gram sample of solid NaHCO 3 is treated with an excess of HCl and heated to remove the water. What is the change in the mass of the solid? NaHCO 3 + HCl NaCl + CO 2 + H 2 O After the reaction, the solid that is present is NaCl. From the equation, we know that 1 mole of NaHCO 3 produces 1 mole of NaCl g NaHCO 3 1 mol NaHCO 3 1 mol NaCl g NaCl = g NaCl g NaHCO 3 1 mol NaHCO 3 1 mol NaCl g NaCl – g NaHCO 3 = g-mass of solid decreases by 3.65 grams.

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Example A 3.75-gram sample of solid is either NaHCO 3 or Na 2 CO 3. When treated w/excess HCl and heated, mass of solid increases by 0.38 gram. Identify original sample. We calculate the quantities of NaCl produced by 3.75 grams of NaHCO 3 and Na 2 CO 3 and we compare the differences in mass with that given in the problem statement g NaHCO 3 1 mol NaHCO 3 1 mol NaCl g NaCl = 2.61 g NaCl g NaHCO 3 1 mol NaHCO 3 1 mol NaCl –2.61 g NaCl – 3.75 g NaHCO 3 = g 3.75 g Na 2 CO 3 1 mol Na 2 CO 3 1 mol Na 2 CO 3 1 mol NaCl = 4.13 g NaCl g Na 2 CO 3 2 mol NaCl g NaCl –4.13 g NaCl – 3.75 G Na 2 CO 3 = So the original sample is Na 2 CO 3 since the mass increases by 0.38 gram

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Homework: Read , pp Q pp , #76, 78, 82, 86, 90 (fun one)

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Limiting Reagents (reactants) One reactant will be completely consumed (limiting reagent/limiting reactant) before other runs out –Reaction stops/no more product made –Determines, or limits, amount product formed Reactant not completely consumed- excess reagent

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Steps to determine limiting reagent: 1.Use equation to calculate stoichiometric mole ratio of reactants 2.Calculate mole ratio of reactants under experimental conditions given 3.Compare two mole ratios: If experimental mole ratio is larger than stoichiometric mole ratio, reactant in denominator is limiting reactant; if smaller, reactant in numerator is limiting reactant 4.In solving any problems involving calculations, always use limiting reactant

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Given grams of Fe(CrO 2 ) 2 and 6.52 grams of K 2 CO 3 are reacted by the equation below, calculate the mass of Fe 2 O 3 produced, then all reactants, products, unreacted substances. Calculate number of moles of each and then compare using balanced equation Divide by coefficients /4 = /8 =

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Therefore K 2 CO 3 is limiting reagent To calculate how much is consumed, figure out yield using limiting reagent, then use that information to see how much of others used Subtract that from what is given

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Theoretical Yield Theoretical Yield amount of product predicted by calculations Calculations under ideal conditions Under laboratory conditions, actual yield usually less than theoretical yield We define percent yield as actual yield divided by theoretical yield multiplied by 100 % yield = actual yield/theoretical yield x 100

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50 g of silver nitrate is mixed with 50 g of hydrochloric acid in a water based solution. A white precipitate forms (silver chloride). The solution is filtered and the white precipitate collected and dried. The dried precipitate is measured to have a mass of 53.6 g. What is the theoretical and percent yield?

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Homework: Read 3.9, pp Q pp , #94 a/c, 96, 100 Do one additional exercise and one challenge problem. Submit quizzes to me by _zumdahl/ace/launch_ace.html?folder_path=/chemist ry/book_content/ _zumdahl/ace&layer=act&s rc=ch03_ace1.xml 321_zumdahl/ace/launch_ace.html?folder_path=/chemist ry/book_content/ _zumdahl/ace&layer=act&s rc=ch03_ace2.xml 321_zumdahl/ace/launch_ace.html?folder_path=/chemist ry/book_content/ _zumdahl/ace&layer=act&s rc=ch03_ace3.xml

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