Presentation on theme: "How much is consumed and produced in chemical reactions"— Presentation transcript:
1How much is consumed and produced in chemical reactions StoichiometryHow much is consumed and produced in chemical reactionsLabs#6 Formula of a Hydrate#7 Empirical Formulas#8 Limiting ReactantChemical EquationsChapter 7
2Atomic Mass Unit Standard/unit needed to measure any mass SI-standard is platinum-iridium cylinderUnit is kilogramMasses of individual atoms cannot be measured with balance but relative masses of atoms of different elements can be measuredAtomic mass scale usedStandard is 1 atom C-12 (assigned mass of exactly 12 amu)1 u = x kgUnit is atomic mass unit (u)
3Masses of other atoms, molecules, ions, and subatomic particles measured by mass spectrometer and values are reported relative to mass of carbon-12Proton x kg = uNeutron x kg = uElectron x kg = uUse 1 for protons/neutrons while electron’s mass is small enough ( u) to be neglected for most purposes
5Formula weights Sum of atomic weights of each atom in chemical formula FW of H2SO4 = 2(AW H) + (AW S) + 4(AW O)= 2(1.0 amu) amu + 4(16.0 amu)= 98.1 amuOf atom = atomic weight of elementOf molecule = molecular weight
6Modern Atomic MassesMost elements mixtures of isotopes-we need to knowWhich isotopes are present naturallyMasses of each isotopesAbundance of each isotope in elementAtomic mass of element is weighted average of masses of naturally occurring isotopes forming elementCalculate weighted average by multiplying mass of each isotope by decimal equivalent of its abundance and then we add each of these products together
7For 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 amuNote 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.
8A 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 69M and 39.90% is present as 71M. The mass value for 69M and 71M are amu and amu, respectively.What is the average atomic mass of the element? amuWhat is the element? Ga
9The element indium exists naturally as two isotopes The element indium exists naturally as two isotopes. 113In has a mass of amu, and 115In 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 –XSubstitute and get 113In = 4.2% and 115In = 95.8%
10The Mole (mol)Mole used because atoms, ions, and simple molecules very smallMust work with very large numbers of them to obtain quantities measurable in lab= #C atoms in exactly 12 grams of pure 12CSI unit defined in relation to mass of C-12 isotopeAmount of substance that contains as many elementary entities as there are atoms in kg of carbon-12In kg of carbon-12 there are x 1023 carbon-12 atoms (experimentally determined )
11Molar 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 element28.09 g Si contains same # of atoms as g C# atoms in a molar mass, called Avogadro's number, is equal to x 1023 atomsQuantity of a substance that contains Avogadro's number of atoms or other entities is called mole1 mole Na+ ions = x 1023 Na+ ions = g Na +1 mole O2 molecules = x 1023 O2 molecules = g O21 mole O3 molecules = x 1023 O3 molecules = g O3
12How many grams does a sample containing 34 atoms of neon weigh? 34 atoms Ne amu Ne g Ne = x g Ne1 atom Ne x amu NeA 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?21.46 g Ag mol Ag = mol Ag107.9 g Agmol Ag x atoms Ag = x atoms Ag1 mol Ag
13The molar mass of a compound is its molecular mass expressed in grams 1 mol NO2 = x 1023 NO2 molecules = g NO2 1 mol N2O5 = x 1023 N2O5 molecules = g N2O5Also note that 1 mole of NO2 consists of 1 mole of N atoms, and 2 moles of O atoms1 mol NO2 contains 1 mol N atoms = g N 1 mol NO2 contains 2 mol O atoms = g O 1 mol NO2 = g NO2Just as molecular mass is sum of atomic masses, molar mass of compound is sum of molar masses of atoms in molecule
14Molecular Mass-Formula Mass Formula mass is an extension of atomic massMultiply # atoms of each element in formula by atomic mass of element and add results togetherFor example, the molecular masses of two nitrogen oxides, NO2 and N2O5, are as follows:molecular mass of NO2 = atomic mass of N + 2(atomic mass of O)= amu + 2(16.00 amu) = amumolecular mass of N2O5 = 2( atomic mass of N) + 5(atomic mass of O)= 2(14.01 amu) + 5(16.00 amu) = amu
16Percent CompositionRelative contribution of mass of element to mass of formula in which it appearsMass composition-when mass of each element in a substance is specified, either as a % or in grams3 steps:Compute molecular mass of compoundCalculate how much of molecular mass comes from each elementDivide each element’s mass contribution by total molecular mass and multiply by 100 to convert to %
17Sodium chloride (NaCl)-The molar mass, 58 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%
18Calculate the mass percent of each element in potassium ferricyanide, K3Fe(CN)6.
19Formulas molecular formula = (empirical formula)n [n = integer] (actual ratio of atoms in compound)molecular formula = C6H6 = (CH)6empirical formula = CH (simplest whole- number ratio of atoms in compound)
20Empirical 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 values4. Multiply each number by integer to obtain all whole numbers
21Molecular Formula Determination Calculated from empirical formula when molar mass is knownDivide molar mass of compound by empirical molar massMultiply empirical formula by quotient obtained from division
22Determine empirical/molecular formulas for a deadly nerve gas that gives the following mass percent analysis:C=39.10% mol C/.0.542H=7.67% mol H/.0.542O=26.11% 1.63 mol O/.0.542P=16.82% mol P/.0.542F=10.30% mol F/.0.542Known molar mass = g.C6H14O3PFEmpirical formula = g. So molecular formula also.
23When CH and CHO are combusted in an excess of O2, the only products formed are CO2 and H2O. By measuring mass of original sample and masses of products, can calculate empirical formula of compoundSince all of C/H appear as CO2 and H2O respectively, masses of these elements can be determinedIf original compound contains oxygen, its mass is determined by subtracting mass of compound from sum of masses of C/HMasses of element are converted to moles and empirical formula is determined.
24Combustion of 11.5 grams of ethanol produces 22.0 grams of CO2 and grams of H2O. Determine the empirical formula of ethanol.
25A 0. 6349 g sample of an unknown produced 1. 603 g of CO2 and 0 A g sample of an unknown produced g of CO2 and g of H2O. Determine the empirical formula of compound.1.603 g CO2 1 mol CO mol C g C = g C44.01 g CO mol CO2 1 mol Cg H2O 1 mol H2O mol H g H = g Hg H2O 1 mol H2O 1 mol H– ( ) = g O0.4374/12.01 = mol C/0.0315/1.01 = mol H/0.166/16.00 = mol O/C3.5H3O = multiply everything by 2C7H6O2
26Homework:Read , ppQ pp , #54a, 56, 58 (you’ll love this one), 65, 68
27Chemical Equation- symbolic way of representing chemical reaction Reactants on the left side of equation.Products on the right side.Symbols Used in Writing Chemical EquationsSymbol Meaning yields, produces⇆ reversible reaction, equilibrium(s) solid phase(l) liquid phase(g) gas phase(aq) aqueous solution+ added to, and
28Types of chemical reactions CombinationA + B CDecompositionC A + BCombustionHydrocarbon (may have -OH)+ O2 CO2 + H2OProduce flameExchange/metathesis reactionsSingle replacementA + BC B + ACDouble replacementAB + CD AD + CB
29Chemical Equation C2H5OH + 3O2 2CO2 + 3H2O Equation balanced 1 mole of ethanol reacts with 3 moles of oxygento produce2 moles of carbon dioxide and 3 moles of water
30Balancing equations- using coefficients Number placed before chemical formula in equation changes amount of substanceMultiplier for formula2H2O means two molecules of waterTwo molecules of water consists of 4 hydrogen atoms and 2 oxygen atoms.3H2O means three molecules of water, which stands for 6 H atoms and 3 O atomsAbsence of coefficient is understood to mean one
31Calculating masses of reactants and products: Balance equationConvert known mass of reactant or product to moles of that substanceUse balanced equation to set up appropriate mole ratiosUse appropriate mole ratios to calculate number of moles of desired reactant or productConvert from moles back to grams if required
32ExampleIf 1.5 moles of C2H6 reacts, how many moles of H2O will be formed?2C2H6 + 7O2 4CO2 + 6H2O1.50 mol C2H mol O g O2 = 168 g O22 mol C2H6 1 mol O2
33ExampleIf 160. grams of O2 reacts, how many grams of CO2 will be formed?160.g O2 1 mol O mol CO2 1 mol CO2= 126 g CO232.0 g O2 7 mol O g CO2
34ExampleA gram sample of solid NaHCO3 is treated with an excess of HCl and heated to remove the water. What is the change in the mass of the solid?NaHCO3 + HCl NaCl + CO2 + H2OAfter the reaction, the solid that is present is NaCl. From the equation, we know that 1 mole of NaHCO3 produces 1 mole of NaCl.12.00 g NaHCO mol NaHCO mol NaCl g NaCl = g NaCl84.01 g NaHCO mol NaHCO mol NaCl8.348 g NaCl – g NaHCO3 = g-mass of solid decreases by 3.65 grams.
35ExampleA 3.75-gram sample of solid is either NaHCO3 or Na2CO3. 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 NaHCO3 and Na2CO3 and we compare the differences in mass with that given in the problem statement.3.75 g NaHCO mol NaHCO mol NaCl g NaCl = 2.61 g NaClg NaHCO mol NaHCO mol NaCl2.61 g NaCl – 3.75 g NaHCO3 = g3.75 g Na2CO mol Na2CO mol Na2CO mol NaCl = 4.13 g NaCl106.0 g Na2CO mol NaCl g NaCl4.13 g NaCl – 3.75 G Na2CO3 = +0.38So the original sample is Na2CO3 since the mass increases by 0.38 gram
37Limiting Reagents (reactants) One reactant will be completely consumed (limiting reagent/limiting reactant) before other runs outReaction stops/no more product madeDetermines, or limits, amount product formedReactant not completely consumed-excess reagent
38Steps to determine limiting reagent: Use equation to calculate stoichiometric mole ratio of reactantsCalculate mole ratio of reactants under experimental conditions givenCompare 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 reactantIn solving any problems involving calculations, always use limiting reactant
39Divide by coefficients Given grams of Fe(CrO2)2 and 6.52 grams of K2CO3 are reacted by the equation below, calculate the mass of Fe2O3 produced, then all reactants, products, unreacted substances.Calculate number of moles of each and then compare using balanced equationDivide by coefficients/4 =0.0472/8 =
40Therefore K2CO3 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 usedSubtract that from what is given
43Theoretical Yield amount of product predicted by calculations Calculations under ideal conditionsUnder laboratory conditions, actual yield usually less than theoretical yieldWe define percent yield as actual yield divided by theoretical yield multiplied by 100% yield = actual yield/theoretical yield x 100
4450 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?