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CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION
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CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATION

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TEMA PARA LA MATERIA DE QUÍMICA...t/t

TEMA PARA LA MATERIA DE QUÍMICA...t/t

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  • 1. STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONQBA Miguel A. Castro Ramírez
  • 2. CHEMICAL EQUATIONS• Chemical reactions are presented in a concise way by chemicalequation• Example: Methane + Oxygen  Carbon Dioxide + Water CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)Reactant Product
  • 3. CHEMICAL EQUATIONS
  • 4. CHEMICAL EQUATIONS Balancing Equation• Balance the equation by determine the coefficients that provideequal numbers of each type of atom on each side of the equation.• Balance equation must contain the smallest possible wholenumber coefficient.• Need to understand the coefficient in front of the formula andsubscript inside the formula.• Two steps in balancing equation: 1. Write the unbalanced equation 2. Adjust the coefficients to get the equal number of each kind of atom on both sides of the arrow.
  • 5. CHEMICAL EQUATIONS Balancing Equation• Example : CH4 (g) + O2 (g) CO2 (g) + H2O (g) (unbalanced) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) (balanced)• Need to understand the coefficient in front of the formula andsubscript inside the formula. (3H2O)• Never change subscripts when balancing an equation.• Don’t forget to put the state of reactants and products in thebalanced equation. (gas-g : liquid-l : solid-s : aqueous(water)-aq)• Δ : indicating heat
  • 6. CHEMICAL EQUATIONSExample : Sodium hydroxide, NaOH and phosphoric acid H3PO4, reactas aqueous solutions to give sodium phosphate, Na3PO4 andwater. The sodium phosphate remains in solution. Write thebalanced equation for this reaction.
  • 7. CHEMICAL EQUATIONSEXAMPLE:This molecular scene depicts an important reaction in nitrogenchemistry (nitrogen is blue, oxygen is red). Write a balancedequation.
  • 8. CHEMICAL EQUATIONSEXERCISE:When aqueous solutions of calcium chloride, CaCl2 and potassiumphosphate K3PO4, are mixed, a reaction occurs in which solidcalcium phosphate, Ca3(PO4)2, separates from the solution. Theother product is KCI(aq). Write the balanced equation.
  • 9. SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY Combination Reaction • Combination reaction: 2 or more substances react to form 1 product. • Example: Magnesium metal burns in air with a dazzling brilliance to produce magnesium oxide.• Mg(s) + O2 (g)  2MgO(s)• Combination reaction between metal & non-metal : Producedionic solid.• In the above reaction: Mg loses e- to form Mg2+ while oxygen gainselectron to produce O2-.
  • 10. SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYDecomposition Reaction • Decomposition Reaction: One substance undergo a reaction to produce two or more other substances. • Heating plays an important roles for this reaction
  • 11. SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY Combustion in Air • These are generally rapid reactions that produce a flame. • Most often involve hydrocarbons reacting with oxygen in the air. • The amount of O2 required in the reaction and the number of CO2 and H2O produced are depend on the composition of the hydrocarbon.• Example: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
  • 12. FORMULA WEIGHTFormula and Molecular Weight• Formula weight of a substance: Sum of the atomic weights ofeach atom in its chemical formula. Example: FW of H2SO4 = 2(AW of H) + (AW of S) + 4(AW of O) 2(1 amu) + 32.1 amu + 4(16 amu) 98.1 amu. • Formula weights are generally reported for ionic compounds. • Molecular weight: Sum of the atomic weights of the atoms in a molecule. Example: MW of C12H22O11 = 12(AW of C) + 22(AW of H) + 11(AW of O) = 342.0 amu
  • 13. FORMULA WEIGHTPercentage Composition from Formulas• One can find the percentage of the mass of a compound thatcomes from each of the elements in the compound by using thisequation• Purpose of calculating: To verify the purity of a compound-Compare the calculated percentage composition of the substancewith that found experimentally. Example: Forensic Chemist• (number of atoms)(atomic weight) % element = x 100 (FW of the compound)
  • 14. FORMULA WEIGHTEXAMPLEA sample of a liquid with a mass of 8.657 g was decomposed intoits elements and gave 5.217 g of carbon, 0.9620 g of hydrogenand 2.478g of oxygen. What is the percentage composition of thiscompound?
  • 15. AVOGADRO’S NUMBER AND THE MOLE•The unit for dealing with the number of atom , ions or moleculesin a common-sized sample is the mole, abbreviated mol.• 1 mole of atoms is the Avogadro number of the atoms.• No Avogadro: 6.02 x 1023 : Defined as wide variety of ways. Butit is simply understand as the conversion factor between amu andgram.• A mole of atoms, a mole of molecules or a mole of anything elseall contain Avogadro’s number of these objects: 1 mol 12C atoms = 6.02 x 1023 12C atoms 1 mol H2O molecules = 6.02 x 1023 H2O molecules
  • 16. AVOGADRO’S NUMBER AND THE MOLE Molar Mass• Molar mass is the mass of 1 mol of a substance in gram - Thus, the mass of a substance (atom, element, moleculeor ions) in amu is numerically equal to the mass in gram of 1 mol ofthat particular substance.1 atom of 12C has a mass of 12 amu - 1 mol 12C has a mass of 12g1 NO3- ion has a mass of 62 amu – 1 mol NO3- has a mass of 62 g1 NaCl unit has a mass of 58.5 amu – 1 mol of NaCl has a mass of58.5 g
  • 17. AVOGADRO’S NUMBER AND THE MOLEInterconverting Masses, Moles and Numbers of Particles• Frequently encountered in calculations using the moles concept. Moles provide a bridge from the molecular scale to the real-world scale.
  • 18. AVOGADRO’S NUMBER AND THE MOLE• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.
  • 19. AVOGADRO’S NUMBER AND THE MOLEEXAMPLE1)Calculate the quantity in moles of carbon dioxide whichcontains 1.505 X 1022 molecules.2)What is number of atoms in 0.6 mole of Li.EXAMPLE1)Calculate the number of sodium ions and oxide ions in 0.4 moleof sodium oxide.
  • 20. AVOGADRO’S NUMBER AND THE MOLEEXERCISEFor a 250g sample of SO2, determinea)The number of moles of SO2 molecules presentb)The number of moles of the sulfur atoms and oxygen atomspresentc)The number of SO2 molecules presentd)The number of sulfur and oxygen atoms presente)The average mass of one SO2 molecule.
  • 21. EMPIRICAL FORMULA FROM ANALYSES Determining Empirical Formula• Empirical formula : The relative number of atoms of eachelement it contains.EXAMPLE1)Determine the empirical formula of the compound formed from2.44 g of Ti and 3.60 g of Cl.2)Determine the empirical formula of an alcohol that has a %composition by mass of 60.0% C, 13.4% H and 26.6% O. What is the name of the alcohol.
  • 22. EMPIRICAL FORMULA FROM ANALYSES Molecular Formula from Empirical Formula• Formula obtained from percentage composition: EmpiricalFormula• Can obtain the molecular formula from the empirical formula ifwe know the molecular weight.• The subscripts in the molecular formula of a substance are alwaysa whole-number multiple of the corresponding subscripts in itsempirical formula.• The multiple is found by comparing the formula weight of theempirical formula with the molecular weight.
  • 23. EMPIRICAL FORMULA FROM ANALYSESEXAMPLEAscorbic acid (vitamin C), found in citrus fruits, tomatoes andgreen vegetables is composed of 40.9% C, 4.6% H and 54.5% O.Determine:a) The empirical formula of ascorbic acidb) The molecular formula if the experimentally determinedmolar mass of an ascorbic acid molecule is 176.13 g/mole.
  • 24. EMPIRICAL FORMULA FROM ANALYSESEXERCISEElemental analysis of a sample of an ionic compound showed2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empiricalformula and name of the compound?
  • 25. EMPIRICAL FORMULA FROM ANALYSES Combustion Analysis• Empirical = Based on observation and experiment• The empirical formula of a compound is based on experimentsthat give the number of moles of each element in a sample.• Compound containing carbon and hydrogen burn, the C willconverted into CO2 while the H will converted into H2O.• Amounts of H2O and CO2 are determined by measuring theincreased volume of the absorber.
  • 26. EMPIRICAL FORMULA FROM ANALYSES• From mass, we can calculate the number of moles of eachelement.•3rd element’s mass : Can be determine by subtracting the massesof C and H from the compounds original mass.
  • 27. EMPIRICAL FORMULA FROM ANALYSESEXAMPLEAcetic acid contains only C,H and O. A 4.24 mg sample of aceticacid is completely burned. It gives 6.21 mg of carbon dioxideand 2.54 mg of water. What is the mass percentage of eachelement in acetic acid with 3 sig. fig?
  • 28. EMPIRICAL FORMULA FROM ANALYSES EXERCISE When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placedin a combustion chamber and burned, the following data areobtained: mass of CO2 absorber after combustion = 85.35 g mass of CO2 absorber before combustion = 83.85 g mass of H2O absorber after combustion = 37.96 g mass of H2O absorber before combustion = 37.55 g What is the molecular formula of vitamin C?
  • 29. QUANTITATIVE INFORMATION FROM BALANCED EQUATION• The coefficients in a chemical equation represent the relativenumbers of molecules in a reaction.• Allow us to convert this information to the masses of thesubstances.• The coefficients in a balanced chemical equation indicate bothrelative numbers of molecules (or formula units) in the reactionand the relative numbers of moles
  • 30. QUANTITATIVE INFORMATION FROM BALANCED EQUATION• The quantities: 2 mol H2, 1 mol O2 and 2 mol H2O :Stoichiometrically equivalent quantities.• The relationship can be represented as 2 mol H2 = 1 mol O2 = 2 mol H2O • = symbol means is stoichiometrically equivalent to. • Can be used to convert between quantities of reactants and products in a chemical reaction.
  • 31. QUANTITATIVE INFORMATION FROM BALANCED EQUATIONExample 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2OCalculate the mass of CO2 produced when 1.00 g of C4H10 isburned.
  • 32. LIMITING REACTANTS• Limiting reactant: The reactant that is entirely consumedwhen a reaction goes to completion.• Excess reactant: Reactant that is not completely consumed.
  • 33. LIMITING REACTANTS • You can make cookies until you run out of one of the ingredients. • Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.
  • 34. LIMITING REACTANTSExampleZinc metal reacts with hydrochloric acid by the followingreaction: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)If 0.30 mol Zn is added to HCl containing 0.52 mol HCl, howmany moles of H2 are produced?
  • 35. LIMITING REACTANTSEXERCISE2 Na3PO4(aq) + 3Ba(NO3)2 (aq)  Ba3(PO4)2 (s) + 6NaNO3 (aq)Suppose that a solution containing 3.50 grams of Na3PO4 is mixedwith a solution containing 6.40 grams of Ba(NO3)2. How manygrams of Ba3(PO4)2 can be formed?
  • 36. LIMITING REACTANTS Theoretical Yields• The theoretical yield is the maximum amount of product that can be made. – In other words it’s the amount of product possible as calculated through the stoichiometry problem.• This is different from the actual yield.• The actual yield is never excess than the theoretical yield.
  • 37. LIMITING REACTANTSOne finds the percent yield by comparing the amountactually obtained (actual yield) to the amount it waspossible to make (theoretical yield). Actual YieldPercent Yield = Theoretical Yield x 100%
  • 38. LIMITING REACTANTSEXERCISEAdipic acid, H2C6H8O4, is used to produce nylon. The acid is madecommercially by a controlled reaction between cyclohexane(C6H12) and O2: 2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(g)(a) Assume that you carry out this reaction starting with 25.0 g ofcyclohexane and that cyclohexane is the limiting reactant. Whatis the theoretical yield of adipic acid?(b) If you obtain 33.5 g of adipic acid from your reaction, what isthe percent yield of adipic acid?

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