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Stoichiometry: Calculations with Chemical Formulas and Equation

Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equation. Chemical Equations. ethanol + oxygen acetic acid + water C 2 H 5 OH + O 2 CH 3 COOH + H 2 O. Or on two steps as.

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Stoichiometry: Calculations with Chemical Formulas and Equation

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  1. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equation

  2. Chemical Equations ethanol + oxygen acetic acid + water C2H5OH + O2 CH3COOH + H2O Or on two steps as 2C2H5OH + O2 2CH3COH + 2H2O 2CH3COH + O2 2CH3COOH 2C2H5OH + 2O2 2CH3COOH + 2H2O Knowing the balanced equation ? grams 10 grams or 10 grams ? grams What Do We Need to Know?

  3. In the grocery store Chemical reactions atoms and molecules package of atoms or molecules (moles) atoms and molecules weight in grams How? oranges and lemon count the numbers atoms and molecules count the numbers Eggs package of eggs (a dozen of eggs) Peanuts and candy weight by grams

  4. ethanol + oxygen acetic acid + water C2H5OH + O2 CH3COOH + H2O Information from Balanced Chemical Equations • Molecular Formula • Or • Formula Unit • Relative number of atoms • and molecules in the reaction

  5. Molecular weight (MW) • MW (substance) = • MW (H2O) = 2(1.0 amu) + 1(16.0 amu) = 18.0 amu • Formula weight (FW) • FW (substance) = • FW (NaCl) = 1(22.99 amu) + 1(35.45 amu) = 58.44 amu 1. Molecular Weight and Formula Weight

  6. Examples Calculate the formula weight of each of the following substances to three significant figures: CH3Cl and Fe2(SO4)3. • FW (CH3Cl) = 1(12.0 amu) + 3(1.0 amu) + 1 (35.45 amu) • = 50.45 amu = 50.4 amu • FW [Fe2(SO4)3] = 2(55.8 amu) + 3(32.1 amu) • + 12(16.0 amu ) • = 399.9 amu = 4.00  102 amu

  7. Problems How can we proceed? In the laboratory • We cannot calculate absolute number of atoms • or molecules in any sample in the lab. Why? 10.0 g of ethanol contains 1.311023 molecules!! It would take 4.15 billion years to count that number With a device that counts million molecules/second!!! • We cannot weigh atoms and molecules in amu unit • or in grams. Why? Mass of an atom  10-23 grams  10-26 Kg

  8. The Mole Concept A mole is defined as the number of atoms in exactly 12-g sample of C-12 The number of atoms in exactly 12-g sample of C-12 equal 6.02213671023. Avogadro’s number (NA) = 6.02213671023 = 6.02 1023 (3 sig. Fig.) 1 mole of any substance contains NA units

  9. Molar Mass(g/mol) The mass of one mole of a substance in grams Mole and Molar Mass 1 mole of any substance contains NA units 1 mole of atoms contains NA atoms 1 mole of molecules contains NA molecules 1 mole of ions contains NA ions 1 mole of $ contains NA $

  10. Formula Weight and Molar Mass Formula Unit Formula Weight amu Molar Mass g. mol-1 H O 16.0 1.0 16.0 1.0 H2O 18.0 18.0 CH3Cl NaCl 58.44 119.4 58.44 119.4 For all substances The molar mass in g/mol is numerically equal to the Formula weight in atomic units.

  11. number of Units (atoms, molecules, ions) nmoles NA Nunit / NA number of moles of atoms, molecules, and ions mass/molar mass nmoles molar mass masses of atoms, molecules, and ions in grams Importance of the Mole Unit

  12. Formula unit Molar mass g /mol number of molec. nmolec. number of moles n / mole Mass m / g H2O 18.0 1 ? mole ? g Mole Calculations 0.166 10-23 2.99  10-23 nmole = nmolec. (molec.) / NA (molec. mol-1) = 1 (molec.) / 6.02 1023 (molec. mol-1) = 0.166 10-23 mol m = n(mol)  M (g. mol-1) = 0.166 10-23 (mol)18.0 (g. mol-1) = 2.99  10-23 g

  13. Formula unit Molar mass g /mol number of molec. nmolec. number of moles n / mole Mass m / g H2O 18.0 ? 1 mole ? g Mole Calculations 6.02 1023 18.0 Nmolec. = n (mol)  NA (molec. mol-1) = 1 (mol)  6.02 1023 (molec. mol-1) = 6.02 1023 molecules m = n  M = 1 (mol)  18.0 (g. mol-1) m = nmole (mol)18.0 (g. mol-1) = 18.0 g

  14. Knowing MW AnBmCk MF Mole ratio: n: m: k we know mass ratio (Formula with simplest mole ratio) we know mole ratio we know empirical formula Determining Chemical Formulas If we know mass percent of each element

  15. AnBmCk MF Knowing the total mass 1. Mass Percentage and Mass Ratio Similarly we calculate mB and mC mass ratio is mA: mB: mC

  16. 2. Mole Ratio and Empirical Formula AnBmCk MF Knowing mA : mB : mC and Knowing MA, MB, MC We can calculate nA : nB : nC, where

  17. Determine the smallest number among nA, nB, nC Assuming that nA is the smallest number, then Divide each of nA, nB, nC by nA The empirical formula 3. The Empirical Formula Get the simplest ratio of nA: nB: nC How?

  18. Simplest mole ratio Element Mass M (g/mol) No. of moles 1 2 Na 17.5 23.0 2 1 2 Cr 39.7 52.0 3.5 (7/2) 7 O 42.8 16.0 An Example for Determining the Empirical Formula Empirical formula is Na2Cr2O7

  19. Elements n of element (mole) M of element (g/mol) Mass % C 1 12.0 H 2 1.0 O 1 16.0 Mass Percentage (Percentage Composition) from the Molecular Formula Calculate the percentage composition in formaldehyde (CH2O) always check that sum of percentage = 100

  20. (Please see the previous example) Calculating the Mass of an Element in a Given Mass of a Compound Calculate the mass of carbon in 83.5 g of formaldehyde (CH2O) In a given sample

  21. A Key Point If we have mass %, assume that the total mass is 100 g. Empirical Formula from % Composition An Example A substance give the following mass%: 17.5% Na, 39.7% Cr, 42.8% O What is the empirical formula?

  22. Empirical Formula Simplest mole ratio Element Mass M (g/mol) No. of moles 1 2 Na 17.5 23.0 2 1 2 Cr 39.7 52.0 3.5 (7/2) 7 O 42.8 16.0 Empirical formula is Na2Cr2O7

  23. Simplest mole ratio Element Mass M (g/mol) No. of moles 1 2 Na 17.5 23.0 2 1 2 Cr 39.7 52.0 3.5 (7/2) 7 O 42.8 16.0 An Example for Determining the Empirical Formula Empirical formula is Na2Cr2O7

  24. (From Formula Weights) In a given sample From masses in grams Calculating the Mass of an Element in a Given Mass of a Compound Calculate the mass of carbon in 83.5 g of formaldehyde (CH2O)

  25. Elements n of element (mole) M of element (g/mol) Mass % C 1 12.0 H 2 1.0 O 1 16.0 Mass Percentage (Percentage Composition) from the Molecular Formula Calculate the percentage composition in formaldehyde (CH2O) always check that sum of percentage = 100

  26. Empirical Formula from % Composition A substance give the following mass%: 17.5% Na, 39.7% Cr, 42.8% O What is the empirical formula? If we have mass %, assume that the total mass is 100 g. Then mNa= 17.5 g mCr = 39.7 g mO = 42.8 g

  27. Empirical Formula Simplest mole ratio Element Mass M (g/mol) No. of moles 1 2 Na 17.5 23.0 2 1 2 Cr 39.7 52.0 3.5 (7/2) 7 O 42.8 16.0 Empirical formula is Na2Cr2O7

  28. Elemental Analysis Example: combustion of acetic acid C, H, O CO2 + H2O 4.24 mg 6.21 mg 2.54 Applying the law of conservation of mass then: mC in reactants = mC in CO2 mH in reactants = mH in H2O mO in reactants = mO in H2O + mO in CO2 mO from air ? Calculating the Percentage Masses of Elements We want to calculate the following: % C, % H, % O

  29. C, H, O CO2 + H2O 4.24 mg 6.21 mg 2.54 sample mass MF mass ratio (in one mole) mass % MF Element No. Moles of element in MF Mass ratio in one mole CO2 C 1 H2O H 2 1. Mass Ratio in one Mole

  30. 2. Mass % sample mass mass % mass ratio (in one mole mass of C in CO2 = mass % C in sample = mass of H in H2O = mass % H in sample = mass % O in sample = 100 – (39.9+6.72) = 54.4%

  31. Simplest mole ratio Element Mass M (g/mol) No. of moles C 39.9 12.0 1 H 6.7 1.01 2 O 53.4 16.0 1 Empirical Formula Assume that the total mass is 100g Empirical formula is CH2O

  32. Molecular Formula from Empirical Formula In the previous example, If we know that the MW is 60.0 amu calculate the molecular formula The sample is acetic acid

  33. Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. • Stoichiometry is based on the following: • 1. The balanced chemical equation. • 2. The relationship between mass and moles. • Such calculations are fundamental • to most quantitative work in chemistry

  34. 3 molecules 2 molecules 1 molecule 1 mole 3 moles 2 moles Molar Interpretation of a Balanced Chemical Equation and the Law of Conservation of Mass

  35. 2 moles 3 moles 2 moles 4 moles 160 g 160 g An Example for Balancing Chemical Equations: Combustion of Methanol Balance the equation

  36. Calculation of Number of Moles of Products from Number of Moles of ReactantsMethod I • Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2. • Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple.

  37. 1 mole 3 moles 2 moles ? moles 6.4 mole Calculation of Number of Moles of Reactants from Number of Moles of ProductsMethod II

  38. 1 mole 3 moles 2 moles ? g 907 Kg Calculation of Masses of Reactants from Masses of Products

  39. Calculation of Masses of Products from Masses of Reactants 1 mole 3 moles 2 moles 5.0 g ? g

  40. 16 tires remain in excess limits the number of assembled cars Limiting Reactant (LR)

  41. Limiting Reactant (or Limiting Reagent) and the Theoretical Yield The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reactant ultimately determines how much product can be obtained. The theoretical yield of product is the maximum amount of product that can be obtained from The limiting reactant.

  42. An Example on the Limiting Reactant and the Theoretical Yield 1 mol 2 moles 1 mol 0.30 mol 0.30 mol 0.52 mol 0.26 mol The theoretical yield HCl The limiting reactant What is the LR? What is the theoretical yield of H2(g) 0.52 mol 0.30 mol Zn is the excess reactant

  43. Example 2: Combustion of Methanol Consider combustion of 10.0 ml of methanol in 20.0L of air. What it is the limiting reactant (LR)? Which reactant remains in excess? What is the theoretical yield of water? How much reactant remains in excess? The density of methanol is 0.791 g/ml, the density of O2 is 1.31 g/L. Answer: work it out on the board Download answer (PDF)

  44. The Experimental (Actual) Yield Is Always Greater Than The Theoretical Yield Why? Experiments are never perfect Consider the previous experiment on combustion of methanol. If the experiment yield of water is 1.95 grams, what is the percent yield of water?

  45. Operational Skills • Calculating the formula weight from a formula or model. • Calculating the mass of an atom or molecule. • Converting moles of substance to grams and vice versa. • Calculating the number of molecules in a given mass. • Calculating the percentage composition from the formula. • Calculating the mass of an element in a given mass of compound. • Calculating the percentages C and H by combustion. • Determining the empirical formula from percentage composition. • Determining the molecular formula from percentage composition and molecular weight. • Relating quantities in a chemical equation. • Calculating with a limiting reactant.

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