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STOICHIOMETRY Chemical Calculations

STOICHIOMETRY Chemical Calculations. Faculty of Biotechnology General Chemistry Lecture 4. Dr. M. Abd-Elhakeem. 1- Molecular weight 2- Mole 3- Avogadro’s number 4- Equivalent weight 5- Chemical equation stoichiometry 6- Gas’s laws 7-Element percent 8- Oxidation number

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STOICHIOMETRY Chemical Calculations

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  1. STOICHIOMETRYChemical Calculations Faculty of Biotechnology General Chemistry Lecture 4 Dr. M. Abd-Elhakeem

  2. 1- Molecular weight 2- Mole 3- Avogadro’s number 4- Equivalent weight 5- Chemical equation stoichiometry 6- Gas’s laws 7-Element percent 8- Oxidation number 9- Solution stoichiometry ( Concentration – Titration)

  3. Atomic weight: The atomic weight of an atom is equal to the number of protons in the nucleus of the atom plus the number of neutrons in the nucleus of the atom (one amu each). Therefore, a carbon atom with six protons and six neutrons has an atomic weight of 12. Molecular weight: The weight of a molecule is the sum of the weights of the atoms of which it is made.

  4. Calculate the molecular weight of each of the following Na2CO3,HCl , NaOH Na2CO3 = (23 X 2) + 12 + (16 X 3) = 106 g Mole: Unit of amount of chemical substance One mole is the molecular weight of a substance in gram. It equals the summation of atomic weight of the forming atoms.

  5. Avogadro's number Avogadro's number, also known as Avogadro's constant, is defined as the quantity of atoms in precisely 12 grams of 12C. The designation is a recognition of Amedeo Avogadro, who was the first to state that a gas' volume is proportional to how many atoms it has. Avogadro's number is given as 6.02214179 x 1023 mol-1.

  6. How many moles in 56 g Na2CO3 Number of moles: weight in grams/ molecular weight Then = 56/106 = 0.528 mole F2 Mole contain …….fluorine atoms = 2 X Avogadro’s number

  7. Keep in your mind Mole -------Wight--------- number of atoms

  8. Chemical equation • is the symbolic representation of a chemical reaction where the reactants are given on the left hand side and the products on the right hand side.

  9. Balancing chemical equations NaOH + HCl NaCl + H2O Na2CO3 + HCl 2 NaCl + H2O + CO2 2

  10. Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH)2 2H2O + BaCl2 1 1 coefficients give MOLAR RATIOS 2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

  11. How many moles of water are produced for each mole of P4O10 (s) when this equation is balanced? _ PH3 (g) + O2 (g) --> P4O10 (s) + H2O (l)

  12. Mole – Mole Conversions When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) 4.3 mol ? mol 2 mol N2O5 4 mol NO2 = moles NO2 8.6 4.3 mol N2O5 ?? mol NO2 b. How many moles of O2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) = mole O2 2.2 4.3 mol ? mol

  13. gram ↔ mole and gram ↔ gram conversions When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of N2O5 were used if 210g of NO2 were produced? 2N2O5(g) 4NO2(g) + O2(g) ? moles 210g (46 x4)g NO2 2 moles N2O5 = moles N2O5 2.28 ?? moles N2O5 210 g NO2 b. How many grams of N2O5 are needed to produce 75.0 grams of O2? 2N2O5(g) 4NO2(g) + O2(g) = grams N2O5 506 75.0 g ? grams

  14. Concentration Extensive quantities such as mass and volume are often used when preparing solutions. We can talk about adding a teaspoon of sugar when we make ice tea, for example, or filling a 2-L pitcher with ice tea. But when we want to compare solutions, we need intensive quantities that tell us how much sugar has been added to a given volume of lemonade or ice tea. We need to know the concentration of the solution, which is the ratio of the amount of solute to the amount of either solvent or solution.

  15. Percent Concentration Will be Studied in practical part It is w/w concentration (g / g) w/vconcentration (g / ml) v/v concentration (ml / ml)

  16. Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) M = number of moles / Volume (L) What is the number of moles in 50 ml of 6 M HCl

  17. Normality Normality is a term used to express concentration. The units of normality are number of equivalents per liter (It is abbreviated as a capital N) N = Number of equivalent / Volume (L)

  18. Equivalent weight • equal the molecular weight / e • No of H in acids • No. of OH in base • No. of acid molecules that react with basic salt, and No. of base molecules that react with acidic salt.

  19. Calculate the equivalent weight of each of the following HCl, H2SO4, CaOH, Li2CO3 HCl = molecular wt/ 1 = (1+35.5)/1 = 36.5

  20. You must learn how to prepare 1- percent solution 2- Molar solution 3- Normal solution 4- Transform between them

  21. 1- w/w % to Molarity or Normality M = 10pd/M. wt. (Eq. wt.) 2- w/v % to Molarity or Normality M = 10 p/M. wt. (Eq. wt.) 3- Molarity to Normality M = N/e

  22. Dilution To dilute any concentrated solution use the relation C X V (before dilution) = C X V (after dilution) What is the needed volume of concentrated HCl (11 M) to prepare 500 ml 1 M solution

  23. Solution Stoichiometry 50.0 ml of 6.0 M H2SO4 were spilled and solid NaHCO3 is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g) 50.0 ml ? g Our Goal 6.0 M = Look! A conversion factor!

  24. Solution Stoichiometry H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g) 50.0 ml ? g Our Goal 6.0 M = NaHCO3 2 mol NaHCO3 84.0 g 50.4 = g NaHCO3 mol NaHCO3 H2SO4 50.0 ml 1 mol H2SO4

  25. Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 First write a balanced Equation.

  26. Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. N X V (acid) = N X V (base) N of NaOH = M X e = 0.102 X 1 N of H2SO4 = M X e = 0.125 X 2 0.102 X V = 0.250 X 35 V = 85.78 ml

  27. Solution Stoichiometry What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2? 1st write out a balanced chemical equation

  28. Solution Stoichiometry 2HCl(aq) + Ba(OH)2(aq)  2H2O(l) + BaCl2 47.1 mL 0.75 M 0.40 M ? mL N X V (acid) = N X V (base)

  29. Solution Stochiometry Problem: 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. We must first write a balanced equation.

  30. Solution Stochiometry Problem: 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. Ca(OH)2(aq) + HNO3(aq)  2 H2O(l) 2 + Ca(NO3)2(aq) 48.0 mL 19.2 mL ? M 0.385 M

  31. Equivalent weight NaOH + HCl NaCl + H2O Na2CO3 + HCl 2 NaCl + H2O + CO2 2

  32. Theoretical yield vs.Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

  33. Try this problem (then check your answer): Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 ml of solution.

  34. Element’s percent • If the bacterial medium additives listed below are priced according to their nitrogen content, which will be the least expensive per 50 kg. bag? NH4Cl, (NH4)2SO4, NH2CONH2 Percent of element in its compounds: Simply to calculate the percent of any element in a compound Atomic weight of the element X number of element X 100 Molecular weight of the compound

  35. Then to calculate the Nitrogen percent in NH4Cl 1- M. wt of salt = 14 + 4+ 35.5 = 53.5 g 2- = 14 X 1X100 = 26.1% 53.5

  36. Gas’s Law • Boyle's Law: for a fixed amount of gas at a constant temperature, the volume of the gas varies inversely with its pressure PV = constant P1 V1 = P2 V2

  37. Charles's Law: the volume of a fixed amount of a gas at constant pressure is directly proportional to its temperature (Kelvin) V/T = constant V1/ T1= V2/ T2

  38. Avogadro's Law: At fixed temperature and pressure, the volume of a gas is directly proportional to the amount of the gas (either No. of moles or No. of molecules) V/n = constant V1/ n1= V2/ n2 One mole of an ideal gas at STP occupies 22.4 liters.

  39. The combined gas law( Ideal gas) • PV/nT = constant = R • P1V1/N1T1 = P2V2/N2T2 • R = 0.082058 L atm /mol K • = 62.364 L Torr/ mol K • = 8.3145 j/mol K

  40. Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O • How many grams of Na2CO3 are needed to neutralize 0.5 mole H2SO4 • How many moles of H2O are produced when we start with 200 g Na2CO3 • How many liters of CO2 are produced when we start with 250 ml of 2N H2SO4 (at STP). • How many moles of Na2CO3 are consumed when we need to produce 200 L of CO2 at ( 20oC and 1 atm). • What is the volume of 6 M H2SO4 was needed to neutralize 106 g Na2CO3

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