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Chapter 3 Stoichiometry

Chapter 3 Stoichiometry. Chemical Stoichiometry. Stoichimetry from Greek “ measuring elements ”. That is “Calculation of quantities in chemical reactions” Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. 3.2 Atomic Masses.

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Chapter 3 Stoichiometry

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  1. Chapter 3 Stoichiometry

  2. Chemical Stoichiometry • Stoichimetry from Greek “measuring elements”. That is “Calculation of quantities in chemical reactions” • Stoichiometry- The study of quantities of materials consumed and produced in chemical reactions.

  3. 3.2 Atomic Masses • Atoms are so small, it is difficult to discuss how much they weigh in grams. • Atomic mass units, amu, were used • Mass of C-atom of C-12 was assigned a mass of exactly 12 amu • Thus, an atomic mass unit (amu) is one twelth (1/12) the mass of a carbon-12 atom. • The masses of all other atoms are given relative to this standard • The decimal numbers on the table are atomic masses in amu.

  4. Atomic Masses by mass spectrometer • Mass Spectrometer • Atoms are passed into a beam of high speed electrons. • Electrons are knocked out and +Ve ions formed • Applied electric field accelerates +Ve ions into a magnetic field.

  5. Atomic Masses by mass spectrometer • Mass Spectrometer • Fast atomic ions (current) bend near magnet • Deflection varies inversely with masses • Multiple isotopes differ in mass and so give multiple beam deflections. • The ratio of the masses of C-13 and C-12 found to be 1.0836129 • The mass of C-13 = mass of C-12 X 1.0836129 = 13.003355 amu • Masses of other atoms can be determined similarly Exact number by definition

  6. Average atomic mass (6.941) Atomic masses are not whole numbers

  7. Atomic masses are not whole numbers • The atomic masses are not whole numbers including Carbon, Why? • Because they are based on averages of atoms and of isotopes. • It could be possible to determine the average atomic mass from the mass of the isotopes and their relative abundance. • add up the percent as decimals times the masses of the isotopes.

  8. Average atomic mass = (%isotope X atomic mass)+(%isotope X atomic mass) 100 Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: (7.42 x 6.015) + (92.58 x 7.016) = ________ amu 100

  9. Atomic Weight of an Element from Isotopic Abundances • Naturally occurring chlorine is 75.78% 35Cl, which has an • atomic mass of 34.969 amu, and 24.22% 37Cl, which has an • atomic mass of 36.966 amu. Calculate the average atomic mass • (that is, the atomic weight) of chlorine. • The average atomic mass is found by multiplying the abundance • of each isotope by its atomic mass and summing these products.

  10. The Mole • The mole (mol) is a number equal to the number of carbon atoms in exactly 12.00 grams of 12C • Techniques such as mass spectrometry were used to count this number • The number was found as 6.02214X1023 • This number was known as “Avogadro’s number” • Thus, one mole of a substance contains of 6.022X1023 units of that substance • So, dozen of eggs is 12; a mole of eggs is Avogadro’s number of eggs.

  11. How the mole is used in chemical calculations? • 12 grams of 12C contain Avogadro’s number of atoms = 6.022X1023 C atoms • 12.01 g of natural C (12C, 13C, 14C) contains 6.022X1023 C atoms • Atomic mass of C atom =12.01 amu

  12. Express amu in grams

  13. Practice • Convert grams of atoms moles of atoms number of atoms • and visa versa

  14. 3.4 Molar mass • Molar mass is the mass of 1 mole of a substance. • Often called molecular weight. • To determine the molar mass of a compound: add up the molar masses of the elements (taking # moles of each element into consideration) that make it up.

  15. Find the molar mass of • CH4 • Mg3P2 • Ca(NO3)3 • Al2(Cr2O7)3 • CaSO4 · 2H2O Convert mass molar mass and visa versa Conversion between mass, molar mass and number of molecules

  16. n x molar mass of element x 100% molar mass of compound 3.5 Percent Composition of Compounds • It is the mass (weight) percent of each element a compound is composed of. • Find the mass of each element, divide by the total mass, multiply by a 100. • Use a mole of the compound for simplicity .

  17. 2 x (12.01 g) 6 x (1.008 g) 1 x (16.00 g) %C = %H = %O = x 100% = 52.14% x 100% = 13.13% x 100% = 34.73% 46.07 g 46.07 g 46.07 g Practice Find the percent composition of each element inC2H6O 52.14% + 13.13% + 34.73% = 100.0%

  18. Determination of percent composition and simplest formula from experiment

  19. Example

  20. 3.6 Determining the formula of a compound Empirical formula: the lowest whole number ratio of atoms in a compound. • molecular formula = (empirical formula)n • [n = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH Molecular formula: the true number of atoms of each element in the formula of a compound.

  21. Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  22. Formulas for molecular compoundsMIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11

  23. Experimental Determination of the formula of a compound by elemental analysis • A compound of unknown composition is decomposed by heat. The elements are carefully trapped and the number of moles of each are analyzed. • A sample of a compound composed of carbon oxygen and hydrogen are combusted in a stream of O2 to produce CO2 and H2O. The H2O and CO2 are trapped and the masses of each are measured.

  24. Experimental Determination of the formula of a compound Pure O2 in CO2 is absorbed Sample Sample is burned completely to form CO2 and H2O O2 and other gases H2O is absorbed

  25. Calculating empirical formula • The sample has a mass of 0.255g. When the reaction is complete, 0.561 g of CO2 and 0.306g of H2O are produced. What is the empirical formula of the compound? • Determine the mass of C in the sample. • For each mole of CO2, there is one mole of C. Convert moles of C to grams of C. • 0.561 g CO2 x (1 mol CO2 / 44.01 g CO2) x (1 mol C / 1 mol CO2) (12.01 g C/ mol C) • = 0.153 g C

  26. 2. Determine the mass of H in the sample. • There are 2 moles of hydrogen per mole of H2O. • 0.306 g H2O x (1 mol H2O / 18.0 g H2O) x • (2 mole H / mole H2O) x (1.01 g H / mol H) • = 0.0343 g H • Mass O = mass sample - mass H - mass C • Mass sample = 0.255 g • Mass O = 0.255 - 0.153 - 0.0343 = 0.068 g O

  27. To get empirical formula, convert g back to moles • 0.153 g C x ( 1 mol C / 12.01 g C ) = 0.0128 mol C • 0.0343 g H x (1 mol H / 1.01 g H) = 0.0340 mol H • 0.068 g O x (1 mol O / 16.0 g O) = 0.0043 mol O • Divide each by 0.0043 to get ratio of each element to O • C: 0.0128 mol C / 0.0043 mol O = 2.98 ~ 3 • There are 3 moles of carbon for each mole of oxygen • H: 0.0340 mol H / 0.0043 mol O = 7.91 ~ 8 • There are 8 moles of hydrogen per mole of oxygen • Empirical Formula C3H8O

  28. g CO2 g H2O mol CO2 mol H2O mol C mol H g C g H Burn 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O

  29. Calculating Empirical formula from the percent composition • We can get a ratio from the percent composition. • Assume you have a 100 g. • The percentages become grams. • Convert grams to moles. • Find lowest whole number ratio by dividing by the smallest value.

  30. Example • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Assume 100 g so • 38.67 g C x 1mol C = 3.220 mole C 12.01 gC • 16.22 g H x 1mol H = 16.09 mole H 1.01 gH • 45.11 g N x 1mol N = 3.219 mole N 14.01 gN Now divide each value by the smallest value

  31. The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N • The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol • = C1H5N1

  32. Determining molecular formula from empirical formula • Since the empirical formula is the lowest ratio, the actual molecule would weigh more. • By a whole number multiple. • Divide the actual molar mass by the empirical formula mass • you get a whole number to increase each coefficient in the empirical formula • Caffeine has a molar mass of 194 g. what is its molecular formula?

  33. molecular formula = (empirical formula)n n = integer] • molecular formula = C6H6 = (CH)

  34. 3.7 Chemical Equations Chemical change involves reorganization of the atoms in one or more substances. Chemical reactions occur when bonds between the outermost parts of atoms are formed or broken Chemical reactions involve changes in matter, the making of new materials with new properties, or energy changes. Atoms cannot be created or destroyed Chemical Reactions are described using a shorthand called a chemical equation

  35. A representation of a chemical reaction: C2H5OH + 3O2 2CO2 + 3H2O reactants products Chemical Equations

  36. The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules: • 2H2 + O2 2H2O Products Reactants

  37. Reading Chemical Equations • The plus sign (+) means “react” and the arrow points towards the substance produce in the reaction. • The chemical formulas on the right side of the equation are called reactants and after the arrow are called product. • The numbers in front of the formulas are called stoichiometric coefficients. • 2Na + 2H2O  2NaOH + H2 • Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Coefficient Reactants Products

  38. Understanding Chemical Equations Coefficients and subscripts included in the chemical formula have different effects on the composition.

  39. Balanced chemical equation • C2H5OH +3O22CO2+3H2O • The equation is balanced. • 1 mole of ethanol reacts to produce with3 moles of oxygen 2 moles of carbon dioxide and3 moles of water

  40. Physical states in chemical equations State Symbol Solid (s) Liquid (l) Gas (g) Dissolved in water (aq) (in aqueous solution) HCl(aq) + NaHCO3(l) CO2(g) + H2O(l) + NaCl(aq) Catalysts, photons (h), or heat () may stand above the reaction arrow ().

  41. 3.8Balancing chemical equations • When balancing a chemical reaction coefficients are added in front of the compounds to balance the reaction, but subscripts should not be changed • Changing the subscripts changes the compound

  42. 2C2H6 C4H12 NOT Balancing Chemical Equations By inspection (Trial and error) • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O • Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

  43. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Balancing Chemical Equations • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3

  44. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Balancing Chemical Equations • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2

  45. 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7

  46. Reactants Products 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7

  47. Stoichiometry • From Greek “measuring elements”. • That is “Calculation of quantities in chemical reactions” • Given an amount of either starting material or product, determining the other quantities. • use conversion factors from • molar mass (g - mole) • balanced equation (mole - mole)

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