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CHEMICAL QUANTITIES: THE MOLE chem i / Ih : Chapter 10

CHEMICAL QUANTITIES: THE MOLE chem i / Ih : Chapter 10. MEASURING MASS. A mole is a quantity of things, just as… 1 dozen = 12 things 1 gross = 144 things 1 mole = 6.02 x 10 23 things “Things” usually measured in moles are atoms, molecules, ions, and formula units.

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CHEMICAL QUANTITIES: THE MOLE chem i / Ih : Chapter 10

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  1. CHEMICAL QUANTITIES: THE MOLEchemi/Ih: Chapter 10

  2. MEASURING MASS • A mole is a quantity of things, just as… 1 dozen = 12 things 1 gross = 144 things 1 mole = 6.02 x 1023 things • “Things” usually measured in moles are atoms, molecules, ions, and formula units

  3. You can measure mass, or volume, or you can count pieces • We measure mass in grams • We measure volume in liters • We count pieces in MOLES

  4. A MOLE… • is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon-12 • 1 mole = 6.02 x 1023of the representative particles • Treat it like a very large dozen 6.02 x 1023is called: Avogadro’s number

  5. Similar Words for an amount: • Pair: 1 pair of shoelaces = 2 shoelaces • Dozen: 1 dozen oranges = 12 oranges • Gross: 1 gross of pencils= 144 pencils • Ream: 1 ream of paper= 500 sheets of paper

  6. What are Representative Particles (“RP”)? • The smallest pieces of a substance: • For a molecular compound: it is the molecule. • For an ionic compound: it is the formula unit (made of ions) • For an element: it is the atom • Remember the 7 diatomic elements? (made of molecules)

  7. Practice Counting Particles • How many oxygen atoms in the following? • CaCO3 3 atoms of oxygen • Al2(SO4)3 12 (4 x 3) atoms of oxygen • How many ions in the following? • CaCl2 • 3 total ions (1 Ca2+ ion and 2 Cl1- ions) • NaOH • 2 total ions (1 Na1+ ion and 1 OH1- ion) • Al2(SO4)3 • 5 total ions (2 Al3+ + 3 SO4 ions)

  8. CONVERSION FACTOR • RPs x ____1 mole___ = MOLES 6.02 x 1023 RPs

  9. EXAMPLES: ATOMS  MOLES • How many moles of B are in 3.15 x 1023 atoms of B? • Conversion: 1 mole B = 6.02 x 1023 atoms B (b/c the atom is the RP of boron) 1 mole B 6.02 x 1023 atoms B = 3.15 x 1023 atoms of B 0.523 mole

  10. EXAMPLES: MOLES  ATOMS • How many atoms of Al are in 1.5 mol of Al? • Conversion: 1 mole = 6.02 x 1023 atoms 1.5 mol of Al 6.02 x 1023 atoms Al 1 mole Al 9.03 x 1023 atoms of Al =

  11. CAUTION: Identify RPs Carefully! • See next slide!

  12. EXAMPLES: MOLECULES  MOLES How many atoms of H are there in 3 moles of H2O? (HINT: Are atoms the RP for water?) Conversions: 1 mole = 6.02 x 1023 molecules (b/c molecules are the RP for H2O) H2O molecule = 2 atoms of Hydrogen 6.02 x 1023molecH2O 1 mole H2O 2 atoms H 1 H2O molecule 3 moles of H2O = 3.612 x 1024 atoms H

  13. MOLAR MASS Def: The mass of a mole of representative particles of a substance. Each element & compound has a molar mass.

  14. 20 Ca 40.08 MOLAR MASS OF AN ELEMENT Determined simply by looking at the periodic table Molar mass (g) = Atomic Mass (amu) * Thus, 1 mol Ca = 40 g 1 atom of Ca weighs 40.08 amu 1 mole of Ca atoms weighs 40.08grams

  15. MOLAR MASS FOR COMPOUNDS • To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound • Then add the masses within the compound Example: H2O H= 1.012 (1.01) + 1 (15.999)= 18.02 g/mol O= 15.999

  16. SOME PRACTICE PROBLEMS • How many atoms of O are in 3.7 mol of O? • 2.2 X 1024 atoms of oxygen • How many atoms of P are in 2.3 mol of P? • 1.4 x 1024 atoms of phosphorus • How many atoms of Ca are there in 2.5 moles of CaCl2? • 1.5 x 1024 atoms Ca • How many atoms of O are there in 1.7 moles of SO4? • 4.1 x 1024 atoms of oxygen

  17. Remember!!!! • The molar mass of any substance (in grams) equals 1 mole • This applies to ALL substance: elements, molecular compounds, ionic compounds • Use molar mass to convert between mass and moles • Ex: Mass, in grams, of 6 mol of MgCl2 ? mass of MgCl2 = 6 mol MgCl2 92.21 g MgCl2 1 mol MgCl2 = 571.26 g MgCl2

  18. VOLUME AND THE MOLE • Volume varies with changes in temperature & pressure • Gases are predictable, under the same physical conditions • Avogadro’s hypothesis helps explain: equal volume of gases, at the same temp and pressure contains equal number of particles • Ex: helium balloon

  19. MOLAR VOLUME • Gases vary at different temperatures, makes it hard to measure • To make it easier to measure gases, we use something called “STP” • “Standard Temperature and Pressure” • Temperature = 0° C • Pressure = 1 atm (atmosphere)

  20. Molar Volume • At STP:1 mole, 6.02 x 1023 atoms, of any gas has a volume of 22.4 L 1 mole gas = 22.4 L gas • Use it to convert between # of moles and vol of a gas @ STP • Ex: what is the volume of 1.25 mol of sulfur gas? 1.25 mol S 22.4 L S = 28.0 L 1 mol S

  21. MOLAR MASS FROM DENSITY • Different gases have different densities • Density of a gas measured in g/L @ a specific temperature • Can use the following formula to solve : grams = grams X 22.4 L mole L 1 mole • Ex: Density of gaseous compound containing oxygen and carbon is 1.964 g/ L, what is the molar mass? • grams = 1.964 g X 22.4 L then you solve mole 1 L 1 mole = 44.o g/mol

  22. Atoms, molecules, etc.

  23. Molarity • Def: the concentration of a solution. How many moles/liter • Can be used to calculate # of moles of a solute • Ex: Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO?

  24. Calculating Percent Composition of a Compound • Like all percent problems: a part ÷ the whole • Find the mass of each of the components (the elements) • Next, divide by the total mass of the compound • Then X 100 % = percent Formula: % Composition = Mass of element X 100% Mass of compound

  25. Method #1: % Comp When Actual Masses are Given A compound is formed when 9.03 g of Mg combines completely with 3.48 g of N. What is the percent composition of the compound? • First add the 2 mass of the 2 compounds to reach the total mass 9.03 g Mg + 3.48 g N = 12.51 g Mg3N2 2. Find the % of each compound % Mg= 9.03 g Mg X 100 = 72.2 % 12.51 g Mg3N2 % N= 3.48 g N X 100 = 27.8 % 12.51 g Mg3N2

  26. Method 2: % Comp When Only The Formula is Known • Can find the percent composition of a compound using just the molar mass of the compound and the element • % mass=mass of the element 1 mol cmpd X100% molar mass of the compound • Example: Find the percent of C in CO2 12.01 g C X 100 = 27.30% C 44.01 g CO2

  27. Using % Composition • Can use % composition as a conversion factor just like the mole • After finding the % comp. of each element in a cmpd. can assume the total compound = 100g • Example: C= 27.3% 27.3 g C O= 72.7 % 72.7 g O • In 100 g sample of compound there is 27.3 g of C & 72.7 g of O How much C would be contained in 73 g of CO2? 73 g CO2 27.3 g C = 19.93 g C 100 g CO2

  28. EMPIRICAL FORMULAS • Empirical formulas are the lowest WHOLE number ratios of elements contained in a compound

  29. REMEMBER… • Molecular formulas tells the actual number of of each kind of atom present in a molecule of the compound • Ex: H2O2 HO Molecular Empirical Formula Formula CO2 CO2 Molecular Empirical For CO2 they are the same Formula Formula

  30. Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = can not be reduced) • Examples: NaCl MgCl2 Al2(SO4)3 K2CO3 Simplest whole number ratio for NaCl

  31. A formula is not just the ratio of atoms, it is also the ratio of moles • In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen • In one molecule of CO2 there is 1 atom of C and 2 atoms of O • Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio) • Molecular: H2O C6H12O6 C12H22O11 (Correct formula) • Empirical: (Lowest whole H2O CH2O C12H22O11 number ratio)

  32. STOP HERE!!!

  33. CALCULATING EMPIRICAL • We can get a ratio from the percent composition • Assume you have a 100 g sample the percentage become grams (75.1% = 75.1 grams) • Convert grams to moles • Find lowest whole number ratio by dividing each number of moles by the smallest value

  34. Example calculations • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N • Assume 100 g sample, so 38.67 g C x 1 mol C = 12.0 g C 16.22 g H x 1 mol H = 1.0 g H 45.11 g N x 1 mol N = 14.0 g N *Now divide each value by the smallest value 3.22 mole C 16.22 mole H 3.22 mole N

  35. …Example 1 • The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N • The ratio is 16.22 mol H = 5 mol H 3.22 mol N 1 mol N C1H5N1 which is = CH5N

  36. MORE PRACTICE • A compound is 43.64 % P and 56.36 % O What is the empirical formula? PO3 • Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O What is its empirical formula? C4H5N2O

  37. EMPIRICAL TO MOLECULAR • Since the empirical formula is the lowest ratio, the actual molecule would weigh more • Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

  38. EXAMPLE • Caffeine has a molar mass of 194 g, what is its molecular formula? • Find the mass of the empirical formula, C4H5N2O • Divide the molar mass by the empirical mass: 194.0 g/mol = 97.1 g/mol • Now multiply the entire empirical formula by 2 2(C4H5N2O) = final molecular formula 2 C8H10N4O2

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