Chemistry The Mole (5) Mrs. Day-Blattner 26/27 February 2018 Feb 26/27 Agenda Activities of analytical chemists Slip Quiz
Analyzing a hydrate - set up and Part 1 of lab procedure
Percent composition - uses of
Empirical Formulas and Molecular formulas
Analyzing a hydrate - Part 2 of lab procedure Slip Quiz
1. Write 4 sentences about the “gum” in gum based on the research you did for homework.
2. State your source. 11.5 The formula for a hydrate H2O Formula Name 1 (NH4)C2O4·H2O Ammonium oxalate monohydrate 2 CaCl2·2H2O Calcium chloride dihydrate 3 NaC2H2O2·3H2O Sodium acetate trihydrate 4 FePO4·4H2O 5 CuSO4·5H2O 6 CoCl2·6H2O 7 MgSO4·7H2O 8 Ba(OH)2·8H2O 10 Na2CO3·10H2O 11.5 The formula for a hydrate H2O Formula Name 1 (NH4)C2O4·H2O Ammonium oxalate monohydrate 2 CaCl2·2H2O Calcium chloride dihydrate 3 NaC2H2O2·3H2O Sodium acetate trihydrate 4 FePO4·4H2O Iron(III) phosphate tetrahydrate 5 CuSO4·5H2O Copper(II) sulfate pentahydrate 6 CoCl2·6H2O Cobalt(II) chloride hexahydrate 7 MgSO4·7H2O Magnesium sulfate heptahydrate 8 Ba(OH)2·8H2O Barium hydroxide octahydrate 10 Na2CO3·10H2O Sodium carbonate decahydrate How do you think chemists found the formulas for hydrates? How do you think chemists found the formulas for hydrates?
To analyze a hydrate we need to drive off the water of hydration (or water of crystallization) to make the crystals anhydrous (without water).
Ex. CoCl2·6H2O (s) → CoCl2 (s) + 6 H2O(g)
Pink blue Checking the formula for cobalt(II) chloride hexahydrate Data to collect and record: Part 1
Mass of empty crucible
Mass of hydrated cobalt(II) chloride in crucible
Mass of cobalt(II) chloride sample in crucible after heating Checking the formula for cobalt(II) chloride hexahydrate
Safety: Checking the formula for cobalt(II) chloride hexahydrate
Safety: All usual safety rules apply. You must wear goggles at all times and be aware of your clothing and arm movements because of the use of open flames today. Do not attempt to move the equipment after heating.
Turn off your bunsen burner and sit down while the equipment cools. Empirical and Molecular Formulas • Empirical formula Formula with the smallest whole number ratio of the elements. Empirical formula may be same as molecular formula like water H2O Or it may not be: hydrogen peroxide empirical
formula is OH and molecular formula H2O2 Calculating Empirical formula • From masses or percent compositions Compound is 40% S 60% O So if we can say that if we had 100g of this compound we would have 40g S and 60g O Calculating Empirical formula • From masses or percent compositions Compound is 40% S 60% O
Convert to moles (multiply by 1/molar mass) 40g S x 1 mole = 40g S x 1 mole mass 1 mole S 32g Calculating Empirical formula Convert to moles 40g S x 1 mole = 40g S x 1 mole = 1.25 mol S mass 1 mole S 32g Calculating Empirical formula 40g S and 60g O
Convert to moles 60g O x 1 mole = 60g O x 1 mole = 3.75 mol O mass 1 mole O 16g Calculating Empirical formula 40g S and 60g O
Now convert to simplest ratio: 1.25 mol S = 3.75 mol O = 1.25 1.25
Calculating Empirical formula 40g S and 60g O
Now convert to simplest ratio: 1.25 mol S = 1 mol S 3.75 mol O = 3 mol O 1.25 1.25
Simplest whole number mole ratio is of S to O is 1:3
Empirical formula SO3 Calculating Empirical formula 40g S and 60g O Now convert to simplest ratio: 1.25 mol S = 1 mol S 3.75 mol O = 3 mol O 1.25 1.25 Simplest whole number mole ratio is of S to O is 1:3
Empirical formula SO3 Calculating Empirical formula
In cases where the calculated mole values are not whole numbers, simply multiply by smallest factor that will make them whole numbers. (See Example Problem 11-11. Page 332) Calculating Empirical formula. Ex. 1 • From masses or percent compositions Compound is 36.84% Nitrogen 63.16 % O We can say that if we had 100g of this compound we would have
Calculating Empirical formula. Ex. 1 • From masses or percent compositions Compound is 36.84% Nitrogen 63.16 % O We can say that if we had 100g of this compound we would have 36.84g N and 63.16g O Ex. 1 Convert to moles (multiply by 1/molar mass) 36.84g N x 1 mole = 2.630 mol N 14.01gN
Convert to moles 63.16g O x Ex. 1 Convert to moles (multiply by 1/molar mass) 36.84g N x 1 mole = 2.630 mol N 14.01gN
Convert to moles 63.16g O x 1 mole == 3.950 mol O 15.99gO
Ex. 1 Now convert to simplest ratio: 2.630 mol N = 3.950 mol O = 2.630 2.630
Simplest whole number ratio
Empirical formula Ex. 1 Now convert to simplest ratio: 2.630 mol N = 1 N 3.950 mol O = 1.5 O 2.630 2.630
Simplest whole number ratio (x 2) is 2 : 3
Empirical formula N2O3 Ex. 2
Propane is a hydrocarbon, a compound composed only of carbon and hydrogen. It is 81.82% carbon (12.01g/mol) and 18.18% hydrogen (1.008 g/mol). What is the emprical formula? Ex. 2 Convert to moles (multiply by 1/molar mass) 81.82g C x 1 mole = 6.818 mol N 12.01gN
Convert to moles 18.18g H x 1 mole = 18.04 mol H 1.008gH
Ex. 2. Now convert to simplest ratio: 6.818 mol C = 1C 18.04 mol H = 2.646 6.818 6.818 2.646 x 3 = 7.9 approx. 8 Simplest whole number ratio x3 = 3: 8
Empirical formula C3H8 Calculating Empirical formula 1. Need to know masses of elements in the compound or percent compositions. 2. Convert percent composition of each element to g then convert to number of moles of that element. 3. Divide each number of moles by smallest value to find simplest ratio of the elements. 4. If not whole numbers, multiply by lowest possible factor to make it so. 5. Write out empirical formula. Calculating Molecular formula Start from calculated Empirical formula And Molar mass (found by experimentation).
Experimentally determined molar mass = n Mass of empirical formula
Then n(empirical formula) = molecular formula See Example Problem 11-12 Ex. 1 Calculating Molecular formula The empirical formula for acetylene and benzene is the same: CH. But acetylene is a gas used in welding and benzene is a liquid used as a solvent. They are obviously very different. The molar mass of the empirical formula is (12.01 + 1.008) = 13.02 g/mol The molar mass of acetylene is experimentally found to be 26.04 g/mol and the molar mass of benzene is 78.12 g/mol. Ex. 1 Calculating Molecular formula Acetylene Exp. determined molar mass = 26.04 g/mol = Mass of empirical formula 13.02 g/mol Ex. 1 Calculating Molecular formula Acetylene Exp. determined molar mass = 26.04 g/mol = 2 Mass of empirical formula 13.02 g/mol This shows that the molar mass of the acetylene is two times the mass represented by the empirical formula. Ex. 1 Calculating Molecular formula Acetylene Exp. determined molar mass = 26.04 g/mol = 2 Mass of empirical formula 13.02 g/mol
Acetylene, molecular formula C2H2 Structural formula: Ex. 1 Calculating Molecular formula Benzene Exp. determined molar mass = 78.12 g/mol = Mass of empirical formula 13.02 g/mol Ex. 1 Calculating Molecular formula Benzene Exp. determined molar mass = 78.12 g/mol = 6 Mass of empirical formula 13.02 g/mol
Molecular formula = 6 (CH)
Molecular formula benzene is C6H6 Benzene structural formulas Re weigh the cooled crucible
Data to collect and record: Part 2
Mass of empty crucible
Mass of cobalt(II) chloride hexahydrate in crucible
Mass of cobalt(II) chloride sample in crucible after heating = Analysis
1) Mass of hydrated cobalt(II) chloride Mass + crucible = g - Mass of crucible = g
= mass of hydrated sample = Analysis
2) Mass of hydrated cobalt(II) chloride Mass after heating + crucible = g - Mass of crucible = g
= mass of anhydrous sample = g Analysis
3) Mass of water lost Mass of hydrated sample = g - mass of anhydrous sample = g = mass of water lost = g Analysis
Convert to moles of each
Molar mass CoCl2
58.93 + 2(35.45) =
Mass g x 1mol = Molar mass Analysis
Convert to moles of each
Molar mass H2O
15.99 + 2(1.008) =
Mass g x 1mol = Molar mass Analysis
Moles of water = =
Moles of CoCl2
Experimental formula for Cobalt(II) chloride - hydrated form:
Next steps: