Formula Stoichiometry. Text pages


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1 Formula Stoichiometry Text pages
2 Formula Mass Review Write a chemical formula for the compound. H 2 CO 3 Look up the average atomic mass for each of the elements. H = C= O = Multiply each element by any subscripts. H = 2 x C= O = 3 x Add everything together. Molar Mass = g
3 Molar Mass as a Conversion Factor 1 mole = formula weight in grams 1 mole = 6.02 x atoms, molecules, formula units, or ions So we can use either of these definitions as a conversion factor.
4 Conversions  Moles to Grams What is the mass in grams of 3.0 mol of Cl 2? 3mol grams 1 mol of Cl 2 = x 2 = grams 3mol Cl 2 x 70.90g Cl 2 = g of Cl 2 1 mol Cl 2
5 Moles to Grams Practice mol of NH g of NH mol of glucose, C 6 H 12 O g of C 6 H 12 O mol of barium chloride, BaCl x 10 3 g BaCl mol of propane, C 3 H g C 3 H 8
6 Moles to Molecules mol C 6 H 14 to molecules of C 6 H 14 1 mol = x molecules mol C 6 H 14 x x molecules C 6 H 14 1 mol C 6 H 14 = 3.01 x 1022 molecules of C 6 H 14
7 Practice Moles to Molecules 4.99 mol CH x molecules CH mol N x molecules N mol PCl x molecules PCl x 105 mol C 6 H 8 O x molecules C 6 H 8 O 6
8 Atoms or Molecules Flowchart Divide by 6.02 X Multiply by 6.02 X Moles Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Mass (grams)
9 Percent Composition What percentage (by mass) is an element in a compound? Plan: find the molar mass of each element and of the compound Divide the molar mass of the element by the molar mass of the compound
10 Example of % Composition Na 2 CO 3 Na g/mol x 2 = g Na/ mol Na 2 CO 3 C g/mol x 1 = g C/ mol Na 2 CO 3 O g/mol x 3 = g O/ mol Na 2 CO 3 Na 2 CO g/mol Percent of Na per mol g Na x 100 = 43.38% Na g Na 2 CO 3
11 % Composition Practice Determine the percentage composition for each element in the compound. Na 2 C 2 O 4 C 2 H 5 OH Al 2 O 3 K 2 SO 4
12 Na 2 C 2 O 4 C 2 H 5 OH Al 2 O 3 K 2 SO 4 Na = 34.31% C = 52.13% Al = 52.92% K = 44.87% C = 17.93% H = 13.15% O = 47.08% S = 18.40% O =47.76% O = 34.72% O = 36.72%
13 Empirical Formulas If you can experimentally determine the percentage composition of a compound you can use that information to determine the smallest whole number ratio of the elements in the compound, the empirical formula. Subscripts in a formula tell you the mole ratios of the elements.
14 Empirical Formula if you are given percentages Example 1. Determine the empirical formula of an unknown compound composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. Find the molar mass of each element For ease of calculation imagine that you are dealing with a sample of 100 g of the substance. Then you may simply change the percentage to grams
15 Example 1. Determine the empirical formula of an unknown compound composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. Find mol for each element Given Conversion factor 36.5g Na x 1 mol Na = 1.59 mol Na g unknown 22.99g Na g unknown 38.1g O x 1 mol O = 2.38 mol O g unknown 16.00g O g unknown 25.4g S x 1 mol S = mol S g unknown 32.07g S g unknown
16 Example 1. Determine the empirical formula of an unknown compound composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. Next find the ratio of moles of each substance to the compound with the least number of mol 1.59 mol Na x 100.0g unknown = 2.01 mol Na g unknown mol sulfur 1 mol S 2.38 mol O x g unknown = 3.01 mol O g unknown mol S 1 mol S mol S x 100.0g unknown = 1.00 mol S g unknown mol S 1 mol S Divide each result by the amount in moles of the least abundant element. So the simplest ratio is 2 mol Na: 3 mol O: 1 mol S These are your subscripts so Na 2 SO 3Does the answer make sense?
17 Empirical formula practice % Cu: 71.6% Br CuBr % K: 12.0% C: 1.01% H: 47.9% O KHCO % Ag: 7.4% P: 15.3% O Ag 3 PO % H: 72.1% I: 27.3% O HIO 3
18 Example 2: Find the empirical formula for an unknown compound with 38.4% K, 23.7% C, 1.66% H, and 36.3%O 36.5g K x 1 mol K = mol K g unknown g K g unknown 23.7g C x 1 mol C = 1.97 mol C g unknown g C g unknown 36.3g O x 1 mol O = 2.27 mol O g unknown 16.00g O g unknown 1.66 g H x 1 mol H = 1.64 mol H g unknown 1.01 g H g unknown
19 Example 2: Find the empirical formula for an unknown compound with 38.4% K, 23.7% C, 1.66% H, and 36.3% O Find the ratio of moles. Divide each result by the amount in moles of the least abundant element mol K x 100.0g unknown = 1.00 mol K g unknown.0982 mol K 1 mol S 1.97 mol C x g unknown = 2.01 mol C g unknown mol K 1 mol S 2.27 mol O x 100.0g unknown = 2.31 mol O g unknown mol K 1 mol S 1.64 mol H x 100.0g unknown = 1.67 mol H g unknown mol K 1 mol S
20 Example 2: Find the empirical formula for an unknown compound with 38.4% K, 23.7% C, 1.66% H, and 36.3% O What to do when there isn t a simple whole number ratio. 1.O0 mol K 1 mol K x 3 3 mol K 2.01 mol C 2 mol C x 3 6 mol C 2.31 mol O 2 1/3 mol O x 3 7 mol O 1.67 mol H 1 2/3 mol H x 3 5 mol H Amount in mol of element per mol K Fraction nearest the decimal value Integer factor K 3 C 6 H 5 O 7 potassium citrate Whole number ratio
21 Empirical Formula Practice a). 36.2% Al, 63.8% S Al 2 S 3 b) % Nb, 6.50% O Nb 5 O 2 c). 57.6% Sr, 13.8% P, 28.6%O Sr 3 P 2 O 8 or Sr 3 (PO 4 ) 2 d). 28.5% Fe, 48.6% O, 22.9% S Fe 2 S 3 O 12 or Fe 2 (SO 4 ) 3
22 Example 3. Given information in mass instead of a percentage. A g sample of contains g of Mn, g N, g of O, Find the empirical formula. For this type of problem you are given exact numbers of grams for each element, so start there. Convert the mass of each element to moles. Determine the mole ratio
23 Example 3. A g sample of contains g of Mn, g N, g of O, Find the empirical formula g Mn 1 mol Mn = mol g g N 1 mol N = mol g g O 1 mol O = mol g
24 Example 3. A g sample of contains g of Mn, g N, g of O, Find the empirical formula. Divide by the smallest number of moles mol Mn = 1.00 mol Mn mol mol N = 2.01 mol N mol MnN 2 O mol O = 6.03 mol O mol
25 Molecular Formula vs Empirical Empirical formula Simplest whole number ratio of elements Formula Molecular Formula The actual formula of a molecular compound. A multiple of the empirical formula You must know the mass of the molecule in order to calculate this (Text p )
26 Calculating Molecular Formulas 1. Find the empirical formula mass 2. Divide the molecular formula mass by the empirical formula mass 3. Multiply the empirical formula by the answer to step 2 Practice 1. Empirical formula = CH 2, molar mass = 28 g/mol 2. Empirical formula = B 2 H 5, molar mass = 54 g/mol 3. Empirical formula = C 2 HCl, molar mass = 179 g/mol 4. Empirical formula = C 6 H 8 O, molar mass = 290 g/mol 5. Empirical formula = C 3 H 2 O, molar mass = 216 g/mol C 2 H 4 B 4 H 10 C 6 H 3 Cl 3 C 18 H 24 O 3 C 12 H 8 O 4
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