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Mullis1 Relationship between mass, moles and molecules in a compound Mass (g) Amount (moles) # molecules or Formula units X molar mass (__g__ mole) X 6.022.

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Presentation on theme: "Mullis1 Relationship between mass, moles and molecules in a compound Mass (g) Amount (moles) # molecules or Formula units X molar mass (__g__ mole) X 6.022."— Presentation transcript:

1 Mullis1 Relationship between mass, moles and molecules in a compound Mass (g) Amount (moles) # molecules or Formula units X molar mass (__g__ mole) X x (units mole) Grams moles = moles gram moles units = units or molecules mole moles grams = grams mole

2 Mullis2 Molar mass Molar mass of a substance = mass in grams of one mole of the substance. A compound’s molar mass is NUMERICALLY equal to its formula mass. 2 mol H x 1.01 g H = 2.02 g H 1 mol H 1 mol O x g O = g O 1 mol Omolar mass H 2 O = g/mol Formula mass H 2 O = amu Molar mass H 2 O = g/mol

3 Mullis3 Molar Mass Example What is the molar mass of K 2 SO 4 ? 2 mol K x g K = g K 1 mol K 1 mol S x g S = g S 1 mol S 4 mol O x g O = g O 1 mol O molar mass K 2 SO 4 = g/mol How many moles of each element are present in this compound? 2 mol K, 1 mol S, 4 mol O

4 Mullis4 What is the molar mass of C 6 H 12 O 6 ? 6 mol C x g C = g C 1 mol C 12 mol H x 1.01 g H = g H 1 mol H 6 mol O x g O = g O 1 mol O molar mass C 6 H 12 O 6 = g/mol How many moles of each element are present in this compound? 6 mol C, 12 mol H, 6 mol O

5 Mullis5 Converting to grams from moles How many moles of glucose are in 4.15x10 -3 g C 6 H 12 O 6 ? 4.15x10 -3 g x 1 mol C 6 H 12 O 6 = 2.30 x mol C 6 H 12 O g How many molecules of glucose are in 4.15x10 -3 g C 6 H 12 O 6 ? 2.30 x mol C 6 H 12 O 6 x x molecules = 1 mol (2.30 x 6.022)( 10 (-5+23) ) = x 10 –18 molecules = 1.39 x 10 –19 molecules

6 Mullis6 What is the mass in grams of 6.25 moles copper (II) nitrate? Cu 2+ NO 3 - : formula is Cu(NO 3 ) 2 Find molar mass of Cu(NO 3 ) 2 first. 1 mol Cu x g Cu = g Cu 1 mol Cu 2 mol N x g N = g N 1 mol N 6 mol O x g O = g O 1 mol O molar mass Cu(NO 3 ) 2 = g/mol Now find mass in grams of 6.25 moles: 6.25 moles x g = 1172 gAns g Cu(NO 3 ) 2 1 mol

7 Mullis7 Atoms and Ions Within Compounds How many carbon atoms are in one mole of C 2 H 6 ? 1 mole C 2 H 6 | 2 moles C | x atoms = x atoms |1 mole C 2 H 6 | 1 mole C How many MOLES of carbon atoms are in one mole of C 2 H 6 ? 1 mole C 2 H 6 | 2 moles C = 2 moles C atoms |1 mole C 2 H 6 How many moles of hydroxide ions are in one mole of calcium hydroxide? How many moles of Ca 2+ ? 1 mole Ca(OH) 2 | 2 moles OH - - =2 moles hydroxide ions |1 mole Ca(OH) 2 1 mole Ca(OH) 2 | 1 mole Ca 2+ =1 mole calcium ions |1 mole Ca(OH) 2

8 Mullis8 Percentage Composition % Composition is the % by mass of each element in a compound. Find the percentage composition of sodium chloride. Na + Cl - : formula is NaCl 1 mol Na x g Na = g Na 1 mol Na 1 mol Cl x g Cl = g Cl 1 mol Cl molar mass NaCl = g/mol g Na x 100 = % Na g NaCl g Cl x 100 = % Cl g NaCl

9 Mullis9 Find the percentage composition of sodium nitrate. Na + NO 3 - : formula is NaNO 3 1 mol Na x g Na = g Na 1 mol Na 1 mol N x g N = g N 1 mol N 3 mol O x g O = g O 1 mol O molar mass NaNO 3 = g/mol g Na x 100 = % Na g NaNO g N x 100 = % N g NaNO g O x 100 = 56.47% O g NaNO 3

10 Mullis10 Percentage Composition Why mass instead of moles? Isn’t 2/3 of the water molecule hydrogen? Moles indicate the amounts of each atom needed to make the molecule stable from an electron standpoint g H x 100 = % H g H 2 O HH O

11 Mullis11 Empirical Formula Use % composition to convert to original formula: 1.Assume 100 g sample, so % = grams 2.Convert grams to moles for each element 3.Divide the number of moles for each element by the smallest number of moles 4.The result for each type of element is its subscript in the empirical formula. 5.The order of elements is usually: Organics: C,H,O,N Inorganics: Metal, nonmetal, oxygen 6.Keep 4 decimal places when dividing numbers. If the result has a decimal between.2 and.8, may need to multiply all numbers by the number needed to get a whole number. Ex: 3.5 should be multiplied by 2 to get 4. Then multiply all other elements by the same number.

12 Mullis12 Example: Empirical and Molecular formula What is the empirical formula of a compound with 54.82% C, 5.624% H, 32.45% O, 7.104% N? g C| 1 mole C = mole C /.5074 = 9 C | 12 g C g H| 1 mole H =5.624 mole H /.5074 = 11 H | 1 g HC 9 H 11 O 4 N g O| 1 mole O = mole O /.5074 = 4 O | 16 g O g N| 1 mole N = mole N /.5074 = 1 N | 14 g N If a compound has this same composition but its molecular weight is 394 g/mol, what is its molecular formula? MW. C 9 H 11 O 4 N = 197 g/mol 394/197 = 2 so molecular formula is C 18 H 22 O 8 N 2

13 Mullis13 Oxidation Numbers Used to indicate the general distribution of electrons among the bonded atoms in molecular compounds or polyatomic ions. Analogous to charges in ionic compounds. An oxidation number is assigned to each element. Assign the ones you know 1 st. Find the others based on the numbers it takes to make the charge equal to the charge of the ion or compound. (A compound has a charge of zero.)

14 Mullis14 Oxidation Numbers: Rules 1.Pure element = 0 2.F = -1 3.O = -2 (except in peroxides and bonds with halogens) 4.H = +1 (except in bonds with metals) 5.The more electronegative element = same (-) charge as its anion 6.The less electonegative element = same (+) charge as its cation 7.The sum of a compound or polyatomic ion’s oxidation numbers is equal to its charge.


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