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The Mole Ch.8

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(8-1) Mole (mol): amt. of substance – # of atoms in 12g of carbon-12 Avogadro’s constant: 6.02 x 10 23 particles / mol –Atoms, molecules (covalent), formula units (ionic)

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Moles # of Atoms 2.66 mol x 6.02 x 10 23 atoms = 1.60 x 10 24 1 molatoms # of Atoms Moles 2.54 x 10 24 atoms x 1 mol = 4.22 mol 6.02 x 10 23 atoms

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Moles to Molecules 1.04 mol x 6.02 x 10 23 molecules = 6.26 x 10 23 1 mol molecules Formula units to moles 3.49 x 10 24 form.u. x 1 mol = 5.80 mol 6.02 x 10 23 form.u.

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Moles to Atoms Practice How many atoms are in 2.5 mol of Si? 1.List known 2.5 mol Si 2.Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 2.5 mol Si x 6.02 x 10 23 atoms 1 mol Si

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Moles to Atoms Practice 3.Cancel units, multiply top #’s & divide by bottom 2.5 mol Si x 6.02 x 10 23 atoms = 1.5 x 10 24 atoms Si 1 mol Si

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Atoms to Moles Practice Convert 3.01 x 10 23 atoms of Si to mols Si 1.List known 3.01 x 10 23 atoms Si 2.Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 3.01 x 10 23 atoms Si x 1 mol Si = 6.02 x 10 23 atoms

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Atoms to Moles Practice 3.Cancel units, multiply top #’s & divide by bottom 3.01 x 10 23 atoms Si x 1 mol Si = 0.500 mol Si 6.02 x 10 23 atoms Si

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Relative Atomic Mass Weighted avg. of each isotope’s mass –Need mass & % abundance (62.94 amu)(.6917) + (64.93 amu)(.3083) = 63.55 amu IsotopeMass (amu) %Decimal Copper-6362.9469.17.6917 Copper-6564.9330.83.3083

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Molar Mass Sum of all the atomic masses of a substance Ex: CO 2 C = 12.01 g/mol, O = 16 g/mol 12.01 g/mol + (2)(16 g/mol) = 44.01 g/mol

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Molar Mass Practice Calculate the molar mass of Ba(NO 3 ) 2 1.List known (at. masses from PT) Ba = 137.33 g/mol, N = 14.01 g/mol, O = 16 g/mol 2.Calculate # of each atom in the formula (outer subscript multiplies w/ inner subscripts) 1 Ba 1 N x 2 = 2 N 3 O x 2 = 6 O

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Molar Mass Practice 3.Multiply each at. mass by the # of atoms & add together (1 Ba)(137.33 g/mol) + (2 N)(14.01 g/mol) + (6 O)(16 g/mol) =261.35 g/mol

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Moles Mass 2.49 mol KF x 58.1 g KF = 145 g KF 1 mol KF Grams Moles 110 g KF x 1 mol KF = 1.89 mol KF 58.1 g KF

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Moles to Mass Practice What’s the mass in grams of 3.50 mol Cu? 1.List known 3.50 mol Cu 2. Find at. mass or calculate molar mass 63.55 g Cu 1 mol Cu

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Moles to Mass Practice 3. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 3.50 mol Cu x 63.55 g Cu 1 mol Cu 4. Cancel units, multiply top #’s, & divide by bottom 3.50 mol Cu x 63.55 g Cu = 222 g Cu 1 mol Cu

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Mass to Moles Practice Determine the # of moles in 237 g of Cu 1.List known 237 g Cu 2. Find at. mass or calculate molar mass 63.55 g Cu 1 mol Cu

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Mass to Moles Practice 3. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 237 g Cu x 1 mol Cu = 63.55 g Cu 4. Cancel units, multiply top #’s, & divide by bottom 237 g Cu x 1 mol Cu = 3.73 mol Cu 63.55 g Cu

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Atoms to Grams 1.2 x 10 24 atoms B x 1 mol B x 10.81 g B = 21.55 g B 6.02 x 10 23 atoms 1 mol B

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(8-2) Percentage Composition % by mass of each element in a cmpd –Ex: H 2 O = 88.7% O, 11.3% H O H

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Determine % Comp. from Chemical Formula % = mass of component x 100 mass of whole To verify answers all components %, should = 100%

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% Comp. Practice Find the % Comp. of Cu 2 S 1.Find at. masses & multiply by # of atoms Cu = (63.55 g/mol) (2) = 127.10 g/mol S = (32.07 g/mol) (1) = 32.07 g/mol

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% Comp. Practice 2. Divide each mass by total mass of cmpd & multiply by 100 127.10 g/mol Cu x 100 = 79.85% Cu 159.17 g/mol Cu 2 S 32.07 g/mol S x 100 = 20.15% S 159.17 g/mol Cu 2 S 3.Add % to verify it’s close to 100% 79.85% + 20.15% = 100%

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Hydrates To determine % comp. of a hydrate, use same format as before % = mass water x 100 mass of whole

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Hydrate % Comp. Practice Determine the % water in Na 2 CO 3 10H 2 O 1.Calculate molar mass of water (coef. is distributed) (20)(1.01 g/mol) + (10)(16 g/mol) = 180.2 g/mol H 2 O 2.Divide water mass by the total hydrate mass 180.2 g/mol H 2 O x 100 = 62.96%H 2 O 286.2 g/mol Na 2 CO 3 10H 2 O

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Empirical Formula Empirical Formula: simplest ratio among the elements of a cmpd –Ex: CH 2 O: CH 2 O, C 2 H 4 O 2, C 6 H 12 O 6

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Determining Emp. Formulas 63.0% Mn, 37.0% O by mass 1.Convert to g assuming you have 100 g 63.0 g Mn, 37.0 g O 2.Convert g to mols using molar mass 63.0 g Mn x 1 mol Mn = 1.15 mol Mn 54.94 g 37.0 g O x 1 mol O = 2.31 mol O 16 g

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Emp. Formula (cont.) 3. Divide by smallest amt (round to whole #) 1.15 mol Mn = 1 mol Mn 1.15 2.31 mol O = 2 mol O 1.15 4. Write empirical formula MnO 2

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Molecular Formula Actual # of atoms in a cmpd Determined by emp.formula & molar mass

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Determining a Molecular Formula emp. formula = P 2 O 5 molar mass of cmpd = 284 g/mol 1.Find emp. formula molar mass (2)(30.97 g/mol) + (5)(16 g/mol) = 141.94 g/mol 2.Solve for n (round to nearest whole #) n = molar mass of cmpd = 284 g/mol = 2 molar mass emp.141.94 g/mol

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Determining a Molecular Formula 3. Use n to get molecular formula n (emp.form.) = molecular form. 2 (P 2 O 5 ) = P 4 O 10 4.To verify answer, calculate molar mass & compare to value given (4)(30.97 g/mol) + (10)(16 g/mol) = 284 g/mol

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