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C. Johannesson Ch. 11 – The Mole Molar Conversions & Calculations.

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Presentation on theme: "C. Johannesson Ch. 11 – The Mole Molar Conversions & Calculations."— Presentation transcript:

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2 C. Johannesson Ch. 11 – The Mole Molar Conversions & Calculations

3 C. Johannesson A. What is the Mole? A counting number (like a dozen) Avogadro’s number (N A ) 1 mol = 6.02  10 23 items A large amount!!!!

4 C. Johannesson 1 mole of hockey pucks would equal the mass of the moon! A. What is the Mole? 1 mole of pennies would cover the Earth 1/4 mile deep! 1 mole of basketballs would fill a bag the size of the earth!

5 C. Johannesson B. Molar Mass Mass of 1 mole of an element or compound. Atomic mass tells the... – atomic mass units per atom (amu) – grams per mole (g/mol) Round to 2 decimal places

6 C. Johannesson B. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol 65.39 g/mol

7 C. Johannesson B. Molar Mass Examples water sodium chloride –H2O–H2O –2(1.01) + 16.00 = 18.02 g/mol –NaCl –22.99 + 35.45 = 58.44 g/mol

8 C. Johannesson B. Molar Mass Examples sodium hydrogen carbonate sucrose –NaHCO 3 –22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol –C 12 H 22 O 11 –12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

9 C. Johannesson C. Percentage Composition the percentage by mass of each element in a compound

10 C. Johannesson  100 = C. Percentage Composition %Cu = 127.10 g Cu 159.17 g Cu 2 S  100 = %S = 32.07 g S 159.17 g Cu 2 S Find the % composition of Cu 2 S.

11 C. Johannesson %Fe = 28 g 36 g  100 = %O = 8.0 g 36 g  100 = Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. C. Percentage Composition

12 C. Johannesson How many grams of copper are in a 38.0-gram sample of Cu 2 S? (38.0 g Cu 2 S)(0.79852) = g Cu Cu 2 S is 79.852% Cu C. Percentage Composition

13 C. Johannesson  100 = %H 2 O = 36.04 g 147.02 g Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? C. Percentage Composition

14 C. Johannesson Practice P. 322 #25; p. 331 #42, 43, 45 –Answers on p. 929 and 930 P. 347 #111, 112; p. 348 #136

15 C. Johannesson D. Molar Conversions molar mass (g/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES 6.02  10 23 (particles/mol) “Particles”=atoms, electrons, molecules, formula units, anything!

16 C. Johannesson D. Molar Conversion Examples How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = mol C

17 C. Johannesson D. Molar Conversion Examples How many molecules are in 2.50 moles of C 12 H 22 O 11 ? 2.50 mol 6.02  10 23 molecules 1 mol

18 C. Johannesson D. Molar Conversion Examples Find the mass of 2.1  10 24 molecules of NaHCO 3. 2.1  10 24 molecules 1 mol 6.02  10 23 molecules 84.01 g 1 mol

19 C. Johannesson E. Empirical Formula C2H6C2H6 CH 3 reduce subscripts Smallest whole number ratio of atoms in a compound

20 C. Johannesson E. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. -go out about 4 or 5 decimals 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

21 C. Johannesson E. Empirical Formula Poem to help you remember… Percent to mass Mass to mole Divide by small Multiply ‘til whole!

22 C. Johannesson E. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.8487 mol N 74.1 g 1 mol 16.00 g = 4.63125 mol O 1.8487 mol

23 C. Johannesson E. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

24 C. Johannesson F. Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

25 C. Johannesson F. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

26 C. Johannesson F. Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = empirical mass = 14.03 g/mol

27 C. Johannesson G. Hydrates Compounds with a certain number of water molecules attached –MgSO 4 ·7H 2 0 = magnesium sulfate heptahydrate –Use a prefix to indicate number of water molecules –MgSO 4 without water is called anhydrous

28 C. Johannesson G. Hydrates To determine the number of water molecules, find the mole ratio of the anhydrous compound to the water Find the moles of each Set up a ratio: moles water/moles salt –Whole number answer is the number that goes in front of H 2 O

29 C. Johannesson G. Hydrates A 5g sample of a hydrate of Cu(NO 3 ) 2 was heated, and only 3.9g of the anhydrous salt remained. What is the formula and name of the hydrate? Step 1: determine moles of the salt and the water –3.9g Cu(NO 3 ) 2 x 1mole = 0.02mol Cu(NO 3 ) 2 –5g-3.9g=1.1g water –1.1gH 2 O x 1mole =.06mol H 2 O 18.02g

30 C. Johannesson Step 2: determine ratio –Moles H 2 O/moles Cu(NO 3 ) 2 –0.06 mol H 2 O/0.02 mol Cu(NO 3 ) 2 = Step 3: formula and name –Cu(NO 3 ) 2 ·3H 2 O –Copper (II) nitrate trihydrate


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