1. T HE M OLE The Chemist’s Package

2. H OW DO WE GROUP THINGS ? Pairs, Dozen, ??? Chemists group chemicals into moles. A mole of any chemical is the same amount. Just like a dozen is always 12…

3. F ROM ONE TO ANOTHER … M OLES TO M OLES We can change from one substance to another using a ratio. Usually a part to a whole This ratio is a mole ratio. It converts from moles of one thing to moles of another. ____moles X = ____moles Y Ex. H 2 O : Part to Whole 2 moles H = 1 mole H 2 O 1 mole O = 1 mole H 2 O

4. H OW MANY ARE THERE ? M OLES TO P ARTICLES Moles can also be converted to tell us how many particles there are. Particles can be atoms (atom), ions (ion), molecules(molec), and formula units (fmu). There are always 6.022 x 10 23 particles in a mole. This number is known as Avagadro’s Number. Ex. H 2 O: 1 mole H 2 O = 6.022 x 10 23 molec H 2 O Ex. Na: 1 mole Na = 6.022 x 10 23 atoms Na

5. H OW MUCH IS IT ? M OLES TO G RAMS The mass of one mole is a chemical substance’s molar mass. Each substance has its own molar mass that can be determined using the periodic table. The molar mass of an element is equal to its atomic mass. Ex. Na: 1 mol Na = 22.99 g Na The molar mass of a compound is equal to the sum of the atomic masses of the elements that are in the compound. Ex. H 2 O: 1 mole H 2 O = 18.02 g H 2 O (2H + 1O)

6. H OW BIG IS IT ? M OLES TO V OLUME The volume of a mole of gas under standard conditions is always the same. This is not true under nonstandard conditions or for liquids and solids. Standard Conditions of Temperature and Pressure (STP) are: 273 K and 101.3 kPa At STP, 1 mole of gas has a volume of 22.4 L. This is called molar volume. Ex. O 2 at STP: 1 mole O 2 = 22.4 L O 2 Ex. CO 2 at STP: 1 mole CO 2 = 22.4 L CO 2

7. C ONVERSION F ACTORS These are all conversion factors. Mole to Mole – Mole Ratio Mole to Particle – Avagadro’s Number Mole to Mass – Molar Mass Mole to Volume – Molar Volume

9. P RACTICE P ROBLEMS M OLE TO M OLE How many moles of carbon are in 2.0 moles of sucrose, C 12 H 22 O 11 ? Known: 2.0 moles sucroseUnknown: ? moles carbon Conversion Factor: Mole Ratio 1 mol C 12 H 22 O 11 = 12 mol C 2.0 mol C 12 H 22 O 11 12 mol C=24 mol C 1 mol C 12 H 22 O 11 How many moles of sucrose will contain 12 moles of hydrogen? K: 12 mol HUK: ? Mol C 12 H 22 O 11 CF: Mole Ratio - 22 mol H = 1 mol C 12 H 22 O 11 12 mol H1 mol C 12 H 22 O 11 =0.55 mol C 12 H 22 O 11 22 mol H

10. P RACTICE P ROBLEMS M OLE TO P ARTICLE How many moles are equal to 2.41 x 10 24 formula units of sodium chloride? K: 2.41 x 10 24 fmu NaClUK: ? mol NaCl CF: Avagadro’s Number - 1 mol NaCl = 6.022 x 10 23 fmu NaCl How many moles are equal to 9.03 x 10 24 atoms of mercury? K: 9.03 x 10 24 atoms HgUK: ? mol Hg CF: 1 mol Hg = 6.022 x 10 23 atoms Hg How many atoms are equal to 4.50 moles of copper? K: 4.50 mol CuUK: ? atoms Cu CF: 1 mol Cu = 6.022 x 10 23 atoms Cu How many molecules are equal to 100.0 moles of carbon dioxide? K: 100.0 moles CO 2 UK: ? molec CO 2 CF: 1 mol CO 2 = 6.022 x 10 23 molec CO 2

11. P RACTICE P ROBLEMS M OLES TO P ARTICLES (2 STEPS ) How many atoms of oxygen are in 3.65 moles of sucrose? K: 3.65 mol C 12 H 22 O 11 UK: ? atoms O CF: Mole Ratio 1 mol C 12 H 22 O 11 = 11 mol O CF: Avagadro’s Number 1 mol O= 6.022 x 10 23 atoms O 3.65 mol C 12 H 22 O 11 11 mol O 6.022 x 10 23 atoms O = 1 mol C 12 H 22 O 11 1 mol O = 2.42 x 10 25 atoms O How many atoms are in 1.00 mole of sucrose?

12. P RACTICE P ROBLEMS M OLAR M ASS Determine the molar mass of the following compounds. Carbon dioxide CO 2 : 1 Carbon + 2 Oxygen = 12.01g + 2(16.00 g) = 44.01 g CO 2 Sulfur trioxide SO 3 : 1 Sulfur + 3 Oxygen = Bromine Br 2 : 2 Bromine = Sodium hydroxide NaOH : 1 Sodium + 1 Oxygen + 1 Hydrogen = Barium nitrate Ba(NO 3 ) 2 : 1 Barium + 2 Nitrogen + 6 Oxygen =

13. P RACTICE P ROBLEMS M OLES TO M ASS Calculate the mass in grams of 0.250 moles of the following compounds. K: 0.250 molesUK: ? grams Sucrose Sodium chloride Potassium permanganate Calcium sulfide Lithium chlorate

14. P RACTICE P ROBLEMS M OLES TO M ASS Calculate the number of moles in 100.0 grams of the following compounds. K: 100.0 gramsUK: ? moles SucroseCF: Sodium chloride CF: Potassium permanganate CF: Calcium sulfide CF: Lithium chlorate CF:

15. P RACTICE P ROBLEMS M OLES TO V OLUME What is the volume in liters of 8.35 moles of sulfur trioxide gas at STP? K: 8.35 mol SO 3 UK: ? L SO 3 CF: 1 mol SO 3 = 22.4 L SO 3 What is the volume in liters of 0.750 moles of carbon dioxide gas at STP? K: 0.750 mol CO 2 UK: ? L CO 2 CF: 1 mol CO 2 = 22.4 L CO 2 How many moles are there in 52.5 liters of oxygen gas at STP? K: 52.5 L O 2 UK: ? mol O 2 CF: 1 mol O 2 = 22.4 L O 2 How many moles are there in 15.0 liters of nitrogen gas at STP? K: 15.0 L N 2 UK: ? mol N 2 CF: 1 mol N 2 = 22.4 L N 2

16. M ULTI -S TEP P ROBLEMS Same conversions… just more than one at a time… Don’t solve them any differently… Identify your known and unknown. Identify the conversion factors you need. Let the Mole Map do the work for you… Solve.

17. M ULTI -S TEP P RACTICE How many grams of carbon dioxide are there in 4.20 L of carbon dioxide at STP? What is the volume of 245 grams of water vapor at STP? How many formula units are there in 63.4 grams of barium hydroxide?

18. M ULTI -S TEP P RACTICE What is the volume of 6.21 x 10 31 molecules of sulfur trioxide gas at STP? What is the mass of 5.23 x 10 18 atoms of gold? How many molecules of silicon dioxide gas does it take to fill a 6.0 L container at STP?

19. M ULTI -S TEP P RACTICE How many atoms of carbon are there in 4.24 grams of carbon tetrabromide? How many liters of carbon dioxide gas are there if you have 4.7 x 10 20 atoms of oxygen? If you have 3.62 x 10 24 atoms of nitrogen, then what is the maximum number of grams of dinitrogen pentoxide that can be formed?

20. E MPIRICAL F ORMULAS Empirical Formulas are the LOWEST whole number ratio of elements in a compound. Sometimes this is the same as the chemical formula Examples: C 2 H 6 → CH 3 C 6 H 12 O 6 → CH 2 O Empirical Formulas can be determined experimentally from the percent composition of a compound.

21. E MPIRICAL F ORMULA D ETERMINATION STEPS Assume you have 100 g of the substance (makes the math easier because everything is a straight percent). Consider the amounts you are given as being in units of grams. Convert the grams to moles for each element. Find the smallest whole number ratio of moles for each element by dividing all values by the smallest value.

22. E XAMPLE : F IND THE EMPIRICAL FORMULA FOR A COMPOUND CONSISTING OF 63% M N AND 37% O Solution for Finding the Empirical Formula Assuming 100 g of the compound, there would be 63 g Mn and 37 g O Convert grams to moles. (Use molar mass from the periodic table.) 63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn 37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. (In this case there is less Mn than O, so divide by the number of moles of Mn) 1.1 mol Mn/1.1 = 1 mol Mn 2.3 mol O/1.1 = 2.1 mol O The best ratio is Mn:O of 1:2 and the formula is MnO 2 The empirical formula is MnO 2

23. E MPIRICAL F ORMULA C ALCULATION W ORKSHEET 1. 88.8% Cu and 11.2% O Assume you have 100 grams. 88.8 g Cu and 11.2 g O Convert grams to moles 88.8g Cu1 mol Cu= 1.40 mol Cu 63.55g Cu 11.2g O1 mol O= 0.700 mol O 16.00g O Find the smallest whole number ratio. 1.40 mol Cu / 0.700 = 2 0.70 mol O /0.700 = 1 So the empirical formula is Cu 2 O.

24. M OLECULAR F ORMULA FROM E MPIRICAL F ORMULA The empirical formula is the lowest ratio of elements and not always the actual ratio of elements in a compound. The actual ratio (molecular formula) can be determined using the molar mass of the actual compound and the molar mass of the empirical formula. The molar mass of the molecular formula is experimentally determined (Given). The molar mass of the empirical formula can be determined from the periodic table. The molecular mass is divided by the empirical mass and equals a whole number. The whole number is multiplied by the subscripts in the molecular formula to give the empirical formula.

25. E XAMPLE : F IND THE MOLECULAR FORMULA FOR A COMPOUND THAT HAS AN EMPIRICAL FORMULA OF CH 2 AND A MOLECULAR MASS OF 41.5 G. The molecular mass is 41.5 grams. Given The empirical mass is 14.03 grams. C + 2H = 12.01 + 2(1.008) = (14.026)14.03 41.5 / 14.03 = (2.96) 3 3 (CH 2 ) = C 3 H 6 The molecular formula is C 3 H 6.

26. M OLECULAR F ORMULA P RACTICE Use Empirical Formulas from Worksheet with the following Molecular Masses. 1. 290 g/mol 2. 178 g/mol 3. 26 g/mol 4. 160 g/mol 5. 34 g/mol 6. 160 g/mol 7. 138 g/mol 8. 240 g/mol 9. 245 g/mol 10. 230 g/mol