Chapters 10 Chemical Quantities

Conversion Factors Conversion factor – A fraction equal to 1 that is used to change one unit into another. (When the numerator = denominator, a fraction equals 1.)

Dimensional Analysis Dimensional Analysis – A problem solving method where conversion factors are used to cancel unwanted units.

Conversion Examples a) Convert $25 to nickels. $25 4 quarters $1 25 * 4 * 5 / 1 / 1 = 500 nickels

Conversion Example 2 Convert 180 days to seconds. 180 days 24 hours 60 minutes 1 hour 60 seconds 1 minute 180 * 24 * 60 * 60 / 1 / 1 / 1 = 15,552,000 seconds or 1.6 x 107 seconds

Common Conversions to Know 1 meter = 100 centimeters 1 meter = 1000 millimeters 1 kilometer = 1000 meters Convert 125 cm to km. 1 m 100 cm 1 km 1000 m 125 cm 125 / 100 / 1000 = 0.00125 km or 1.25 x 10-3 km

Conversion Example 4 Convert 15 m/s to km/hr. 15 m 1 s 1 km 1000 m 60 seconds 1 minute 60 minutes 1 hour 15 * 60 * 60 / 1000 = 54 km/hr

Measuring Matter How do we describe how much of something we have? By count, by mass, by volume. We use words like “dozen” to talk about an amount. In chemistry, we use the MOLE.

Mole Mole – SI unit for measuring an amount of a substance. Avogadro’s Number = 6.02 x 1023 Representative particles = smallest unit that still has all the characteristics of that substance. 1 mole = 6.02 x 1023 representative particles

Representative Particles What is the representative particle of : Element (ex. Cu): ___________ Exception: The representative particle of the 7 diatomic elements is a molecule. (ex. H2) Covalent compound (ex. H2O): _________ Ionic Compound (ex. NaCl): ___________ atom molecule formula unit

Conversions 4 moles Ca = atoms Ca. 4 moles Ca 6.02 x 1023 atoms Ca 1 mole Ca = 2.41 x 1024 atoms Ca

Conversions 5 x 1018 atoms Cu = moles Cu. 5 x 1018 atoms Cu 1 mole Cu = 8.3 x 10-6 moles Cu

Conversions 9.2 moles F2 = molecules F2? 9.2 moles 6.02 x 1023 molecules F2 1 mole = 5.5 x 1024 molecules F2

Conversions 9.2 moles F2 = atoms F? 9.2 moles F2 6.02 x 1023 mlcls F2 2 atoms F 1 mole 1 molecule F2 = 1.1 x 1025 atoms F

Conversions 3.4 moles C2H4 = total atoms? 3.4 moles C2H4 6.02 x 1023 mlcls C2H4 6 atoms 1 mole C2H4 1 mlcl C2H4 = 1.22 x 1025 atoms

Molar Mass Molar Mass – The mass of one mole of an element or compound. Molar mass of a compound = the sum of the masses of the atoms in the formula. Use the atomic masses in grams/mol on the periodic table.

Molar Mass Find the molar mass of each: 87.6 g/mol Sr MgBr2 Ba3(PO4)2 87.6 g/mol (24.3) + (2 x 79.9) = 184.1 g/mol (3 x 137.38) + (2 x 30.97) + (8 x 16) = 602.1 g/mol

Mole–Gram Conversions 1 mole = molar mass (in grams) 5.3 moles LiOH = ___________ grams LiOH (Molar mass LiOH : 7 + 16 + 1 = 24 g/mol) 5.3 moles LiOH 24 g LiOH 1 mole LiOH = 127.2 grams LiOH

Gram-Mole Conversions 68 grams F2 = moles F2? 68 grams F2 1 mole F2 38 grams 68 / 38 = 1.8 moles F2

STP STP = Standard Temperature & Pressure Standard Temp  0oC Standard Press  1 atm (See Reference Tables)

Molar Volume of a Gas Avogadro’s Hypothesis: equal volumes of gases at the same temperature and pressure contain equal numbers of particles. At STP, 1 mole of any gas occupies a volume of 22.4 L. 1 mole = 22.4 L (of a gas at STP)

Mole-Volume Conversions 5.4 moles He = L He at STP? 5.4 moles He 22.4 L He 1 mole He 5.4 x 22.4 = 120.96 L He

Mole-Volume Conversions 5.4 moles CH4 = L CH4 gas at STP? 5.4 moles CH4 22.4 L CH4 1 mole CH4 5.4 x 22.4 = 120.96 L CH4

Volume-Mole Conversion 560 L SO3 = mol SO3 at STP 560 L SO3 1 mole SO3 22.4 L SO3 560 / 22.4 = 25 mole SO3

Molar Mass-Density Conversions grams liters grams mole Density = Molar Mass = A gaseous compound composed of sulfur and oxygen has a density of 3.58 g/L at STP. What is the molar mass of this gas? 3.58 g 22.4 L L 1 mole 3.58 x 22.4 = 80.2 g/mole

Molar Mass-Density Conversion What is the density of Krypton gas at STP? 83.8 grams Kr 1 mole mole 22.4 Liters 83.8 / 22.4 = 3.74 g/L Kr

22.4 Liters at STP (gases only) Molar Mass 6.02 x 1023 particles 22.4 Liters at STP (gases only) 1 mole 1 mole Grams (use Per.Tble) 6.02 x 1023 particles

Multi-step Problem: Example 1 If you had 5.0 L of CO2 how many grams would that be? Step 1: L  moles Step 2: moles  grams 5.0 L CO2 1 mole CO2 44.0 g CO2 = 22.4 L CO2 1 mole CO2 9.8 g CO2

Multi-step Problem: Example 2 How many molecules are in 60.0 grams of water? Step 1: grams  moles Step 2: moles  molecules 60.0 g H2O 1 mole H2O 6.02 x 1023 mlcls = 18.0 g H2O 1 mole H2O = 2.0 x 1024 molecules of H2O

Percent Composition Percent Composition - % by mass of each element in a compound Percent = Part Whole x 100

Percent Composition Percent Comp = Example: Find the mass percent composition of Al2(SO4)3 Mass of 1 element Mass of compound x 100 54 342 x 100 = 15.8% % Al: Al: 2 x 27 = 54 S: 3 x 32 = 96 O: 12 x 16 = 192 Total Comp. = 342 96 342 % S: x 100 = 28.1% 192 342 %O: x 100 = 56.1%

Percent Example Find the percent composition of NiSO3. Ni: 58.7 g %Ni: 58.7 S: 32 g 138.7 O: (3 x 16) = 48 g Total Comp. 138.7 x 100 = 42.3 % 32 138.7 %S: x 100 = 23.1 % 48 138.7 %O: x 100 = 34.6%

More Percents Which of the following shows a compound that is 92.3%C and 7.7%H? a) C2H4 b) C3H6 c) CH4 d) C6H6

Empirical Formulas Empirical Formula – The simplest formula. Shows the smallest whole number ratio of elements in a compound. Covalent formulas will not always be empirical. Example: CH Molecular Formula – The actual formula. For ionic compounds – it will be the simplest ratio. For molecular compounds – it will NOT always be the simplest ratio. Example: C6H6

To Calculate Empirical Formula Calculate the empirical formula of a 2.5 gram compound containing 0.90g Ca and 1.60g Cl. Step 1: Convert GRAMS to MOLES. Ca: 0.90g 1 mole = 0.0224 mole Ca 40.1 g Cl: 1.60g 1 mole = 0.0451 mole Cl 35.5 g

Calculating Empirical Formula Step 2: DIVIDE the # of moles of each substance by the smallest number to get the simplest mole ratio. Ca: 0.0224 = 1 Cl: 0.0451 = 2.01 ~ 2 0.0224 0.0224 CaCl2

Calculating Empirical Formulas Step 3: If the numbers are whole numbers, use these as the subscripts for the formula. If the numbers are not whole numbers, multiply each by a factor that will make them whole numbers. Look for these fractions: 0.5  x 2 0.33  x 3 0.25  x 4

Empirical Formula Example What is the empirical formula of a compound that is 66% Ca and 34% P? (Assume you have 100 grams of a compound and replace % with grams.)

Empirical Formula Example Step 1: grams  moles Ca: 66g 1 mole = 1.646 mole Ca 40.1 g P 34g 1 mole = 1.097 mole P 31.0 g Step 2: Divide by the smallest. Ca: 1.646 = 1.5 P: 1.097 = 1 1.097 1.097

Empirical Formula Example Step 3: Multiply until you get whole numbers. (If you multiply one factor by a number, you have to multiply ALL the factors by that number!) Ca: 1.5 x 2 = 3 P: 1 x 2 = 2 Ca3P2

Determining Molecular Formulas A compound has an empirical formula of CH2O. Its molecular mass is 180g/mol. What is its molecular formula? Step 1: Find the mass of the empirical formula. C: 1 x 12 = 12 H: 2 x 1 = 2 O: 1 x 16 = 16 Total: 30

Determining Molecular Formula Step 2: Divide the molecular mass by the mass of the empirical formula to get the “multiplying factor”. Step 3: Multiply each of the subscripts in the empirical formula by this factor to get the molecular formula. 6 (CH2O)  C6H12O6 180 30 = 6

Determining Molecular Formula Find the molecular formula of ethylene glycol (CH3O) if its molar mass is 62 g/mol. Step 2: 62 / 31 = 2 Step 3: 2 (CH3O)  C2H6O2 Step 1: 12 + (3 x 1) + 16 = 31 g/mol

Empirical/Molecular Example The percent composition of methyl butanoate is 58.8% C, 9.8% H, and 31.4 % O and its molar mass is 204 g/mol. What is its empirical formula? What is its molecular formula?

Empirical/Molecular Example 58.8 g C 1 mol C 12.0 g C 4.9 = 2.5 1.96 9.8 = 5 1.96 = 1 x 2 = 5 x 2 = 10 x 2 = 2 = 4.9 mol C 9.8 g H 1 mol H 1.0 g H = 9.8 mol H 31.4 g O 1 mol O 16.0 g O = 1.96 mol O Empirical Formula = C5H10O2

Empirical/Molecular Example Empirical Formula  C5H10O2 Mass = 5(12) + 10(1) + 2(16) = 102 g/mole Molecular mass  204 g/mol = 2 Empirical mass  102 g/mol So molecular formula is 2 x emp. form: 2(C5H10O2) = C10H20O4