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1 Chapter 10 “Chemical Quantities” Chemistry Pioneer High School Mr. David Norton

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2 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe methods of measuring the amount of something.

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3 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Define Avogadro’s number as it relates to a mole of a substance.

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4 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Distinguish between the atomic mass of an element and its molar mass.

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5 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe how the mass of a mole of a compound is calculated.

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6 What is a Mole? What is a Mole? n You can measure mass, n or volume, n or you can count pieces. n We measure mass in grams. n We measure volume in liters. n We count pieces in MOLES.

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7 Moles (abbreviated: mol) n Defined as the number of carbon atoms in exactly 12 grams of carbon-12. n 1 mole is 6.02 x 10 23 representative particles. n Treat it like a very large dozen n 6.02 x 10 23 is called: Avogadro’s number.

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8 Representative particles n The smallest pieces of a substance: –For a molecular compound: it is the molecule. –For an ionic compound: it is the formula unit (made of ions). –For an element: it is the atom. »Remember the 7 diatomic elements (made of molecules)

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9 Types of questions n How many oxygen atoms in the following? –CaCO 3 –Al 2 (SO 4 ) 3 n How many ions in the following? –CaCl 2 –NaOH –Al 2 (SO 4 ) 3

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10 Some practice problems: n How many molecules of CO 2 are there in 4.56 moles of CO 2 ? n How many moles of water is 5.87 x 10 22 molecules? n How many atoms of carbon are there in 1.23 moles of C 6 H 12 O 6 ? n How many moles is 7.78 x 10 24 formula units of MgCl 2 ?

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11 Measuring Moles n Remember relative atomic mass? n The amu was one twelfth the mass of a carbon-12 atom. n Since the mole is the number of atoms in 12 grams of carbon-12, n the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

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12 Gram Atomic Mass (gam) n Equals the mass of 1 mole of an element in grams n 12.01 grams of C has the same number of pieces as 1.008 grams of H and 55.85 grams of iron. n We can write this as: 12.01 g C = 1 mole C (this is also the molar mass) n We can count things by weighing them.

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13 Examples n How much would 2.34 moles of carbon weigh? n How many moles of magnesium is 24.31 g of Mg? n How many atoms of lithium is 1.00 g of Li? n How much would 3.45 x 10 22 atoms of U weigh?

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14 What about compounds? n in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms n To find the mass of one mole of a compound –determine the moles of the elements they have –Find out how much they would weigh (from the periodic table) –add them up

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15 Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

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16 Section 10.2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: –Describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass.

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17 Section 10.2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: –Identify the volume of a quantity of gas at STP.

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18 Molar Mass n is the generic term for the mass of one mole of any substance (expressed in grams/mol) n Molar mass is the generic term for the mass of one mole of any substance (expressed in grams/mol) n The same as: 1) Gram Molecular Mass, 2) Gram Formula Mass, and 3) Gram Atomic Mass n just a much broader term.

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19 Examples n Calculate the molar mass of the following and tell what type it is: n Na 2 S nN2O4nN2O4nN2O4nN2O4 nCnCnCnC n Ca(NO 3 ) 2 n C 6 H 12 O 6 n (NH 4 ) 3 PO 4

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20 Molar Mass is… n The number of grams of 1 mole of atoms, ions, or molecules. n We can make conversion factors from these. –To change grams of a compound to moles of a compound.

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21 For example n How many moles is 5.69 g of NaOH?

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22 For example n How many moles is 5.69 g of NaOH?

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23 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles

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24 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH

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25 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g

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26 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

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27 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

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28 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

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29 The Mole-Volume Relationship n Many of the chemicals we deal with are gases. –They are difficult to weigh (or mass). n Need to know how many moles of gas we have. n Two things effect the volume of a gas –Temperature and pressure n We need to compare them at the same temperature and pressure.

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30 Standard Temperature and Pressure n 0ºC and 1 atm pressure n abbreviated “STP” n At STP 1 mole of any gas occupies 22.4 L –Called the molar volume n 1 mole = 22.4 L of any gas at STP

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31 Practice Examples n What is the volume of 4.59 mole of CO 2 gas at STP? n How many moles is 5.67 L of O 2 at STP? n What is the volume of 8.8 g of CH 4 gas at STP?

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32 Density of a gas n D = m / V –for a gas the units will be g / L n We can determine the density of any gas at STP if we know its formula. n To find the density we need: 1) mass and 2) volume. n If you assume you have 1 mole, then the mass is the molar mass n And, at STP the volume is 22.4 L.

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33 Practice Examples n Find the density of CO 2 at STP. n Find the density of CH 4 at STP.

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34 Another way: n Given the density, we can find the molar mass of the gas. n Again, pretend you have 1 mole at STP, so V = 22.4 L. n m = D x V n m is the mass of 1 mole, since you have 22.4 L of the stuff. n What is the molar mass of a gas with a density of 1.964 g/L? n How about a density of 2.86 g/L?

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35 Summary n These four items are all equal: a) 1 mole b) molar mass (in grams/mol) c) 6.02 x 10 23 representative particles d) 22.4 L of gas at STP Thus, we can make conversion factors from them.

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36 Section 10.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Describe how to calculate the percent by mass of an element in a compound.

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37 Section 10.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Interpret an empirical formula.

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38 Section 10.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Distinguish between empirical and molecular formulas.

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39 Calculating Percent Composition of a Compound n Like all percent problems: Part whole Part whole n Find the mass of each of the components (the elements), n Next, divide by the total mass of the compound, then x 100 x 100 %

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40 Example n Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

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41 Getting it from the formula n If we know the formula, assume you have 1 mole. n Then you know the mass of the pieces and the whole (these values come from the periodic table).

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42 Examples n Calculate the percent composition of C 2 H 4 ? n How about Aluminum carbonate? n Sample Problem 10.10, p.307 n We can also use the percent as a conversion factor n Sample Problem page 308

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43 Formulas molecular formula = (empirical formula) n [n = integer] molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

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44 Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

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45 Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

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46 Calculating Empirical n Just find the lowest whole number ratio –C 6 H 12 O 6 –CH 4 N n It is not just the ratio of atoms, it is also the ratio of moles of atoms. n In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. n In one molecule of CO 2 there is 1 atom of C and 2 atoms of O.

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47 Calculating Empirical n We can get a ratio from the percent composition. n Assume you have a 100 g. –The percentages become grams. n Convert grams to moles. n Find lowest whole number ratio by dividing by the smallest value.

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48Example n Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. n Assume 100 g so n 38.67 g C x 1mol C = 3.220 mole C 12.01 gC n 16.22 g H x 1mol H = 16.09 mole H 1.01 gH n 45.11 g N x 1mol N = 3.219 mole N 14.01 gN Now divide each value by the smallest value

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49 Example n The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N n The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N = C 1 H 5 N 1 n A compound is 43.64 % P and 56.36 % O. What is the empirical formula? n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

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50 Empirical to molecular n Since the empirical formula is the lowest ratio, the actual molecule would weigh more. –By a whole number multiple. n Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula n Caffeine has a molar mass of 194 g. what is its molecular formula?

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