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1 Chapter 7 Chemical Quantities Charles Page High School Dr. Stephen L. Cotton

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2 Section 7.1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe how Avogadros number is related to a mole of any substance.

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3 Section 7.1 The Mole: A Measurement of Matter n OBJECTIVES: –Calculate the mass of a mole of any substance.

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4 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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5 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 particles. n Treat it like a very large dozen n 6.02 x is called Avogadros number.

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6 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 (ions). –For an element: it is the atom. »Remember the 7 diatomic elements (made of molecules)

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7 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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8 Types of questions 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 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 formula units of MgCl 2 ?

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9 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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10 Gram Atomic Mass (gam) n Equals the mass of 1 mole of an element in grams n grams of C has the same number of pieces as grams of H and grams of iron. n We can write this as g C = 1 mole C n We can count things by weighing them.

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

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12 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 –add them up

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13 What about compounds? n What is the mass of one mole of CH 4 ? 1 mole of C = g 4 mole of H x 1.01 g = 4.04g 1 mole CH 4 = = 16.05g n The Gram Molecular Mass (gmm) of CH 4 is 16.05g –this is the mass of one mole of a molecular compound.

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14 Gram Formula Mass (gfm) n The mass of one mole of an ionic compound. n Calculated the same way as gmm. n What is the GFM of Fe 2 O 3 ? 2 moles of Fe x g = g 3 moles of O x g = g The GFM = g g = g

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15 Section 7.2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: –Use the molar mass to convert between mass and moles of a substance.

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16 Section 7.2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: –Use the mole to convert among measurements of mass, volume, and number of particles.

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17 Molar Mass n Molar mass is the generic term for the mass of one mole of any substance (in grams) n The same as: 1) gram molecular mass, 2) gram formula mass, and 3) gram atomic mass- just a much broader term.

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18 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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19 Molar Mass 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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20 For example n How many moles is 5.69 g of NaOH?

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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? l need to change grams to moles

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23 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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24 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 = g 1 mole of H = 1.01 g

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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 = g 1 mole of H = 1.01 g l 1 mole NaOH = 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 = g 1 mole of H = 1.01 g l 1 mole NaOH = 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 = g 1 mole of H = 1.01 g l 1 mole NaOH = g

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28 Examples n How many moles is 4.56 g of CO 2 ? n How many grams is 9.87 moles of H 2 O? n How many molecules is 6.8 g of CH 4 ? n 49 molecules of C 6 H 12 O 6 weighs how much?

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29 Gases n Many of the chemicals we deal with are gases. –They are difficult to weigh. 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 gas occupies 22.4 L n Called the molar volume n 1 mole = 22.4 L of any gas at STP

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31 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 the mass and the volume. n If you assume you have 1 mole, then the mass is the molar mass (from PT) n At STP the volume is 22.4 L.

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

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34 The other 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 g/L? n 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) c) 6.02 x representative particles d) 22.4 L at STP Thus, we can make conversion factors from them.

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36 Section 7.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Calculate the percent composition of a substance from its chemical formula or experimental data.

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37 Section 7.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Derive the empirical formula and the molecular formula of a compound from experimental data.

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38 Calculating Percent Composition of a Compound n Like all percent problems: Part whole Part whole n Find the mass of each component, n then divide by the total mass. x 100 %

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39 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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40 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.

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41 Examples n Calculate the percent composittion of C 2 H 4 ? n How about Aluminum carbonate? –Sample Problem 7-11, p.191 n We can also use the percent as a conversion factor –Sample Problem 7-12, p.191

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42 The Empirical Formula n The lowest whole number ratio of elements in a compound. n The molecular formula = the actual ratio of elements in a compound. n The two can be the same. n CH 2 is an empirical formula n C 2 H 4 is a molecular formula n C 3 H 6 is a molecular formula n H 2 O is both empirical & molecular

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43 Calculating Empirical n Just find the lowest whole number ratio n C 6 H 12 O 6 n 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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44 Calculating Empirical n We can get a ratio from the percent composition. n Assume you have a 100 g. n The percentages become grams. n Convert grams to moles. n Find lowest whole number ratio by dividing by the smallest.

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45 Example n Calculate the empirical formula of a compound composed of % C, % H, and %N. n Assume 100 g so n g C x 1mol C = mole C gC n g H x 1mol H = mole H 1.01 gH n g N x 1mol N = mole N gN

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46 Example n The ratio is mol C = 1 mol C molN 1 mol N n The ratio is mol H = 5 mol H molN 1 mol N n = C 1 H 5 N 1 n A compound is % P and % 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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47 Empirical to molecular n Since the empirical formula is the lowest ratio, the actual molecule would weigh more. n By a whole number multiple. n Divide the actual molar mass by the empirical formula mass. n Caffeine has a molar mass of 194 g. what is its molecular formula?

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48 Example n A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be g. What is its molecular formula? n Sample Problem 7-14, p.194

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