Presentation on theme: "1 Chapter 10 “Chemical Quantities” Chemistry Pioneer High School Mr. David Norton."— Presentation transcript:
1 Chapter 10 “Chemical Quantities” Chemistry Pioneer High School Mr. David Norton
2 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe methods of measuring the amount of something.
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.
4 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Distinguish between the atomic mass of an element and its molar mass.
5 Section 10.1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe how the mass of a mole of a compound is calculated.
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.
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.
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)
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
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 ?
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.
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.
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?
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
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
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.
17 Section 10.2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: –Identify the volume of a quantity of gas at STP.
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.
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
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.
21 For example n How many moles is 5.69 g of NaOH?
22 For example n How many moles is 5.69 g of NaOH?
23 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles
24 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH
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
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
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
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
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.
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
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?
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.
33 Practice Examples n Find the density of CO 2 at STP. n Find the density of CH 4 at STP.
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?
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.
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.
37 Section 10.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Interpret an empirical formula.
38 Section 10.3 Percent Composition and Chemical Formulas n OBJECTIVES: –Distinguish between empirical and molecular formulas.
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 %
40 Example n Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.
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).
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
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.
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
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
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.
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.
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
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?
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?