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Section 7.1 The Mole: A Measurement of Matter. OBJECTIVES: Describe how Avogadro’s number is related to a mole of any substance. Calculate the mass of a mole of any substance. What is a Mole?. You can measure mass , or volume , or you can count pieces . We measure mass in grams .
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Section 7.1The Mole: A Measurement of Matter • OBJECTIVES: • Describe how Avogadro’s number is related to a mole of any substance. • Calculate the mass of a mole of any substance.
What is a Mole? • You can measure mass, • or volume, • or you can count pieces. • We measure mass in grams. • We measure volume in liters. • We can count pieces by counting atoms, formula units or molecules.
Moles (abbreviated: mol) • Defined as the number of carbon atoms in exactly 12 grams of carbon-12 (C-12 an isotope that has mass of 12). • 1 mole is 6.02 x 1023 particles. • Treat it like a very large dozen • 6.02 x 1023 is called Avogadro’s number.
Representative particles • 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)
Types of questions • How many oxygen atoms in the following? • CaCO3 • 3 • Al2(SO4)3 • 12 • How many ions in the following? • CaCl2 • 3 • NaOH • 2 • Al2(SO4)3 • 5
Types of questions (Particles) • How many moles of water is 5.87 x 1022 molecules? • How many moles is 7.78 x 1024 formula units of MgCl2? Units MUST be the same so they can cancel out. Units for final answer. 5.87 x 1022 molecules 1 mole = 0.0975 mol 1 6.02 x 1023 molecules 7.78 x 1024 formula units 1 mole = 12.9 mol 1 6.02 x 1023 formula units
Types of Questions (to Moles) 4.56 mol 6.02 x 1023 molecules = 2.75 x 1024 molecules 1 1 mole How many molecules of CO2 are there in 4.56 moles of CO2?
Measuring Moles • Remember relative atomic mass? • The amu was one twelfth the mass of a carbon-12 atom. • Since the mole is the number of atoms in 12 grams of carbon-12, • the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.
Gram Atomic Mass (gam) • Equals the mass of 1 mole of an element in grams • 12.01 grams of C has the same number of pieces as 1.008 grams of H and 55.85 grams of iron. • We can write this as 12.01 g C = 1 mole C • We can count things by massing them.
Examples • How much would 1.00 moles of carbon weigh? • 12.01 grams • How many moles of magnesium is 24.31 g of Mg? • 1 mol • How many atoms of lithium is 1.00 g of Li? • 8.67 x 1022 atoms • How much would 3.45 x 1022 atoms of U weigh? • 13.64 grams
What about compounds? • in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms • 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
What about compounds? • What is the mass of one mole of CH4? 1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH4 = 12.01 + 4.04 = 16.05g • The Gram Molecular Mass (gmm) of CH4 is 16.05g • this is the mass of one mole of a molecular compound.
Gram Formula Mass (gfm) • The mass of one mole of an ionic compound. • Calculated the same way as gmm. • What is the GFM of Fe2O3? 2 moles of Fe x 55.85 g = 111.70 g 3 moles of O x 16.00 g = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70 g
Note! Gram Atomic Mass, Gram Molecular Mass, Gram Formula Mass are all generally called Molar Mass. As it sounds, molar mass is the mass of exactly 1 mole (in grams) of that substance.
Section 7.2Mole-Mass and Mole-Volume Relationships • OBJECTIVES: • Use the molar mass to convert between mass and moles of a substance.
Section 7.2Mole-Mass and Mole-Volume Relationships • OBJECTIVES: • Use the mole to convert among measurements of mass, volume, and number of particles.
Molar Mass • Molar mass is the generic term for the mass of one mole of any substance (in grams) • The same as: 1) gram molecular mass, 2) gram formula mass, and 3) gram atomic mass- just a much broader term.
Examples • Calculate the molar mass of the following and tell what type it is: • Na2S • N2O4 • C • Ca(NO3)2 • C6H12O6 • (NH4)3PO4 • 78.04 g/mol ionic compound • 92.02 g/mol molecular compound • 12.01 g/mol atom • 164.1 g/mol ionic compound • 180.18 g/mol molecular compound • 149.12 g/mol ionic compound
Molar Mass • The number of grams of 1 mole of atoms, ions, or molecules. • We can make conversion factors from these. • To change grams of a compound to moles of a compound.
For example • How many moles is 5.69 g of NaOH?
For example • How many moles is 5.69 g of NaOH?
For example • How many moles is 5.69 g of NaOH? • need to change grams to moles
For example • How many moles is 5.69 g of NaOH? • need to change grams to moles • for NaOH
For example • How many moles is 5.69 g of NaOH? • need to change grams to moles • for NaOH • 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g
For example • How many moles is 5.69 g of NaOH? • need to change grams to moles • for NaOH • 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g • 1 mole NaOH = 40.00 g
For example • How many moles is 5.69 g of NaOH? • need to change grams to moles • for NaOH • 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g • 1 mole NaOH = 40.00 g
For example • How many moles is 5.69 g of NaOH? • need to change grams to moles • for NaOH • 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g • 1 mole NaOH = 40.00 g
Examples • How many moles is 4.56 g of CO2? • 0.1 mol • How many grams is 9.87 moles of H2O? • 177.86 g • How many molecules is 6.8 g of CH4? • 2.55 x 1023 molecules • 49 molecules of C6H12O6 weighs how much? • 1.47 x 10 -20 grams
Gases • Many of the chemicals we deal with are gases. • They are difficult to weigh. • Need to know how many moles of gas we have. • Two things effect the volume of a gas • Temperature and pressure • We need to compare them at the same temperature and pressure.
Standard Temperature and Pressure • 0ºC and 1 atm pressure • abbreviated STP • At STP 1 mole of gas occupies 22.4 L • Called the molar volume • 1 mole = 22.4 L of any gas at STP
Types of Questions (Volume) 3.97 L 1 mole = 0.177 mol 1 22.4 L How many moles are 3.97 L of O2 @ STP (@STP 1 mole of ALL gases take up 22.4 L of space)? Remember: STP Standard Temperature and Pressure (0°C or 273K temperature and 101.3kPa or 1 atmosphere pressure).
Examples • What is the volume of 4.59 mole of CO2 gas at STP? • 102.816 L of CO2 • How many moles is 5.67 L of O2 at STP? • 0.25 mol of O2 • What is the volume of 8.8 g of CH4 gas at STP? • 12.287 L of CH4
Density of a gas • D = m / V • for a gas the units will be g / L • We can determine the density of any gas at STP if we know its formula. • To find the density we need the mass and the volume. • If you assume you have 1 mole, then the mass is the molar mass (from PT) • At STP the volume is 22.4 L.
Examples • Find the density of CO2 at STP. • 1.965 g/L of CO2 • Find the density of CH4 at STP. • 0.716 g/L of CH4
The other way • Given the density, we can find the molar mass of the gas. • Again, pretend you have 1 mole at STP, so V = 22.4 L. • m = D x V • m is the mass of 1 mole, since you have 22.4 L of the stuff.
The other way • What is the molar mass of a gas with a density of 1.964 g/L? • 43.994 g/mol • 2.86 g/L? • 64.064 g/mol
Summary • These four items are all equal: a) 1 mole b) molar mass (in grams) c) 6.02 x 1023 representative particles d) 22.4 L at STP Thus, we can make conversion factors from them.
Section 7.3Percent Composition and Chemical Formulas • OBJECTIVES: • Calculate the percent composition of a substance from its chemical formula or experimental data.
Section 7.3Percent Composition and Chemical Formulas • OBJECTIVES: • Derive the empirical formula and the molecular formula of a compound from experimental data.
Calculating Percent Composition of a Compound • Like all percent problems: Part whole • Find the mass of each component, • then divide by the total mass. x 100 %
Example • Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.
Getting it from the formula • If we know the formula, assume you have 1 mole. • Then you know the mass of the pieces and the whole.
Examples • Calculate the percent composittion of C2H4? • How about Aluminum carbonate? • Sample Problem 7-11, p.191 • We can also use the percent as a conversion factor • Sample Problem 7-12, p.191
The Empirical Formula • The lowest whole number ratio of elements in a compound. • The molecular formula = the actual ratio of elements in a compound. • The two can be the same. • CH2 is an empirical formula • C2H4 is a molecular formula • C3H6 is a molecular formula • H2O is both empirical & molecular
Calculating Empirical • Just find the lowest whole number ratio • C6H12O6 • CH4N • It is not just the ratio of atoms, it is also the ratio of moles of atoms. • In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. • In one molecule of CO2 there is 1 atom of C and 2 atoms of O.
Calculating Empirical • We can get a ratio from the percent composition. • Assume you have a 100 g. • The percentages become grams. • Convert grams to moles. • Find lowest whole number ratio by dividing by the smallest.
Example • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Assume 100 g so • 38.67 g C x 1mol C = 3.220 mole C 12.01 gC • 16.22 g H x 1mol H = 16.09 mole H 1.01 gH • 45.11 g N x 1mol N = 3.219 mole N 14.01 gN
Example • The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N • The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N • = C1H5N1 • A compound is 43.64 % P and 56.36 % O. What is the empirical formula? • Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?
Empirical to molecular • Since the empirical formula is the lowest ratio, the actual molecule would weigh more. • By a whole number multiple. • Divide the actual molar mass by the empirical formula mass. • Caffeine has a molar mass of 194 g. what is its molecular formula?
Example • A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be 98.96 g. What is its molecular formula? • C2H4Cl2 • Sample Problem 7-14, p.194