Chapter 7 Chemical Quantities
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Transcript of Chapter 7 Chemical Quantities
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Chapter 7Chemical Quantities
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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.
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Section 7.1The Mole: A Measurement of Matter
• OBJECTIVES:– Calculate the mass of a mole of any substance.
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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 count pieces in MOLES.
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Moles (abbreviated: mol)
• Defined as the number of carbon atoms in exactly 12 grams of carbon-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.
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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)
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Types of questions
• How many oxygen atoms in the following?– CaCO3
– Al2(SO4)3
• How many ions in the following?– CaCl2– NaOH– Al2(SO4)3
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Types of questions
• How many molecules of CO2 are there in 4.56 moles of CO2 ?
• How many moles of water is 5.87 x 1022 molecules?
• How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
• How many moles is 7.78 x 1024 formula units of MgCl2?
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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 mass number on the periodic table is
also the mass of 1 mole of those atoms in grams.
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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 weighing them.
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Examples
• How much would 2.34 moles of carbon weigh?
• How many moles of magnesium is 24.31 g of Mg?
• How many atoms of lithium is 1.00 g of Li?
• How much would 3.45 x 1022 atoms of U weigh?
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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
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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.
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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
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Section 7.2Mole-Mass and Mole-Volume Relationships
• OBJECTIVES:– Use the molar mass to convert between mass
and moles of a substance.
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Section 7.2Mole-Mass and Mole-Volume Relationships
• OBJECTIVES:– Use the mole to convert among measurements
of mass, volume, and number of particles.
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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.
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Examples
• Calculate the molar mass of the following and tell what type it is:
• Na2S
• N2O4
• C
• Ca(NO3)2
• C6H12O6
• (NH4)3PO4
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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.
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For example
• How many moles is 5.69 g of NaOH?
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For example
• How many moles is 5.69 g of NaOH?
5 69. g
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For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles
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For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles for NaOH
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For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
g
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
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For example
• How many moles is 5.69 g of NaOH?
5 69. g mole
g
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
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For example
• How many moles is 5.69 g of NaOH?
5 69. g 1 mole
40.00 g
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
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For example
• How many moles is 5.69 g of NaOH?
5 69. g 1 mole
40.00 = 0.142 mol NaOH
g
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
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Examples
• How many moles is 4.56 g of CO2?
• How many grams is 9.87 moles of H2O?
• How many molecules is 6.8 g of CH4?
• 49 molecules of C6H12O6 weighs how much?
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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.
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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
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Examples
• What is the volume of 4.59 mole of CO2 gas at STP?
• How many moles is 5.67 L of O2 at STP?
• What is the volume of 8.8 g of CH4 gas at STP?
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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.
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Examples
• Find the density of CO2 at STP.
• Find the density of CH4 at STP.
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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.
• What is the molar mass of a gas with a density of 1.964 g/L?
• 2.86 g/L?
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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.
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Section 7.3Percent Composition and Chemical Formulas
• OBJECTIVES:– Calculate the percent composition of a
substance from its chemical formula or experimental data.
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Section 7.3Percent Composition and Chemical Formulas
• OBJECTIVES:– Derive the empirical formula and the molecular
formula of a compound from experimental data.
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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 %
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Example
• Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.
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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.
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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
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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
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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.
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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.
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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
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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?
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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?
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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?
• Sample Problem 7-14, p.194