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## Chapter 6

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**Chapter 6**Chemical Quantities**Homework**• Recommended • “Questions” 1-15 (odd) • Required • “Problem” 17-97 (odd) • “Cumulative Problems” 99-123 (odd) • Recommended • “Highlight Problems” 125, 127**Counting Particles By Weighing**• If a person requests 500 quarter inch hexagonal nuts for purchase • How would you count 500 hex nuts? • You could count the hex nuts individually • Or, count the number of hex nuts by weight • First, find the average weight by weighing out 10 hex nuts and obtaining the total weight (10 hex nuts weigh 105 g) Average Weight**Counting Particles By Weighing**• What size (wt.) will contain 500 hex nuts? • Calculate what weight will contain 500 hex nuts • So, weigh out 5.25 kg of the hex nuts**Counting Atoms by the Gram**• Many chemical calculations require counting atoms and molecules • There are two common methods for specifying the quantity of material in a sample of a substance: in terms of mass and in terms of amount • It is difficult to do chemical calculations in terms of atoms or formula units • Since atoms are so small, extremely large numbers are needed in calculations • Need to use a special counting unit just as used for other items • A ream of paper • One dozen donuts • A pair of shoes**Counting Atoms by the Gram**• It is more convenient to use a special counting unit for such large quantities of particles • Mole: A unit that contains 6.022 х 1023 objects • It is used due to the extremely small size of atoms, molecules, and ions • 6.022x1023 particles in 1 mole • Called Avogadro’s Number • Periodic Table • The average atomic massinamu(one atom) • The weight of 1 mole of the element in grams • Avogadro’s number provides the connecting relationship between molar masses and atomic masses**Converting Between Moles and Number of Atoms**• Avogadro’s number relates the number of particles present to moles present • Use when solving chemical calculations involving atoms, molecules, or ions present in a given amount of a substance. Avogadro’s Number 6.022 × 1024 1.204 × 1024 6.022 × 1023 1 mole 2 moles 10 moles Number of Atoms (Particles) Moles of substance**Converting Between Moles and Number of Molecules**• How many molecules of bromine are present in 0.045 mole of bromine gas? Avogadro’s number Given: 0.045 mol Br2 Find: molecules of Br2 Equality: Conversion factors: Solution:**Converting Between Grams and Moles of an Element:The Atomic**Mass Unit • Atoms are too small to conveniently count individually • The usual standards of mass (g or lbs.) are not convenient for use with atoms • To avoid such small numbers scientists use a mass value for atoms relative to a standard instead of an actual mass value • The standard is the atomic mass unit (amu) with all masses measured relative to carbon-12 • The mass of this isotope is set at 12.00 amu The mass of one atom of uranium-238 = 3.95 × 10-22 g**Converting Between Grams and Moles of an Element:The Atomic**Mass Unit • One amu is defined as the mass of one-twelfth the mass of a carbon-12 atom (nearly equal to one proton) • The masses of all other atoms are determined relative to carbon-12 • The masses of these particles in amu brings the mass of the proton and neutron close to 1.**Converting Between Grams and Moles of an Element:The Atomic**Mass Unit • Atomic masses are determined on a relative scale • The standard scale references the carbon-12 isotope = 12.00 amu • All other atomic masses are determined relative to carbon-12**Converting Between Grams and Moles of an Element: The Atomic**Mass Unit • Atoms of the same element do not necessarily have exactly the same mass • Each element exists as a mixture of isotopes so use an average mass • To calculate the average mass of a sample of atoms • Use a “weighted average” for the atomic mass • Number on the bottom of each square in the periodic table is the average weight of the element (in amu)**Converting Between Grams and Moles of an Element: The Atomic**Mass Unit • Using Atomic Mass to Count Atoms • Calculating the number of atoms in a specific mass • If you have a sample of an element, can calculate the number of atoms in that sample • From the atomic mass per one atom a conversion factor can be made • For example: One nitrogen atom has an atomic mass of 14.01 amu**Calculating Atomic Mass in amu(Sample Calculation)**• Calculate the mass (in amu) of 1.0 х 104 carbon atoms 1)Given: 2) Plan: Convert from atoms to amu 3) CF 4) Set Up Problem**Converting Between Grams and Moles of an Element**• The molar mass of an element (monatomic) is the mass of one mole of atoms of the element • The molar mass of one mole of carbon-12 atoms 12.01 g • The molar mass (in grams) of any element has the same numerical value as the atomic mass (in amu) of the same element • This mass contains 6.022 х 1023 particles of that element • Use the periodic table to obtain the molar mass of any element**Converting Between Grams and Moles of an Element**• When the number of grams (weighed out) of a substance equals the formula mass of that substance, Avogadro’s number of atoms or molecules of that substance are present Formula mass of water = 18.02 g Formula mass of I2 = 253.8 g**Converting Between Grams and Moles of an Element**• The three quantities most often calculated in chemical problems • Number of particles • Number of moles • Number of grams • Using molar mass as a conversion factor is one of the most useful in chemistry • Can be used for gram to mole and mole to gram conversions**Converting Between Grams and Moles of an Element**• The molar mass provides a relationship between the number of grams of an element and the number of moles of the same element • The molar mass becomes part of the conversion factor used to convert from grams to moles or moles to grams Solution map Molar Mass Moles of Element Grams of Element**Converting Between Grams of an Element and Number of**Particles • The grams, moles, and number of particles of a substance are interrelated through the conversion factors of molar mass and Avogadro’s number • These two relationships can be combined into a single diagram which is useful to convert grams of a substance to the number of particles (atoms) of a substance Molar Mass Grams of substance Moles of substance Avogadro’s Number Particles of substance Moles of substance**Converting Between Grams of an Element and Number of Atoms**(Sample Calculation) • How many lead atoms are there in a lead weight with a mass of 1.2 g? Find: Pb atoms Given: 1.2 g Pb Conversion Factors 1 mol Pb= 6.022 × 1023 Pb atoms 1 mol Pb= 207.19 g Solution Map Molar mass Avog. Number mol Pb 1.2 g Pb number of Pb atoms Solution 3.5 × 1021 Pb atoms**Converting Between Grams and Moles of a Compound**• The molar mass provides a relationship between the number of grams of a substance and the number of moles of the same substance • The molar mass of any substance is always numerically equal to its formula mass in amu Solution map Molar Mass Moles of substance Grams of substance**Converting Between Grams and Moles of a Compound**• Calculate the mass (in grams) of 0.555 mol of Al2O3 Find: g Al2O3 Given: 0.555 mol Al2O3 Conversion Factors 1 mol Al2O3= 101.96g Solution Map Molar mass g Al2O3 mol Al2O3 Solution 56.6 g Al2O3**Converting Between Grams of an Compound and Number of**Molecules • The molar mass provides a relationship between the number of grams of a substance and the number of moles of the same substance • Avogadro’s number provides a relationship between the number of particles of a substance and the number of moles of the same substance • These two relationships can be combined into a single diagram which is useful to convert grams of a substance to the number of particles (molecules) of a substance Molar Mass Moles of substance Grams of substance Avogadro’s Number Moles of substance Particles of substance**Converting Between Grams of an Element and Number of**Molecules (Sample Calculation) • How many molecules of CO2 are in a 5.61 g sample of CO2? Find: CO2 molecules Given: 5.61 g CO2 Conversion Factors 1 mol CO2= 44.01 g 1 mol CO2= 6.022 × 1023 CO2molec Solution Map Molar mass Avog’s No. mol CO2 5.61 g CO2 number of CO2 molecules Solution 7.68 × 1022CO2 molecules**Converting Between Grams of an Element and Number of**Molecules (Sample Calculation) • Ethylene glycol has the chemical formula: C2H6O2. Find the mass of 1.24 × 1024 molecules of ethylene glycol (antifreeze) Find: g of C2H6O2 Given: 1.24 × 1024 molecules C2H6O2 Conversion Factors 1 mol C2H6O2= 62.05 g 1 mol C2H6O2= 6.022 × 1023 C2H6O2molec Solution Map Avog No. molar mass molecules C2H6O2 g of C2H6O2 mol C2H6O2 Solution 128 g C2H6O2**Chemical Formulas as Conversion Factors**• The subscripts in a chemical formula indicate the number of atoms of each element present in a single molecule or formula unit of a compound • The subscripts in a chemical formula can also indicate the number of moles of atoms of each element present in one mole of a compound • e.g. In one molecule of glucose (C6H12O6) there are 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen • e.g. In one mole of glucose (C6H12O6) there are 6 moles of carbon, 12 moles of hydrogen, and 6 moles of oxygen**Converting Between Moles of a Compound and Moles of a**Constituent Element • Some problems require a calculation of the moles of a particular element in a chemical formula of the compound • The subscript of the particular element in the compound becomes part of the conversion factor used to convert moles of compound to moles of element within the compound Solution map Formula subscript Moles of Element (within compound) Moles of compound**Converting Between Moles of a Compound and Moles of a**Constituent Element (Sample Calculation) • How many moles of carbon atoms are present in 1.85 moles of glucose? Find: mol C Given: 1.85 mol C6H12O6 Conversion Factors 1 mol C6H12O6= 6 mol C Solution Map Formula Subscript mol C6H12O6 mol C Solution 11.1 mol C**Converting Between Grams of a Compound and Moles of a**Constituent Element (Sample Calculation) • International Foods Coffee contains 3.0 mg of sodium chloride per cup of coffee. How many moles of sodium are in each cup of coffee? Find: mol Na Given: 3.0 mg NaCl Conversion Factors 1 mg NaCl = 10-3 g NaCl 1 mol Na =1 mol NaCl 1 mol NaCl= 58.44 g NaCl Solution Map mol Na g NaCl mg NaCl mol NaCl Solution 5.13 × 10-5 mol Na**Converting Between Grams of a Compound and Grams of a**Constituent Element • In chemical problems, the three quantities most often calculated for a substance are number of particles (atoms, molecules, or formula units), number of moles, number of grams • All of these quantities are interrelated and the conversion factors which deal with these relationships are Avogadro’s number, molar mass, and chemical formula subscripts • The three concepts can be combined into a single diagram that is useful for problem solving mol Na mol NaCl 1 mol Na = 1 mol NaCl 1 mol = 58.44 g NaCl 0.003 g NaCl**Converting Between Grams of a Compound and Grams of a**Constituent Element • Throughout Chapter 6, there are calculations for which the initial information is given: moles, grams, or particles. • Additional information is then requested concerning a component of that same substance (find): moles, grams, or particles • Now, use concepts illustrated in previous problems to solve problems in which a given mass of a sample substance is given and we need to calculate the mass of a constituent element Molar Mass Formula subscript Molar mass Grams of compound Moles of element Grams of element Moles of compound**Converting Between Grams of a Compound and Grams of a**Constituent Element • Caffeine is the stimulant found in coffee and tea. It has the chemical formula C8H10N4O2. How many grams of carbon are present in a 50.0 g sample? Find: g of C Given: 50.0 g C8H10N4O2 Conversion Factors 1 mol C8H10N4O2= 194 g C8H10N4O2 1 mol C= 12.01 g C 8 mol C= 1 mol C8H10N4O2 Solution Map Mol-mol ratio Molar Mass Molar Mass 50.0 g C8H10N4O2 g of C mol of C mol C8H10N4O2 Solution 24.8 g C**Mass Percent Composition of Compounds**• The mass percent of an element may be obtained by: • Decomposition of a sample of a known mass • Then, the masses of the constituent elements present in each sample are then obtained • Each element’s mass, and the original sample mass, is used to calculate the percent composition • Mass data obtained from the combination of elements which combine to form a compound • Each element’s mass, and the product’s mass, is used to calculate the percent composition**Mass Percent Composition of Compounds**• The composition of compounds is determined by their formulas (e.g. the formula of carbon dioxide tells us it’s composed of carbon and oxygen (2 oxygen atoms and 1 carbon atom) • Another way to describe the composition of compounds is in terms of the mass percentof each element in the compound • The mass percent composition is the percent by mass of each element in a compound**Mass Percent Composition from a Chemical Formula**• The mass percent composition is the percent by mass of each element in a compound • To determine a compound’s mass percent composition: • its chemical formula is needed • the molar mass of each element that composes the compound • the subscript numbers in the formula are equal to the number of moles of each element in one mole of the compound (its molar mass)**Mass Percent Composition of Compounds**• Percent numbers are always in terms of parts per hundred • For example, the percent composition of water is 88.81% oxygen and 11.19% hydrogen • The percent composition for each element is determined by dividing the mass contributed by each element, by the molar mass and multiplied by 100 • The sum of the percents of all of the elements in a compound equals 100% • If the mass percent of a particular element in a compound is known, you can calculate the amount of the element present in any sample size of that compound**Mass Percent Composition from a Chemical**Formula • Calculate the mass percent composition of each element in NH4OH? Determine the mass of 1 mol of compound Molar mass 14.01 g N 5.040 g H 16.00 g O Determine the contribution of each element**Calculating Empirical Formulas for Compounds**• Chemical formulas are determined by calculation using experimentally obtained information • Depending on the amount of experimental data available, two types of formulas can be obtained • Empirical formulas (simplest formula) • Molecular formulas (true formula)**Calculating Empirical Formulas for Compounds**• The empirical formula gives the simplest ratio of elements in a compound • It uses the smallest possible whole number ratio of atoms present in a formula unit of a compound • The subscripts in the formula cannot be reduced any further to a simpler ratio by division with a small integer • If the percent composition of a compound is known, an empirical formula can be calculated**Calculating an Empirical Formula from Experimental Data**To Determine the empirical formula: • Write down or calculate the given number of grams of each element present in a sample of the compound • Convert the mass of each element to moles of element using the appropriate molar mass • Using the calculated moles of each of the elements: • Write down a pseudoformulausing the moles of each element as a subscript • Find the lowest whole number mole ratio • Divide all the subscripts in the formula by the smallest subscript**Obtaining an Empirical Formula from Experimental Data**Converting Decimal Numbers to Whole Numbers • The subscripts in a “true” formula are expressed as whole numbers, not as decimals • The resulting decimal numbers from a calculation represent each element’s subscript • Arithmetic may need to be applied to convert decimals to whole numbers • If the number(s) are NOT whole numbers, multiply each number by the same small integer (2, 3, 4, 5, or 6) until a whole number is obtained for each subscript number**Obtaining an Empirical Formula from Experimental Data**• Lactic acid has a molar mass of 90.08 g and has this percent composition: • 40.0% C, 6.71% H, 53.3% O • What is the empirical and molecular formula of lactic acid? • Assume a 100.0 g sample size • Convert percent numbers to grams**Obtaining an Empirical Formula from Experimental Data**Step 1: Mass of each element in a 100 g sample of the compound 40.0 g C 6.71 g H 53.3 g O Step 2: Convert mass to moles Step 3: Pseudoformula C3.33H6.66 O3.33 Step 4: Divide by smallest subscript CH2O Empirical formula is**Calculating Molecular Formulas for Compounds**• Molecular formulas are the formulas of molecules • They illustrate all of the atoms of each element in the molecule • It does not necessarily have the smallest set of subscript numbers: It can be the same as or a whole number multiplier of its empirical formula • To determine a molecular formula, the compound’s molecular mass is needed**Relating Empirical and Molecular Formulas**• Empirical Formula • Smallest possible set of subscript numbers • Smallest whole number ratio • All ionic compounds are given as empirical formulas • Molecular Formulas • The actual formulas of molecules • When an empirical formula is known, the molar mass is needed to determine if the empirical formula and the molecular formula are the same • It shows all of the atoms present in a molecule • It may be the same as the EF or a whole-number multiple of its EF Molecular formula = n х Empirical formula**Calculating Molecular Formulas for Compounds**• n represents a whole number multiplier from 1 to as large as necessary • Calculate the empirical formula and the mass of the empirical formula • Divide the given molecular mass by the calculated empirical formula molar mass • Answer is a whole number multiplier**Relating Empirical and Molecular Formulas**• Multiply each subscript in the empirical formula by the whole number multiplier to get the molecular formula • For molecular compounds, the two types of formulas may or may not be the same**Relating Empirical and Molecular Formulas**• For ionic compounds • The empirical and molecular formula are the same • For molecular compounds • The two types of formulas may be the same but not necessarily**Calculating Molecular Formulas for Compounds**• What is the molecular formula of lactic acid? • Obtain the value of n (whole number multiplier) • Multiply the empirical formula by the multiplier Empirical formula is CH2O Empirical formula molar mass = 12.01 g + 2 (1.008 g) + 16.00 g = 30.03 g/mol Molecular formula = n х empirical formula C3H6O3 Molecular formula = 3 (CH2O)