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Chapter 6 Chemical Quantities

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**Homework Recommended Required “Questions” 1-15 (odd)**

“Problem” (odd) “Cumulative Problems” (odd) “Highlight Problems” 125, 127

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**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

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**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

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**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 Because it requires working with large numbers. See Stoker page 281

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**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 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 mass in amu (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 Chemist use a special counting units Objects can be anything “Objects” can be anything Particles in the sense of atoms and molecules

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**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 10 moles 2 moles 1 mole 1.204 × 1024 6.022 × 1024 6.022 × 1023 Moles of substance Number of Atoms (Particles)

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**Converting Between Moles and Number of Molecules**

How many molecules of bromine are present in mole of bromine gas? Avogadro’s number Given: mol Br2 Find: molecules of Br2 Equality: Conversion factors: Solution:

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**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 amu The mass of one atom of uranium-238 = 3.95 × g

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**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. The atomic mass unit brings the mass of the proton and neutron numerically close to 1.

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**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 = amu All other atomic masses are determined relative to carbon-12

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**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) All masses of other atoms are measured relative to carbon-12

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**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 amu

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**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 Find: total mass of C atoms (amu) 3) CF 4) Set Up Problem

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**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 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 х 1023 particles of that element Use the periodic table to obtain the molar mass of any element It was determined experimentally that Avogadro’s number of atoms gives the same numerical value between amu and mol This is connection between the microscopic and macroscopic values

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**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 Molar mass of water = g Formula mass of water molecule = amu Formula mass of I2 molecule = amu Molar mass of I2 = g

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**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 Three most commonly calculated quantities for chemical problems Molar mass can be used to calculate concentrations of solution, to calculate yields of products formed in reactions based on the limiting reagent

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**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 Grams of Element Moles of Element

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**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 Moles of substance Particles of substance

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**How many lead atoms are there in a lead weight with a mass of 1.2 g?**

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? Given: 1.2 g Pb Find: Pb atoms Conversion Factors 1 mol Pb= g 1 mol Pb= × 1023 Pb atoms Solution Map Molar mass Avog. Number 1.2 g Pb mol Pb number of Pb atoms Solution 3.5 × 1021 Pb atoms

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**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 Grams of substance Moles of substance

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**Converting Between Grams and Moles of a Compound**

Calculate the mass (in grams) of mol of Al2O3 Given: mol Al2O3 Find: g Al2O3 Conversion Factors 1 mol Al2O3 = g Solution Map Molar mass mol Al2O3 g Al2O3 Solution 56.6 g Al2O3

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**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 Grams of substance Moles of substance Avogadro’s Number Moles of substance Particles of substance

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**How many molecules of CO2 are in a 5.61 g sample of CO2?**

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? Given: 5.61 g CO2 Find: CO2 molecules Conversion Factors 1 mol CO2= g 1 mol CO2 = × 1023 CO2 molec Solution Map Molar mass Avog’s No. 5.61 g CO2 mol CO2 number of CO2 molecules Solution 7.68 × 1022 CO2 molecules

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**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) Given: 1.24 × molecules C2H6O2 Find: g of C2H6O2 Conversion Factors 1 mol C2H6O2= g 1 mol C2H6O2= × 1023 C2H6O2 molec Solution Map Avog No. molar mass molecules C2H6O2 mol C2H6O2 g of C2H6O2 Solution 128 g C2H6O2

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**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 First case is the microscopic level per molecule Second case is the macroscopic level per mole

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**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 To quantify a particular element, the subscript becomes part of the conversion factor. Solution map Formula subscript Moles of compound Moles of Element (within compound)

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**How many moles of carbon atoms are present in 1.85 moles of glucose?**

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? Given: 1.85 mol C6H12O6 Find: mol C Conversion Factors 1 mol C6H12O6= 6 mol C Solution Map Formula Subscript mol C6H12O6 mol C Solution 11.1 mol C

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**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? Given: 3.0 mg NaCl Find: mol Na Conversion Factors 1 mg NaCl = 10-3 g NaCl 1 mol NaCl= g NaCl 1 mol Na =1 mol NaCl Solution Map mg NaCl g NaCl mol NaCl mol Na Solution 5.13 × 10-5 mol Na

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**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 NaCl 1 mol Na = 1 mol NaCl mol Na 1 mol = g NaCl 0.003 g NaCl

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**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 compound Moles of element Grams of element

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**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? Given: 50.0 g C8H10N4O2 Find: g of C Conversion Factors 1 mol C8H10N4O2= 194 g C8H10N4O2 1 mol C= g C 8 mol C= 1 mol C8H10N4O2 Solution Map Mol-mol ratio Molar Mass Molar Mass 50.0 g C8H10N4O2 mol C8H10N4O2 mol of C g of C Solution 24.8 g C

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**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 Stoker page 280

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**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 percent of each element in the compound The mass percent composition is the percent by mass of each element in a compound

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**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) “In one mole” of a compound is a key fact that allows the calculation of percent composition from chemical formulas. The formula of carbon dioxide indicates it is composed of one atom of carbon and two atoms of oxygen. Now we will express it in terms of mass percent

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**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 The number of items of a specified type in a group of 100 items

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**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 On a per mol basis Determine the contribution of each element

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**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)

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**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 Formulas are determined by calculation of data experimentally obtained Smallest possible set of subscript numbers The subscript numbers in formulas can be read in terms of the number of moles of each element

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**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 pseudoformula using 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

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**Arithmetic may need to be applied to convert decimals to whole numbers **

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 Cannot have 1.5 atoms in a formula

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**Obtaining an Empirical Formula from Experimental Data**

Lactic acid has a molar mass of 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 g sample size Convert percent numbers to grams Lactic acid builds up in muscles after aerobic exercise Since percent represents parts of one element per 100 parts of the total compound

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**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 Empirical formula is CH2O

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**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

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**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 Ionic compounds do not exist as molecules Molecular formulas account for every atom in the structure Molecular formula = n х Empirical formula

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**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

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**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

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**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

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**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 = g + 2 (1.008 g) g = g/mol The small interger is the name of the small number multiplier Molecular formula = n х empirical formula C3H6O3 Molecular formula = 3 (CH2O)

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**end Ionic compounds do not exist as molecules**

Molecular formulas account for every atom in the structure

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