Chemical Composition PDF

6: CHEMICAL COMPOSITION  CHAPTER OVERVIEW 6: Chemical Composition Chapter 6 is concerned with the amounts of substances which participate in chemical reactions, the quantities of heat given off or absorbed when reactions occur, and the volumes of solutions which react exactly with one another. The...

6: CHEMICAL COMPOSITION  CHAPTER OVERVIEW 6: Chemical Composition Chapter 6 is concerned with the amounts of substances which participate in chemical reactions, the quantities of heat given off or absorbed when reactions occur, and the volumes of solutions which react exactly with one another. These seemingly unrelated subjects are discussed together because many of the calculations involving them are almost identical in form. The same is true of the density calculations, and of the calculations involving molar mass and the Avogadro constant. 6.1: Prelude to Chemical Composition - How Much Sodium? 6.2: Counting Nails by the Pound 6.3: Counting Atoms by the Gram 6.4: Counting Molecules by the Gram 6.5: Chemical Formulas as Conversion Factors 6.6: Mass Percent Composition of Compounds 6.7: Mass Percent Composition from a Chemical Formula 6.8: Calculating Empirical Formulas for Compounds 6.9: Calculating Molecular Formulas for Compounds 6: Chemical Composition is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 1 6.1: PRELUDE TO CHEMICAL COMPOSITION - HOW MUCH SODIUM? Why is knowledge of composition important? Everything in nature is either chemically or physically combined with other substances. To find the amount of a material in a sample, you need to know what fraction of the sample it is. Some simple applications of composition are: the amount of sodium in sodium chloride for a diet, the amount of iron in iron ore for steel production, the amount of hydrogen in water for hydrogen fuel, and the amount of chlorine in freon to estimate ozone depletion. How much salt is in this salt packet? (CC-BY-SA; 3.0; Swilliams). 6.1: Prelude to Chemical Composition - How Much Sodium? is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.1.1 https://chem.libretexts.org/@go/page/47485 6.2: COUNTING NAILS BY THE POUND COUNTING BY WEIGHING AND AVOGADRO'S NUMBER The size of molecule is so small that it is physically difficult, if not impossible, to directly count out molecules (Figure 6.2.1 ). However, we can count them indirectly by using a common trick of "counting by weighing". Figure 6.2.1: A comparison of the scales of various biological and technological objects. (CC BY-SA 3.0; Wikipedia) Consider the example of counting nails in a big box at a hardware store. You need to estimate the number of nails in a box. The weight of an empty box is 213 g and the weight of the box plus a bunch of big nails is 1340 g. Assume that we know that the weight of one big nail is 0.450 g. Hopefully it's not necessary to tear open the package and count the nails. We agree that mass of big nails = 1340 g − 213 g = 1227 g Therefore 1227 g Number of big nails in box = = 2, 726.6 big nails = 2, 730 big nails. (6.2.1) 0.450 g/big nail You have just counted the number of big nails in the box by weighing them (rather than by counting them individually). Figure 6.2.2: Galvanized nails. Individually counting nails in a box would would require significant effort. Alternatively, we can count them by weighing. (Public Domain; Wikipedia). Now consider if the box of nails weighed the same, but the box were filled with small nails with an individual mass of 0.23 g/small nail instead? You would do the same math, but use a different denominator in Equation 6.2.1 : 1227 g Number of small nails in box = = 5, 334.7 small nails = 5, 335 small nails. (6.2.2) 0.230 g/small nail The individual mass is the conversion factor used in the calculation and changes, based on the nature of the nail (big or small). Let's ask a different question: how many dozens of nails are there in the same box of small nails described above? 6.2.1 https://chem.libretexts.org/@go/page/47487 If we know the information from Equation 6.2.2 , we can just use the conversion of how many nails are in a dozen: 5, 335 small nails = 444.6 dozen small nails (6.2.3) 12 small nails/dozen If we want to get this value from weighing, we use the "dozen mass" instead of individual mass: 12 × 0.23g = 2.76 g/dozen small nails. (6.2.4) So following Equation 6.2.2 , we get: 1227 g Number of dozens of small nails = = 444.6 dozen small nails (6.2.5) 2.76 g/dozen small nails and this is the same result as Equation 6.2.3. These calculations demonstrate the difference between individual mass (i.e., per individual) and collective mass (e.g., per dozen or per gross). The collective mass of most importance to chemistry is molar mass (i.e., mass per mole or mass per 6.022 × 10 ). 23  AVOGADRO'S NUMBER Avogadro's number is an accident of nature. It is the number of particles that delivers a mole of a substance. Avogadro's number = 6.022 × 10 23. The reason why the value is an accident of nature is that the mole is tied to the gram mass unit. The gram is a convenient mass unit because it matches human sizes. If we were a thousand times greater in size (like Paul Bunyan) we would find it handy to use kilogram amounts. This means the kilogram mole would be convenient. The number of particles handled in a kilogram mole is 1000 times greater. The kilo Avogadro number for the count of particles in a kilomole is 6.022 × 10. 26 If humans were tiny creatures (like Lilliputians) only 1/1000 our present size, milligrams would be more convenient. This means the milligram mole would be more useful. The number of particles handled in a milligram mole (millimole) would be 1/1000 times smaller. The milli Avogadro number for the count of particles in a millimole is 6.022 × 10. 20 What do you think would happen to Avogadro's number if the American system was used and amounts were measured in pound moles? Remember 1 pound = 454 grams. Avogadro's number would be larger by a factor of 454. A pound mole of hydrogen would weigh 1 pound, which is 454 grams. A gram mole of hydrogen weighs 1 gram and contains 6.022 × 10 H atoms. 23 MOLAR MASS FOR ELEMENTS You are able to read the periodic table and determine the average atomic mass for an element like carbon. The average mass is 12.01 amu. This mass is a ridiculously tiny number of grams. It is too small to handle normally. The molar mass of carbon is defined as the mass in grams that is numerically equal to the average atomic weight. This means 1g/molecarbon = 12.01 g carbon this is commonly written 1 mol carbon = 12.01 grams carbon. This is the mass of carbon that contains 6.022 × 10 23 carbon atoms. Avogadro's number is 6.022 × 10 23 particles. This same process gives us the molar mass of any element. For example: 1 mol neon = 20.18 g neon N e 1 mol sodium = 22.99 g sodium N a MOLAR MASS FOR COMPOUNDS  EXAMPLE 6.2.1: MOLAR MASS OF WATER The formulas for compounds are familiar to you. You know the formula for water is H 2 O. It should be reasonable that the weight of a formula unit can be calculated by adding up the weights for the atoms in the formula. Solution The formula weight for water weight from hydrogen + weight from oxygen The formula weight for water 6.2.2 https://chem.libretexts.org/@go/page/47487  2 H atoms x 1.008 amu + 1 O atom x 16.00 amu = 18.016 amu The molar mass for water 18.016 grams water or 18 grams to the nearest gram  EXAMPLE 6.2.2: MOLAR MASS OF METHANE The formula for methane, the major component in natural gas, is CH. 4 Solution The formula weight for methane weight from hydrogen + weight from carbon The formula weight for methane 4 H atoms x 1.008 amu + 1 C atom x 12.01 amu = 16.04 amu The molar mass for methane 16.04 grams per mole of methane  EXAMPLE 6.2.3: MOLAR MASS OF ETHYL CHLORIDE What is its molar mass for ethyl chloride CH 3 CH Cl 2 ? Solution The formula weight weight from hydrogen + weight from carbon + weight from chlorine The formula weight 5 H atoms x 1.008 amu + 2 C atom x 12.01 amu + 35.5 amu = 64.5 amu The molar mass for ethyl chloride 64.5 grams per mole of ethyl chloride  EXAMPLE 6.2.4: MOLAR MASS OF SULFUR DIOXIDE What is the molar mass for sulfur dioxide, SO 2 , a gas used in bleaching and disinfection processes? (g) Solution Look up the atomic weight for each of the elements in the formula. 1 sulfur atom = 32.07 amu 1 oxygen atom = 16.00 amu Count the atoms of each element in the formula unit. one sulfur atom two oxygen atoms The formula weight weight from sulfur + weight from oxygen The formula weight 1 sulfur atom x (32. 07 amu ) + 2 oxygen atoms x (16.00 amu) The formula weight SO 2 = 32. 07 amu + 32.00 amu = 64.07 amu = 64 amu SO 2 The molar mass for SO is 2 64.07 grams of SO ; 1 mol SO = 64 grams per mole of SO 2 2 2 6.2.3 https://chem.libretexts.org/@go/page/47487   EXERCISE 6.2.1 What is the formula weight and molar mass for alum, KAl(SO )2 ∙ 12 H2 O 4 ? Answer 1. Check the periodic table for the atomic masses for each atom in the formula. 2. Count the number of each type of atom in the formula. 3. Multiply the number of atoms by the atomic mass for each element. 4. Add up the masses for all of the elements. Table 6.2.1: Masses of each element in alum, KAl(SO 4 ) 2 ∙ 12 H O 2 element average atomic mass number of atoms in formula rounded to nearest one unit for simplicity potassium k 39.1 amu 1 39. amu aluminum 26.98 amu 1 27. amu sulfur 32.07 amu 2 64. amu oxygen 16.00 amu 8 + 12 = 20 320. amu hydrogen 1.008 amu 2 x 12 = 24 24. amu Molar mass is 474 grams (add up the amu of each element to find the total of 474 amu). This is a mass in grams that is numerically (474) the same as the formula weight. 1 mole alum KAl(SO 4 )2 ∙ 12 H2 O = 474 grams alum KAl(SO 4 )2 ∙ 12 H2 O This page titled 6.2: Counting Nails by the Pound is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Delmar Larsen, Marisa Alviar-Agnew, Henry Agnew, & Henry Agnew. 6.2.4 https://chem.libretexts.org/@go/page/47487 6.3: Counting Atoms by the Gram  Learning Objectives Use Avogadro's number to convert to moles and vice versa given the number of particles of an element. Use the molar mass to convert to grams and vice versa given the number of moles of an element. When objects are very small, it is often inconvenient, inefficient, or even impossible to deal with the objects one at a time. For these reasons, we often deal with very small objects in groups, and have even invented names for various numbers of objects. The most common of these is "dozen" which refers to 12 objects. We frequently buy objects in groups of 12, like doughnuts or pencils. Even smaller objects such as straight pins or staples are usually sold in boxes of 144, or a dozen dozen. A group of 144 is called a "gross". This problem of dealing with things that are too small to operate with as single items also occurs in chemistry. Atoms and molecules are too small to see, let alone to count or measure. Chemists needed to select a group of atoms or molecules that would be convenient to operate with. Avogadro's Number and Mole In chemistry, it is impossible to deal with a single atom or molecule because we can't see them, count them, or weigh them. Chemists have selected a number of particles with which to work that is convenient. Since molecules are extremely small, you may suspect this number is going to be very large, and you are right. The number of particles in this group is 6.02 × 10 particles and 23 the name of this group is the mole (the abbreviation for mole is mol). One mole of any object is 6.02 × 10 of those objects. 23 There is a particular reason that this number was chosen and this reason will become clear as we proceed. When chemists are carrying out chemical reactions, it is important that the relationship between the numbers of particles of each reactant is known. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them. The mole is used for this purpose. The mole (symbol: mol) is the base unit of amount of substance ("number of substance") in the International System of Units or System International (SI), defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons. The current definition was adopted in November 2018, revising its old definition based on the number of atoms in 12 grams of carbon-12 (12C) (the isotope of carbon with relative atomic mass 12 Daltons, by definition). For most purposes, 6.022 × 1023 provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 1023 atoms, 1 mole of eggs contains 6.022 × 1023 eggs. This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain. It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 1023. Converting Between Number of Atoms to Moles and Vice Versa We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. If we are given the number of atoms of an element X, we can convert it into moles by using the relationship 23 1 mol X = 6.022 × 10 X atoms.  Example 6.3.1: Moles of Carbon The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is 4.72 × 10 24 atoms of carbon? Solution Solutions to Example 6.3.1 6.3.1 https://chem.libretexts.org/@go/page/47489  The element carbon exists in two primary forms: graphite and Steps for Problem Solving diamond. How many moles of carbon atoms is 4.72 × 10 24 atoms of carbon? Identify the "given" information and what the problem is asking you Given: 4.72 × 10 C atoms 24 to "find." Find: mol C List other known quantities. 1 mol = 6.022 × 10 23 C atoms Prepare a concept map and use the proper conversion factor. 1 mol C 24 Cancel units and calculate. 4.72 × 10 C atoms × 23 = 7.84 mol C 6.02 × 10 C atoms The given number of carbon atoms was greater than Avogadro's number, so the number of moles of C atoms is greater than 1 mole. Think about your result. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures. Molar Mass Molar mass is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of the element lithium is 6.94g, the molar mass of zinc is 65.38g, and the molar mass of gold is 196.97g. Each of these quantities contains 6.022 × 10 atoms of that particular element. The units for molar mass are grams per 23 mole or g/mol. 1.00 mol of carbon-12 atoms has a mass of 12.0 g and contains 6.022 × 10 atoms. 1.00 mole of any element has 23 a mass numerically equal to its atomic mass in grams and contains 6.022 × 10 particles. The mass, in grams, of 1 mole of 23 particles of a substance is now called the molar mass (mass of 1.00 mole). Converting Grams to Moles of an Element and Vice Versa We can also convert back and forth between grams of an element and moles. The conversion factor for this is the molar mass of the substance. The molar mass is the ratio giving the number of grams for each one mole of the substance. This ratio is easily found by referring to the atomic mass of the element using the periodic table. This ratio has units of grams per mole or g/mol. Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure 6.4.1 illustrates what conversion factor is needed and two examples are given below. Figure 6.3.1 : A Simple Flowchart for Converting Between Mass and Moles of an Element. Flowchart: to convert moles of Element A to mass of Element A, use g/mol, and to convert vice versa, use mol/g.  Example 6.3.2: Chromium Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium. 6.3.2 https://chem.libretexts.org/@go/page/47489 Solution Solutions to Example 6.3.2 Chromium metal is used for decorative electroplating of car Steps for Problem Solving bumpers and other surfaces. Find the mass of 0.560 moles of chromium. Identify the "given" information and what the problem is asking you Given: 0.560 mol Cr to "find." Find: g Cr List other known quantities. 1 mol Cr = 52.00g Cr Prepare a concept map and use the proper conversion factor. 52.00 g Cr Cancel units and calculate. 0.560 mol Cr × = 29.1 g Cr 1 mol Cr Since the desired amount was slightly more than one half of a mole, Think about your result. the mass should be slightly more than one half of the molar mass. The answer has three significant figures because of the 0.560 mol  Example 6.3.3: Silicon How many moles are in 107.6g of Si? Solution Solutions to Example 6.3.3 Steps for Problem Solving How many moles are in 107.6g of Si. Identify the "given" information and what the problem is asking you Given: 107.6g Si to "find." Find: mol Si List other known quantities. 1 mol Si = 28.09g Si Prepare a concept map and use the proper conversion factor. 1 mol Si Cancel units and calculate. 107.6 g Si × = 3.83 mol Si 28.09 g Si Think about your result. Since 1 mol of Si is 28.09g, 107.6 should be about 4 moles.  Exercise 6.3.1 a. How many moles are present in 100.0 g of Al? b. What is the mass of 0.552 mol of Ag metal? Answer a: 3.706 mol Al Answer b: 59.5 g Ag 6.3.3 https://chem.libretexts.org/@go/page/47489 Summary A mole is defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons. There are 6.02214076 × 10 particles in 1.00 mole. This number is called Avogadro's number. 23 The molar mass of an element can be found by referring to the atomic mass on a periodic table with units of g/mol. Using dimensional analysis, it is possible to convert between grams, moles, and the number of atoms or molecules. Further Reading/Supplemental Links learner.org/resources/series61.html - The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos, but there is no charge. The website has one video that relates to this lesson called The Mole. Using Avogadro's law, the mass of a substance can be related to the number of particles contained in that mass. The Mole: (www.learner.org/vod/vod_window.html?pid=803) Vision Learning tutorial: The Mole http://visionlearning.com/library/mo...p?mid-53&1=&c3= Contributions & Attributions Wikipedia 6.3: Counting Atoms by the Gram is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.3.4 https://chem.libretexts.org/@go/page/47489 6.4: COUNTING MOLECULES BY THE GRAM  LEARNING OBJECTIVES Define molecular mass and formula mass. Perform conversions between mass and moles of a compound. Perform conversions between mass and number of particles. MOLECULAR AND FORMULA MASSES The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example 6.4.1.  EXAMPLE 6.4.1: ETHANOL Calculate the molecular mass of ethanol, whose condensed structural formula is CH 3 CH OH 2. Among its many uses, ethanol is a fuel for internal combustion engines Solution Solutions to Example 6.4.1 Calculate the molecular mass of ethanol, whose condensed Steps for Problem Solving structural formula is CH CH OH 3 2 Given: Ethanol molecule (CH3CH2OH) Identify the "given"information and what the problem is asking you to "find." Find: molecular mass The molecular formula of ethanol may be written in three different ways: CH3CH2OH (which illustrates the presence of an ethyl group Determine the number of atoms of each element in the molecule. CH3CH2−, and an −OH group) C2H5OH, and C2H6O; All show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom. 1 C atom = 12.011 amu Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of 1 H atom = 1.0079 amu each element by the number of atoms of that element. 1 O atom = 15.9994 amu 2C: (2 atoms)(12.011amu/atom) = 24.022 amu 6H: (6 atoms)(1.0079amu/atom) = 6.0474amu Add the masses together to obtain the molecular mass. +1O: (1 atoms)(15.9994amu/atom) =15.9994amu C2H6O : molecular mass of ethanol = 46.069amu  EXERCISE 6.4.1: FREON Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, which has a condensed structural formula of CCl 3. F Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows: Figure 6.4.1: Molecular structure of freon-11, CCl 3 F. Answer 137.37 amu 6.4.1 https://chem.libretexts.org/@go/page/47490 Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units. Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.  EXAMPLE 6.4.2: CALCIUM PHOSPHATE Calculate the formula mass of Ca (PO ) 3 4 2 , commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk. Solution Solutions to Example 6.4.2 Steps for Problem Solving Calculate the formula mass of Ca 3 (PO ) 4 2 , commonly called calcium phosphate. Given: Calcium phosphate [Ca3(PO4)2] formula unit Identify the "given" information and what the problem is asking you to "find." Find: formula mass The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43− ions. Determine the number of atoms of each element in the molecule. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms. 1 Ca atom = 40.078 amu Obtain the atomic masses of each element from the periodic table and multiply 1 P atom = 30.973761 amu the atomic mass of each element by the number of atoms of that element. 1 O atom = 15.9994 amu 3Ca: (3 atoms) (40.078 amu/atom)=120.234amu 2P: (2 atoms) (30.973761amu/atom)=61.947522amu Add together the masses to give the formula mass. + 8O: (8 atoms)(15.9994amu/atom)=127.9952amu Formula mass of Ca3(PO4)2=310.177amu  EXERCISE 6.4.2: SILICON NITRIDE Calculate the formula mass of Si N , commonly called silicon nitride. It is an extremely hard and inert material that is used to make 3 4 cutting tools for machining hard metal alloys. Figure 6.4.2: Si 3 N 4 bearing parts. (Public Domain; David W. Richerson and Douglas W. Freitag; Oak Ridge National Laboratory). Answer 140.29 amu MOLAR MASS The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of 6.4.2 https://chem.libretexts.org/@go/page/47490 grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively. The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole. The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol: Table 6.4.1: Molar Mass of Select Substances Substance (formula) Basic Unit Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol) carbon (C) atom 12.011 (atomic mass) 12.011 ethanol (C2H5OH) molecule 46.069 (molecular mass) 46.069 calcium phosphate [Ca3(PO4)2] formula unit 310.177 (formula mass) 310.177 CONVERTING BETWEEN GRAMS AND MOLES OF A COMPOUND The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride (CaCl ). Since calcium chloride is a solid, it would be convenient to use a 2 balance to measure the mass that is needed. Dimensional analysis will allow you to calculate the mass of CaCl that you should measure as 2 shown in Example 6.4.3.  EXAMPLE 6.4.3: CALCIUM CHLORIDE Calculate the mass of 3.00 moles of calcium chloride (CaCl2). Figure 6.4.3: Calcium chloride is used as a drying agent and as a road deicer. Solution Solutions to Example 6.4.3 Steps for Problem Solving Calculate the mass of 3.00 moles of calcium chloride Given: 3.00 moles of CaCl Identify the "given" information and what the problem is asking you to "find." 2 Find: g CaCl 2 List other known quantities. 1 mol CaCl = 110.98 g CaCl 2 2 Prepare a concept map and use the proper conversion factor. 110.98 g CaCl 2 Cancel units and calculate. 3.00 mol CaCl 2 × = 333 g CaCl 2 1 mol CaCl 2 Think about your result. 6.4.3 https://chem.libretexts.org/@go/page/47490   EXERCISE 6.4.3: CALCIUM OXIDE What is the mass of 7.50 mol of (calcium oxide) CaO? Answer 420.60 g  EXAMPLE 6.4.4: WATER How many moles are present in 108 grams of water? Solution Solutions to Example 6.4.4 Steps for Problem Solving How many moles are present in 108 grams of water? Given: 108 g H O Identify the "given" information and what the problem is asking you to "find." 2 Find: mol H O 2 List other known quantities. 1 mol H O = 18.02 g 2 H2O Prepare a concept map and use the proper conversion factor. 1 mol H O 2 Cancel units and calculate. 108 g H O × 2 = 5.99 mol H O 2 18.02 g H O 2 Think about your result.  EXERCISE 6.4.4: NITROGEN GAS What is the mass of 7.50 mol of Nitrogen gas N ? 2 Answer 210 g CONVERSIONS BETWEEN MASS AND NUMBER OF PARTICLES In "Conversions Between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to moles. To convert from mass to number of particles or vice- versa, it will first require a conversion to moles as shown in Figure 6.4.1 and Example 6.4.5. Figure 6.4.4: Conversion from number of particles to mass, or from mass to number of particles, requires two steps. To convert from number of particles to moles, use mol/Avogrado's #, an d to convert from moles to mass, use g/mol.  EXAMPLE 6.4.5: CHLORINE How many molecules is 20.0 g of chlorine gas, Cl ? 2 Solution 6.4.4 https://chem.libretexts.org/@go/page/47490  Solutions to Example 6.4.5 Steps for Problem Solving How many molecules is 20.0 g of chlorine gas, Cl ? 2 Identify the "given" information and what the problem is Given: 20.0 g Cl 2 asking you to "find." Find: # Cl molecules 2 1 mol Cl = 70.90 g Cl , List other known quantities. 2 2 1mol Cl = 6.022 × 10 2 23 Cl 2 molecules Prepare a concept map and use the proper conversion factor. The conversion factors are 1 mole Cl 2 over 70.90 grams Cl 2 , and 2 6. 0 2 2 × 1 0 3 Cl 2 molecules over 1 mole Cl. 2 1 mol Cl 6.02 × 10 23 molecules Cl 2 2 20.0 g Cl × × Cancel units and calculate. 2 70.90 g Cl 1 mol Cl 2 2 23 = 1.70 × 10 molecules Cl 2 Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is Think about your result. less than half of Avogadro's number.  EXERCISE 6.4.5: CALCIUM CHLORIDE How many formula units are in 25.0 g of CaCl ? 2 Answer 1.36 x 1023 CaCl formula units 2 SUMMARY Calculations for formula mass and molecular mass are described. Calculations involving conversions between moles of a material and the mass of that material are described. Calculations are illustrated for conversions between mass and number of particles. This page titled 6.4: Counting Molecules by the Gram is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.4.5 https://chem.libretexts.org/@go/page/47490 6.5: Chemical Formulas as Conversion Factors  Learning Objectives Use chemical formulas as conversion factors. Figure 6.5.1 shows that we need 2 hydrogen atoms and 1 oxygen atom to make one water molecule. If we want to make two water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make five molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom. Figure 6.5.1 Water Molecules. The ratio of hydrogen atoms to oxygen atoms used to make water molecules is always 2:1, no matter how many water molecules are being made. Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of water (H2O) we can construct the relationships given in (Table 6.5.1). Table 6.5.1 : Molecular Relationships for Water 1 Molecule of H 2O Has 1 Mol of H 2O Has Molecular Relationships 2 mol H atoms or 2 H atoms 2 mol of H atoms 1 mol H2 O molecules 1 mol H2 O molecules 2 mol H atoms 1 mol O atoms or 1 O atom 1 mol of O atoms 1 mol H2 O molecules 1 mol H2 O molecules 1 mol O atoms  The Mole is big A mole represents a very large number! The number 602,214,129,000,000,000,000,000 looks about twice as long as a trillion, which means it’s about a trillion trillion. (CC BY-SA NC; what if? [what-if.xkcd.com]). A trillion trillion kilograms is how much a planet weighs. If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times. Table 6.5.2 : Molecular and Mass Relationships for Ethanol 1 Molecule of C2 H6 O Has 1 Mol of C 2 H6 O Has Molecular and Mass Relationships 2 mol C atoms or 1 mol C2 H6 O molecules 2 C atoms 2 mol of C atoms 1 mol C2 H6 O molecules 2 mol C atoms 6 mol H atoms or 1 mol C2 H6 O molecules 6 H atoms 6 mol of H atoms 1 mol C2 H6 O molecules 6 mol H atoms 6.5.1 https://chem.libretexts.org/@go/page/47491  1 Molecule of C 2 H6 O Has 1 Mol of C 2 H6 O Has Molecular and Mass Relationships 1 mol O atoms or 1 mol C2 H6 O molecules 1 O atom 1 mol of O atoms 1 mol C2 H6 O molecules 1 mol O atoms 24.02 g C or 2 (12.01 amu) C 2 (12.01 g) C 1 mol C2 H6 O molecules 24.02 amu C 24.02 g C 1 mol C2 H6 O molecules 24.02 g C 6.048 g H or 6 (1.008 amu) H 6 (1.008 g) H 1 mol C2 H6 O molecules 6.048 amu H 6.048 g H 1 mol C2 H6 O molecules 6.048 g H 16.00 g O or 1 (16.00 amu) O 1 (16.00 g) O 1 mol C2 H6 O molecules 16.00 amu O 16.00 g O 1 mol C2 H6 O molecules 16.00 g O The following example illustrates how we can use the relationships in Table 6.5.2 as conversion factors.  Example 6.5.1: Ethanol If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms does it have? Solution Solutions to Example 6.5.1 If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles Steps for Problem Solving of carbon atoms does it have? Identify the "given" information and what the problem is asking you to Given: 2.5 mol C2H6O "find." Find: mol C atoms List other known quantities. 1 mol C2H6O = 2 mol C Prepare a concept map and use the proper conversion factor. Note how the unit mol C2H6O molecules cancels algebraically. Cancel units and calculate. 2.5 mol C2 H6 O molecules × 2 mol C atoms = 5.0 mol C atoms 1 mol C2 H6 O molecules There are twice as many C atoms in one C2H6O molecule, so the final Think about your result. amount should be double.  Exercise 6.5.1 If a sample contains 6.75 mol of Na2SO4, how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have? Answer 13.5 mol Na atoms, 6.75 mol S atoms, and 27.0 mol O atoms The fact that 1 mol equals 6.022 × 1023 items can also be used as a conversion factor.  Example 6.5.2: Oxygen Mass Determine the mass of Oxygen in 75.0g of C2H6O. Solution Solutions to Example 6.5.2 6.5.2 https://chem.libretexts.org/@go/page/47491  Steps for Problem Solving Determine the mass of Oxygen in 75.0g of C2H6O Identify the "given" information and what the problem is asking you to Given: 75.0g C2H6O "find." Find: g O 1 mol O = 16.0g O List other known quantities. 1 mol C2H6O = 1 mol O 1 mol C2H6O = 46.07g C2H6O Prepare a concept map and use the proper conversion factor. The conversion factors are 1 mol C2H6O over 46.07 g C2H6O, 1 mol O over 1 mol C2H6O, and 16.00 g O over 1 mole O. 1 mol C2 H6 O 1 mol O 16.00 g O Cancel units and calculate. 75.0 g C2 H6 O × × × = 26.0 g 46.07 g C2 H6 O 1 mol C2 H6 O 1 mol O Think about your result.  Exercise 6.5.2 a. How many molecules are present in 16.02 mol of C4H10? How many C atoms are in 16.02 mol? b. How many moles of each type of atom are in 2.58 mol of Na2SO4? Answer a: 9.647 x 1024 C4H10 molecules and 3.859 x 1025 C atoms Answer b: 5.16 mol Na atoms, 2.58 mol S atoms, and 10.3 mol O atoms Summary In any given formula, the ratio of the number of moles of molecules (or formula units) to the number of moles of atoms can be used as a conversion factor. 6.5: Chemical Formulas as Conversion Factors is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.5.3 https://chem.libretexts.org/@go/page/47491 6.6: Mass Percent Composition of Compounds  Learning Objectives Determine percent composition of each element in a compound based on mass. Packaged foods that you eat typically have nutritional information provided on the label. The label on a jar of peanut butter reveals that one serving size is considered to be 32 g. The label also gives the masses of various types of compounds that are present in each serving. One serving contains 7 g of protein, 15 g of fat, and 3 g of sugar. By calculating the fraction of protein, fat, or sugar in one serving size of peanut butter and converting to percent values, we can determine the composition of peanut butter on a percent by mass basis. Percent Composition Chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound. It is calculated in a similar way to that of the composition of the peanut butter. mass of element % by mass = × 100% mass of compound The sample problem below shows the calculation of the percent composition of a compound based on mass data.  Example 6.6.1: Percent Composition from Mass Data A certain newly synthesized compound is known to contain the elements zinc and oxygen. When a 20.00 g sample of the compound is decomposed, 16.07 g of zinc remains. Determine the percent composition of the compound. Solution Solutions to Example 6.6.1 When a 20.00 g sample of the zinc-and-oxygen compound is Steps for Problem Solving decomposed, 16.07 g of zinc remains. Determine the percent composition of the compound. Given : Mass of compound = 20.00 g Identify the "given" information and what the problem is asking you Mass of Zn = 16.07 g to "find." Find: % Composition (% Zn and %O) Subtract to find the mass of oxygen in the compound. Divide each element's mass by the mass of the compound to find the percent by List other known quantities. mass. Mass of oxygen = 20.00 g - 16.07 g = 3.93 g O 16.07 g Zn % Zn = × 100% = 80.35% Zn 20.00 g 3.93 g O %O = × 100% = 19.65% O Cancel units and calculate. 20.00 g Calculate the percent by mass of each element by dividing the mass of that element by the mass of the compound and multiplying by 100%. The calculations make sense because the sum of the two percentages Think about your result. adds up to 100%. By mass, the compound is mostly zinc.  Exercise 6.6.1 Sulfuric acid, H2SO4 is a very useful chemical in industrial processes. If 196.0 g of sulfuric acid contained 64.0g oxygen and 4.0 g of hydrogen, what is the percent composition of the compound? 6.6.1 https://chem.libretexts.org/@go/page/47492  Answer 2.04% H, 32.65% S, and 65.3% O Summary Processes are described for calculating the percent composition of a compound based on mass. 6.6: Mass Percent Composition of Compounds is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar- Agnew & Henry Agnew. 6.6.2 https://chem.libretexts.org/@go/page/47492 6.7: Mass Percent Composition from a Chemical Formula  Learning Objectives Determine the percent composition of each element in a compound from the chemical formula. The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. This is divided by the molar mass of the compound and multiplied by 100%. mass of element in 1 mol % by mass = × 100% molar mass of compound The percent composition of a given compound is always the same, given that the compound is pure.  Example 6.7.1 Dichlorine heptoxide (Cl O ) is a highly reactive compound used in some organic synthesis reactions. Calculate the percent 2 7 composition of dichlorine heptoxide. Solution Solutions to Example 6.7.1 Calculate the percent composition of dichlorine heptoxide Steps for Problem Solving (Cl O ). 2 7 Identify the "given" information and what the problem is asking you Given : Cl2O7 to "find." Find: % Composition (% Cl and %O) Mass of Cl in 1 mol Cl2O7 , 2 Cl : 2 x 35.45 g = 70.90 g List other known quantities. Mass of O in 1 mol Cl2O7 , 7 O: 7 x 16.00 g = 112.00 g Molar mass of Cl2O7 = 182.90 g/mol 70.90 g Cl %Cl = × 100% = 38.76% Cl 182.90 g 112.00 g O %O = × 100% = 61.24% O Cancel units and calculate. 182.90 g Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by 100%. Think about your result. The percentages add up to 100%. Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorine heptoxide is 38.76% Cl and 61.24% O. Suppose that you needed to know the masses of chlorine and oxygen present in a 12.50 g sample of dichlorine heptoxide. You can set up a conversion factor based on the percent by mass of each element. 38.76 g Cl 12.50 g Cl O × = 4.845 g Cl 2 7 100 g Cl O 2 7 61.24 g O 12.50 g Cl O × = 7.655 g O 2 7 100 g Cl O 2 7 The sum of the two masses is 12.50 g, the mass of the sample size. 6.7.1 https://chem.libretexts.org/@go/page/47493   Exercise 6.7.1 Barium fluoride is a transparent crystal that can be found in nature as the mineral frankdicksonite. Determine the percent composition of barium fluoride. Answer a: 78.32% Ba and 21.67% F Summary Processes are described for calculating the percent composition of a compound based on the chemical formula. 6.7: Mass Percent Composition from a Chemical Formula is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.7.2 https://chem.libretexts.org/@go/page/47493 6.8: Calculating Empirical Formulas for Compounds  Learning Objectives Define empirical formula. Determine empirical formula from percent composition of a compound. In the early days of chemistry, there were few tools for the detailed study of compounds. Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials. The "new" field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely. The relative amounts of elements could be determined, but so many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios. We did not know exactly how many of these atoms were actually in a specific molecule. Determining Empirical Formulas An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios because if we know the molar amounts of each element in a compound, we can determine the empirical formula. In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it. These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula.  Empirical Formula: In Steps Steps to determine empirical formula: 1. Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams. 2. Use each element's molar mass to convert the grams of each element to moles. 3. In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest. 4. If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element. 5. In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply each of the moles by the smallest whole number that will convert each into a whole number. Write the empirical formula.  Example 6.8.1 A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen. Find the empirical formula of the compound. Solution Solutions to Example 6.8.1 Find the empirical formula of a compound of 69.94% iron and Steps for Problem Solving 30.06% oxygen. Given: Identify the "given" information and what the problem is asking you % of Fe = 69.94% to "find." % of O = 30.06% Find: Empirical formula = Fe ? O? Calculate 6.8.1 https://chem.libretexts.org/@go/page/47494  Find the empirical formula of a compound of 69.94% iron and Steps for Problem Solving 30.06% oxygen. 69.94 g Fe a. Assume a 100 g sample, convert the same % values to grams. 30.06 g O 1 mol Fe 69.94 g Fe × = 1.252 mol Fe 55.85 g Fe b. Convert to moles. 1 mol O 30.06 g O × = 1.879 mol O 16.00 g O 1.252 mol Fe : 1.252 1.879 mol c. Divide both moles by the smallest of the results. O : 1.252 The "non-whole number" empirical formula of the compound is Fe O1.5 1 Fe:O = 2 (1:1.5) = 2:3 Multiply each of the moles by the smallest whole number that will Since the moles of O is still not a whole number, both moles can be convert each into a whole number. multiplied by 2, while rounding to a whole number. Write the empirical formula. The empirical formula of the compound is Fe 2 O 3. The subscripts are whole numbers and represent the mole ratio of the Think about your result. elements in the compound. The compound is the ionic compound iron (III) oxide.  Exercise 6.8.1 Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula? Answer HgCl2 Summary A process is described for the calculation of the empirical formula of a compound, based on the percent composition of that compound. 6.8: Calculating Empirical Formulas for Compounds is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.8.2 https://chem.libretexts.org/@go/page/47494 6.9: Calculating Molecular Formulas for Compounds  Learning Objectives Understand the difference between empirical formulas and molecular formulas. Determine molecular formula from percent composition and molar mass of a compound. Below, we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar. Some people can distinguish them on the basis of taste, but it's not a good idea to go around tasting chemicals. The best way to tell glucose and sucrose apart is to determine the molar masses—this approach allows you to easily tell which compound is which. Figure 6.9.1 : (A) the molecular structure of glucose and (B) the molecular structure of sucrose. Molecular Formulas Molecular formulas give the kind and number of atoms of each element present in the molecular compound. In many cases, the molecular formula is the same as the empirical formula. The chemical formula will always be some integer multiple (n ) of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula). Molecular Formula = n(Empirical formula) therefore Molecular Formula n = Empirical Formula The integer multiple, n, can also be obtained by dividing the molar mass, MM , of the compound by the empirical formula mass, EF M (the molar mass represented by the empirical formula). M M (molarmass) n = EF M (empiricalf ormulamolarmass) Table 6.9.1 shows the comparison between the empirical and molecular formula of methane, acetic acid, and glucose, and the different values of n. The molecular formula of methane is CH and because it contains only one carbon atom, that is also its 4 empirical formula. Sometimes, however, the molecular formula is a simple whole number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is C H O. Glucose is a simple sugar that 2 4 2 cells use as a primary source of energy. Its molecular formula is C H O. The structures of both molecules are shown in Figure 6 12 6 6.9.2. They are very different compounds, yet both have the same empirical formula of CH O. 2 Table 6.9.1 : Molecular Formula and Empirical Formula of Various Compounds. Name of Compound Molecular Formula Empirical Formula n Methane CH 4 CH 4 1 Acetic acid C H O 2 4 2 CH O 2 2 Glucose C H 6 12 O 6 CH O 2 6 6.9.1 https://chem.libretexts.org/@go/page/47495  Figure 6.9.2 : Acetic acid (left) has a molecular formula of C 2 H O 4 2 , while glucose (right) has a molecular formula of C H 6 12 O 6. Both have the empirical formula CH O. 2 Empirical formulas can be determined from the percent composition of a compound as discussed in section 6.8. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps: 1. Calculate the empirical formula molar mass (EFM). 2. Divide the molar mass of the compound by the empirical formula molar mass. The result should be a whole number or very close to a whole number. 3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.  Example 6.9.1 The empirical formula of a compound of boron and hydrogen is BH. Its molar mass is 27.7 g/mol. Determine the molecular 3 formula of the compound. Solution Solutions to Example 6.9.1 Steps for Problem Solving Determine the molecular formula of BH. 3 Given: Identify the "given" information and what the problem is asking you Empirical formula = BH 3 to "find." Molar mass = 27.7 g/mol Find: Molecular formula =? Calculate the empirical formula mass (EFM). Empirical formula molar mass (EFM) = 13.84 g/mol Divide the molar mass of the compound by the empirical formula 27.7g/mol molar mass mass. The result should be a whole number or very close to a whole = = 2 EFM 13.84g/mol number. Multiply all the subscripts in the empirical formula by the whole BH ×2 = B H number found in step 2. The result is the molecular formula. 3 2 6 Write the molecular formula. The molecular formula of the compound is B 2 H 6. The molar mass of the molecular formula matches the molar mass of Think about your result. the compound.  Exercise 6.9.1 Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What are the empirical and chemical formulas for ascorbic acid? Answer Empirical Formula C3H4O3 Answer Molecular Formula C6H8O6 6.9.2 https://chem.libretexts.org/@go/page/47495 Summary A procedure is described that allows the calculation of the exact molecular formula for a compound. 6.9: Calculating Molecular Formulas for Compounds is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 6.9.3 https://chem.libretexts.org/@go/page/47495