Chemistry
Empirical and Molecular Formula
The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of each type of atom in a molecule. The empirical formula can be determined from the molecular formula by dividing the subscripts by the greatest common factor. Both formulas provide important information about the composition of a compound.
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eBook - PDFIntroductory Chemistry
An Active Learning Approach
- Mark Cracolice, Edward Peters, Mark Cracolice(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
To find the amount of any element in a known amount of compound, use percentage as a con- version factor, based on the equivalency: grams of the element per 100 grams of the compound. Goal 11 Distinguish between an empirical formula and a molecular formula. An empirical formula shows the simplest whole-number ratio of atoms of the elements in a compound. Empirical formulas are calculated from percentage composition data. They may also be calculated from the mass of each element in a sample of a compound. Empirical formulas may or may not be the actual molecular formulas of compounds. The molar mass of the com- pound is needed to determine molecular formulas from empirical formulas. Goal 12 Given data from which the mass of each element in a sample of a compound can be determined, find the empirical formula of the compound. To determine an empirical formula: 1. Determine the percentage composition by mass or the mass of each element in a sample of the compound. 2. Convert the masses into amount in moles of atoms of the different elements. 3. Determine the amount ratio of moles of atoms. 4. Express the amounts of atoms as the smallest possible ratio of integers. 5. Write the empirical formula, using the number for each atom in the integer ratio as the subscript in the formula. Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 257 Small-Group Discussion Questions Goal 13 Given the molar mass and empirical formula of a compound, or information from which they can be found, determine the molecular formula of the compound.
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Molecular Formulas from Empirical Formulas and Molecular Masses The empirical formula is the accepted formula unit for ionic compounds. For molecular com- pounds, however, chemists prefer molecular formulas because they give the number of atoms of each type in a molecule, rather than just the simplest ratio of moles of elements in a com- pound as the empirical formula does. Sometimes an Empirical and Molecular Formula are the same. Two examples are H 2 O and NH 3 . When they are different, the subscripts of the molecular formula are integer multiples of those in the empirical formula. The subscripts of the molecular formula P 4 O 10 , for example, are each two times those in the empirical formula, P 2 O 5 , as you saw earlier. The molecular mass of P 4 O 10 is likewise two times the empirical formula mass of P 2 O 5 . This observation pro- vides us with a way to find out the molecular formula for a compound provided we have a way of determining experimentally the molecular mass of the compound. If the experimental molecular mass equals the calculated empirical formula mass, the empirical formula is the same as the molecular formula. Otherwise, the molecular mass will be a whole-number mul- tiple of the empirical formula mass. Whatever the integer is, it’s a common multiplier for the subscripts of the empirical formula. NOTE There are many methods for determining molecular masses. Some are discussed in Chapters 10 and 12. Instruments such as the mass spectrometers described in Chapter 2 can also be used. 136 CHAPTER 3 The Mole and Stoichiometry EXAMPLE 3.12 Determining a Molecular Formula from an Empirical Formula and a Molecular Mass Styrene, the raw material for polystyrene foam plastics, such as packing peanuts or insulation, has an empirical formula of CH.
eBook - PDF- Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2015(Publication Date)
- Openstax(Publisher)
Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text. Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n: molecular or molar mass ⎛ ⎝ amu or g mol ⎞ ⎠ empirical formula mass ⎛ ⎝ amu or g mol ⎞ ⎠ = n formula units/molecule The molecular formula is then obtained by multiplying each subscript in the empirical formula by n, as shown by the generic empirical formula A x B y : (A x B y ) n = A nx B nx For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom).
eBook - PDF- Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2019(Publication Date)
- Openstax(Publisher)
These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are: Thus, this compound may be represented by the formula C 0.142 H 0.284 . Per convention, formulas contain whole- number subscripts, which can be achieved by dividing each subscript by the smaller subscript: (Recall that subscripts of “1” are not written but rather assumed if no other number is present.) The empirical formula for this compound is thus CH 2 . This may or not be the compound’s molecular formula 3.2 • Determining Empirical and Molecular Formulas 131 as well; however, additional information is needed to make that determination (as discussed later in this section). Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of: In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl 2 O 7 as the final empirical formula. In summary, empirical formulas are derived from experimentally measured element masses by: 1. Deriving the number of moles of each element from its mass 2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula 3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained Figure 3.11 outlines this procedure in flow chart fashion for a substance containing elements A and X. FIGURE 3.11 The empirical formula of a compound can be derived from the masses of all elements in the sample.
eBook - PDF- Morris Hein, Susan Arena, Cary Willard(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
The molecular formula is the true formula, representing the total number of atoms of each element present in one molecule of a compound. It is entirely possible that two or more substances will have the same percent composition, yet be distinctly different com- pounds. For example, acetylene (C 2 H 2 ) is a common gas used in welding; benzene (C 6 H 6 ) is an important solvent obtained from coal tar and is used in the synthesis of styrene and nylon. Both acetylene and benzene contain 92.3% C and 7.7% H. The smallest ratio of C and H corresponding to these percentages is CH (1:1). Therefore, the empirical formula for both acetylene and benzene is CH, even though the molecular formulas are C 2 H 2 and C 6 H 6 , respectively. Often the molecular formula is the same as the empirical formula. If the molecular formula is not the same, it will be an integral (whole-number) multiple of the empirical formula. For example, CH = empirical formula (CH) 2 = C 2 H 2 = acetylene ( molecular formula ) (CH) 6 = C 6 H 6 = benzene ( molecular formula ) Table 7.1 compares the formulas of these substances. Table 7.2 shows Empirical and Molecular Formula relationships of other compounds. TABLE 7.1 Molecular Formulas of Two Compounds Having an Empirical Formula with a 1:1 Ratio of Carbon and Hydrogen Atoms Composition Formula % C % H Molar mass CH (empirical) 92.3 7.7 13.02 (empirical) C 2 H 2 (acetylene) 92.3 7.7 26.04 (2 × 13.02) C 6 H 6 (benzene) 92.3 7.7 78.12 (6 × 13.02) CHECK YOUR UNDERSTANDING 7.9 Molecular Formula TABLE 7.2 Some Empirical and Molecular Formulas Substance Empirical formula Molecular formula Substance Empirical formula Molecular formula Acetylene CH C 2 H 2 Diborane BH 3 B 2 H 6 Benzene CH C 6 H 6 Hydrazine NH 2 N 2 H 4 Ethylene CH 2 C 2 H 4 Hydrogen H H 2 Formaldehyde CH 2 O CH 2 O Chlorine Cl Cl 2 Acetic acid CH 2 O C 2 H 4 O 2 Bromine Br Br 2 Glucose CH 2 O C 6 H 12 O 6 Oxygen O O 2 Hydrogen chloride HCl HCl Nitrogen N N 2 Carbon dioxide CO 2 CO 2 Iodine I I 2
eBook - PDFGeneral Chemistry I as a Second Language
Mastering the Fundamental Skills
- David R. Klein(Author)
- 2015(Publication Date)
- Wiley(Publisher)
Both 4 and 10 can be divided by 2, giving C 2 H 5 . Therefore, the empirical formula is C 2 H 5 . Notice that an empirical formula uses the smallest whole numbers that express the ratio. So, your answer is not CH 2.5 even though that expresses the correct ratio. For each of the following molecular formulas below, write the empirical for- mula next to it. In some cases, the empirical formula might be the same as the mo- lecular formula. 2.2. C 2 H 6 2.3. C 3 H 8 2.4. CO 2 2.5. H 2 SO 4 2.6. C 6 H 12 O 6 2.7. Hg 2 I 2 2.8. Al 2 O 3 2.9. C 5 H 5 N Empirical formulas are mostly an historical artifact. In the old days, chemists determined the formulas of compounds through combustion analysis. This technique allowed for chemists to determine empirical formulas, rather than molecular for- mulas. But in applying modern-day technology, we have all kinds of fancy tech- niques (NMR, Mass Spec, X-Ray Diffraction, etc.) that allow us to easily determine the molecular formula. It is somewhat of a mystery why textbooks continue to teach empirical formulas. For whatever reason, empirical formulas have not yet left the textbooks, so you will need to know what they are. But for the most part, you will not use empirical formulas as you move along in this course. The focus will be on molecular formulas. In the rest of this chapter, most of the problems will start by giving a molec- ular formula. You will soon see that the molecular formula is a very valuable piece of information, because you can usually extract the information you need from the molecular formula. It is a common mistake for students to miss this point. When a problem says: “22.4 g of Al 2 O 3 . . .” you have actually been given two pieces of information: 22.4 g and Al 2 O 3 . If you fail to see the second piece of information, you will not have enough information to solve the problem. We will see many ex- amples of this soon. But first, we have to go over a few more basics. 2.1 Empirical and Molecular FormulaS 29
eBook - PDF- Steven Murov(Author)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
The determination of the molecular formula from the empirical formula requires an additional measurement such as the molecular mass. If a molecular mass measurement for the above compound results in 150 ± 2 g/mole, the molecular formula would be calculated as C 5 H 10 O 5 . This means the compound could be a five carbon sugar such as ribose. molecular formula = molecular mass x subscripts of empirical formula = 150 x CH 2 O = C 5 H 10 O 5 empirical form. mass 30 The Empirical and Molecular Formulas can go a long way in helping to identify a compound. Almost all ionic compounds are solids at room temperature. The crystal structure of the ionic solid consists of a repeating array, and it is not straightforward to define the smallest unit (a molecule) of the substance that still would have the properties of the substance. Thus it is better to refer to the formula mass of an ionic substance rather than its molecular mass. For the same reasons, the formula calculations for ionic compounds should be reduced to the lowest whole number ratio or the empirical formula. Except for a few unusual cases, the term molecular formula is not applicable for ionic compounds. For metals with only one common oxidation state, it is usually possible to predict with a high degree of confidence, the empirical formula that will result from the reaction of a metal with a non-metal. For instance, based on common oxidation numbers, sodium chloride and sodium oxide should be and are NaCl and Na 2 O respectively. Aluminum chloride and aluminum oxide should be and are AlCl 3 and Al 2 O 3 . When there are multiple oxidation states of a metal, more than one compound is possible. For instance, copper and oxygen form Cu 2 O and CuO and more information is needed to distinguish between the two. When naming a copper oxide, it is very important that the name distinguish between the two possibilities [copper(I) oxide or cuprous oxide and copper(II) oxide or cupric oxide].
eBook - PDF- Donald Pavia, Gary Lampman, George Kriz, James Vyvyan, Donald Pavia, Gary Lampman, George Kriz, James Vyvyan(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
MOLECULAR FORMULAS AND WHAT CAN BE LEARNED FROM THEM B efore attempting to deduce the structure of an unknown organic compound from an exam-ination of its spectra, we can simplify the problem somewhat by examining the molecular formula of the substance. The purpose of this chapter is to describe how the molecular formula of a compound is determined and how structural information may be obtained from that formula. The chapter reviews both the modern and classical quantitative methods of determining the molecular formula. While use of the mass spectrometer (Chapter 3) can supplant many of these quantitative analytical methods, they are still in use. Many journals still require that a satis-factory quantitative elemental analysis (Section 1.1) be obtained prior to the publication of research results. 1 C H A P T E R 1 1.1 ELEMENTAL ANALYSIS AND CALCULATIONS The classical procedure for determining the molecular formula of a substance involves three steps: 1. A qualitative elemental analysis to find out what types of atoms are present: C, H, N, O, S, Cl, and so on. 2. A quantitative elemental analysis (or microanalysis ) to find out the relative numbers (per-centages) of each distinct type of atom in the molecule. 3. A molecular mass (or molecular weight ) determination . The first two steps establish an empirical formula for the compound. When the results of the third procedure are known, a molecular formula is found. Virtually all organic compounds contain carbon and hydrogen. In most cases, it is not neces-sary to determine whether these elements are present in a sample: their presence is assumed. However, if it should be necessary to demonstrate that either carbon or hydrogen is present in a compound, that substance may be burned in the presence of excess oxygen. If the combustion produces carbon dioxide, carbon must be present; if combustion produces water, hydrogen atoms must be present.
eBook - PDFChemistry
The Molecular Nature of Matter
- James E. Brady, Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
© Envision/Corbis Images The Mole and Stoichiometry Chapter Outline 3.1 | The Mole and Avogadro’s Number 3.2 | The Mole, Formula Mass, and Stoichiometry 3.3 | Chemical Formula and Percentage Composition 3.4 | Determining Empirical and Molecular Formulas 3.5 | Stoichiometry and Chemical Equations 3.6 | Limiting Reactants 3.7 | Theoretical Yield and Percentage Yield 3 108 After reading this chapter, you should be able to: I n this chapter we will learn the fundamentals of chemical calculations called stoichiometry (stoy-kee-AH-meh-tree), which loosely translates as “the measure of the elements.” These calculations are key for success in the chemistry laboratory. You will also find this chapter to be important for future courses in organic chemistry, biochemistry, and almost any other advanced laboratory course in the sciences. Stoichiometry involves converting chemical formulas and equations that represent individual atoms, molecules, and formula units to the laboratory scale that uses milligrams, grams, and even kilograms of these substances, just as the number of M&Ms in the bowl can be measured by weighing the M&Ms.To do this, we introduce the concept of the mole. The mole allows the chemist to scale up from the atomic/molecular level to the laboratory scale, much as a fast-food restaurant scales up the amount of ingredients from a single hamburger to mass-production as in Figure 3.1. Our stoichiometric calculations are usually conversions from one set of units to another using dimensional analysis. To be successful using dimensional analysis calculations we need two things: a knowledge of the equalities that can be made into conversion factors and a logical sequence of steps to guide us from the starting set of units to the desired units. Figure 3.6, at the end of this chapter, is a flowchart that organizes the sequence of conversion steps and the conversion factors used in stoichiometric calculations in this chapter.
- Frank H. Clarke(Author)
- 2013(Publication Date)
- Academic Press(Publisher)
II. CALCULATIONS A. Program 11, Percentage Composition For program 11, the percentage of an element present in a compound is provided by Eq (1-1): _ _. 4 Weight sum of the element 1ΛΛ (Λ 1Λ % Element = *rz-.: :—— r-r-x 100 (1-1) Molecular weight 8 1. MOLECULAR FORMULAS The weight sum is the atomic weight of the element multiplied by the number of gram-atoms present. For water, the molecular weight of water is multiplied by the number of moles of water and substituted as its weight sum. The atomic weights are those provided in the Merck Index, Ninth Edition, 1976. B. Program 12, Empirical Formula For program 12, the atomic ratio of each element in a compound is obtained by dividing the percentage of the element present in the com-pound by its atomic weight. When these ratios are normalized, the result is the empirical formula. The molecular formula may actually be a multi-ple of the empirical formula. Program 12 assumes that carbon, hydrogen, and nitrogen analyses are available. Unless oxygen is the only other element present, only carbon, hydrogen, and nitrogen contributions to the empirical formula are ob-tained. When oxygen is present, and there are no additional elements, the percentage of oxygen is calculated as the difference between 100% and the sum of the percentages of carbon, hydrogen, and nitrogen. The empirical formula is a multiple of the least-abundant element for each of the other elements, respectively. This is a simple calculation when only carbon, hydrogen, and nitrogen are involved. One simply divides the atomic ratios of carbon and hydrogen by that of nitrogen and rounds off the numbers thus obtained. When oxygen is the fourth element and is least abundant, one is tempted to simply divide each of the other ratios by that of oxygen. This does not give a satisfactory result in most situations because the ratio for oxygen is based on an analysis that was obtained indirectly.
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