Molar Mass Calculations

12 Key excerpts on "Molar Mass Calculations"

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  eBook - PDF

  Foundations of College Chemistry

  • Morris Hein, Susan Arena, Cary Willard(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  The area of chemistry that deals with quantitative relationships among reactants and products is known as stoichiometry (stoy-key-ah-meh-tree). Solving problems in stoichiometry requires the use of moles in the form of mole ratios. The chemist also finds it necessary to calculate amounts of products or reactants by using a balanced chemical equation. With these calculations, the chemist can control the amount of product by scaling the reaction up or down to fit the needs of the laboratory and can thereby minimize waste or excess materials formed during the reaction. A Short Review Molar mass The sum of the atomic masses of all the atoms in an element or com- pound is called molar mass. The term molar mass also applies to the mass of a mole of any formula unit—atoms, molecules, or ions; it is the atomic mass of an atom or the sum of the atomic masses in a molecule or an ion (in grams). Relationship between molecule and mole. A molecule is the smallest unit of a molecular substance (e.g., Br 2 ), and a mole is Avogadro’s number (6.022 × 10 23 ) of molecules of that substance. A mole of bromine (Br 2 ) has the same number of mol- ecules as a mole of carbon dioxide, a mole of water, or a mole of any other molecu- lar substance. When we relate molecules to molar mass, 1 molar mass is equivalent to 1 mol, or 6.022 × 10 23 molecules. The term mole also refers to any chemical species. It represents a quantity (6.022 × 10 23 particles) and may be applied to atoms, ions, electrons, and formula units of nonmolecular substances. In other words, 1 mole = { 6.022 × 10 23 molecules 6.022 × 10 23 formula units 6.022 × 10 23 atoms 6.022 × 10 23 ions Other useful mole relationships are molar mass = grams of a substance number of moles of the substance molar mass = grams of a monatomic element number of moles of the element number of moles = number of molecules 6.022 × 10 23 molecules mole Balanced equations.

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  Basic Concepts of Chemistry

  • Leo J. Malone, Theodore O. Dolter(Authors)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  The molar mass of a compound is the formula weight of the compound expressed in grams and is the mass of Avogadro’s number of molecules or ionic formula units. One important purpose of this chapter is for you to become comfortable with the conversions among moles, mass, and numbers of atoms or molecules. The two conversion factors that are used for this are the molar mass and Avogadro’s number. The formula of a compound implies a mole ratio. It can be used to determine the number of moles, the masses, and the percent composition of each element in a compound. In the reverse calculation, the empirical formula of a compound can be determined from its mass composition or percent composition. One needs to know the molar mass of a compound to determine the molecular formula from the empirical formula. The molecular formula is determined from the percent composition and the molar mass. C H A P T E R S U M M A RY Chapter Problems 173 a. C 4 H 5 b. C 8 H 10 c. 1.44 * 10 23 d. 9.50% ANSWERS TO CHAPTER 5 SYNTHESIS PROBLEM C H A P T E R P R O B L E M S Relative Masses of Particles (SECTION 5-1) 5-1. One penny weighs 2.47 g and one nickel weighs 5.03 g. What mass of pennies has the same number of coins as there are in 12.4 lb of nickels? 5-2. A small glass bead weighs 310 mg and a small marble weighs 8.55 g. A quantity of small glass beads weighs 5.05 kg. What does the same number of marbles weigh? 5-3. A piece of pure gold has a mass of 145 g. What is the mass of the same number of silver atoms? 5-4. A large chunk of pure aluminum has a mass of 212 lb. What is the mass of the same number of carbon atoms? 5-5. A piece of copper wire has a mass of 16.0 g; the same number of atoms of a precious metal has a mass of 49.1 g. What is the metal? 5-6. In the compound CuO, what mass of copper is present for each 18.0 g of oxygen? 5-7. In the compound NaCl, what mass of sodium is present for each 425 g of chlorine? 5-8.

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  Foundations of Chemistry

  An Introductory Course for Science Students

  • Philippa B. Cranwell, Elizabeth M. Page(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  3 Amount of Substance At the end of this chapter, students should be able to: • Determine relative atomic and molecular masses using the periodic table of the elements • Explain the meaning of the mole and use Avogadro ’ s number to calculate numbers of atoms of elements or molecules of substances in a certain amount in moles • Use the equation relating the amount of substances in moles to mass and molar mass to calculate any one of these parameters knowing the other two • Calculate theoretical and actual percentage yields in chemical reactions • Calculate percentage composition by mass of elements in compounds from the molecular formula and relative atomic masses • Calculate percentage purity of compounds from analytical data relating to percentage composition • Determine the empirical and molecular formula of a compound from its percentage composition by mass • Calculate the concentration of a solution in various units, knowing its composition • Perform calculations to determine how to dilute a concentrated solution to a given molarity • Carry out calculations using titration data to determine the concentration of a solution • Use the molar gas volume to work out the mass of a certain volume of a known gas under standard conditions of temperature and pressure Foundations of Chemistry: An Introductory Course for Science Students , First Edition. Philippa B. Cranwell and Elizabeth M. Page. © 2021 John Wiley & Sons Ltd. Published 2021 by John Wiley & Sons Ltd. Companion website: www.wiley.com/go/Cranwell/Foundations 3.1 Masses of atoms and molecules Atoms join together to make molecules, and molecules react together to make different molecules or join together to make bigger molecules. In making these new substances, scientists need to know how much of one molecule or com-pound will react with another and how much product will be obtained as a result. To achieve this, we need to be able to count molecules or groups of mole-cules.

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  General Chemistry for Engineers

  • Jeffrey Gaffney, Nancy Marley(Authors)
  • 2017(Publication Date)
  • Elsevier

    (Publisher)

  The molar mass is used to convert the mass of a substance in grams to the number of moles or to convert the number of moles of a substance to its mass in grams. For example, if the mass of substance A is known, the number of moles of substance A can be calculated by;

  grams A

  molar mass A

  g / mol

  = moles A

    (2)

  Similarly, if the number of moles of substance B is known, the mass of substance B can be calculated by;

  moles B × molar mass B

  g / mol

  = grams B

    (3)

  Example 4.1: Determining the Number of Moles of a Compound From Its Mass How many moles are there in 15.0 g of sodium chloride? How many moles of sodium ions are there in the same sample? How many moles of chloride ions?

  1.  Determine the molar mass of NaCl.

  The atomic mass of Na = 23.0 amu, the atomic mass of Cl = 35.5 amu The molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol

  2.  Convert grams of NaCl to moles of NaCl.

  15.0 g

  58.5 g / mol

  = 0.256 moles Nacl

  3.  Determine the number of moles of ions.

  NaCl has one sodium ion and one chloride ion. 15.0 g NaCl = 0.256 moles of sodium ions and 0.256 moles of chloride ions.

  Example 4.2: Determining the Mass of a Compound From the Number of Moles

  How many grams are in 0.700 moles of hydrogen peroxide (H2 O2 )?

  1.  

  Determine the molar mass of H2 O2 :

  The atomic mass of H = 1.01 amu, the atomic mass of O = 16.0 amu.

  The molar mass of H2 O2  = (1.01 × 2) + (16.0 × 2) = 34.0 g/mol

  2.  Convert moles to grams:

  0.700 mol × 34.0 g/mol = 23.8 grams of H2 O2 in 0.700 moles.

  4.2 The Empirical Formula

  While the molecular formula of a compound shows the number of each type of atom in a molecule, the empirical formula shows the simplest positive integer ratio of atoms present in the compound. As such, the empirical formula does not necessarily represent the actual numbers of atoms present in a single molecule of the compound. For example, the molecular formula for sulfur monoxide is SO and the molecular formula for disulfur dioxide is S2 O2

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  eBook - PDF

  Chemistry 2e

  • Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
  • 2019(Publication Date)
  • Openstax

    (Publisher)

  This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022 10 23 /mol. Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 3.5). FIGURE 3.5 Each sample contains 6.022 10 23 atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott) The molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12 C atom weighs 12 amu (its atomic mass is 12 amu). A mole of 12 C weighs 12 g (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12 C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu ( Figure 3.6). 3.1 • Formula Mass and the Mole Concept 121 FIGURE 3.6 Each sample contains 6.02 10 23 molecules or formula units—1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of C 8 H 17 OH (1-octanol, formula mass 130.2 amu), 454.4 g of HgI 2 (mercury(II) iodide, formula mass 454.4 amu), 32.0 g of CH 3 OH (methanol, formula mass 32.0 amu) and 256.5 g of S 8 (sulfur, formula mass 256.5 amu).

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  Chemistry

  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.

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  eBook - ePub

  Survival Guide to General Chemistry

  • Patrick E. McMahon, Rosemary McMahon, Bohdan Khomtchouk(Authors)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  compound is defined as the mass of one mole of molecules or formula units.

  The molar mass (MM) of any compound is numerically equal to the molecular mass or formula unit mass of the compound expressed in gram units; the units of MM = grams/moles.

  Molar mass is most often used directly in calculations. The molar mass (MM) = sum of the molar masses of each atom (in gram units) in the molecule or each atom (in gram units) in the simplest whole number ratio formula.

  Example:

  MM of CuCl 2

   

  ( ionic compound )

  : 1 Cu × 63 .54 g/mole = 63 .54 grams/mole

   + 2 CI × 35 .45 g/mole = 

  70 .90 grams/mole

  _

  total = 

  1 3 4

  . 4 4 g / m o l e

  Example:

  MM of CO 2

    ( molecule ) : 1 C × 12 .01 g/mole = 12 .01 grams/mole

   + 2 0 × 16 .00 g/mole = 

  32 .00 grams/mole

  _

  total = 

  4 4

  .

  0 1

    g /

  m o l e

  Counting all the atoms in formulas containing polyatomic ions requires care.

  Example:

  MM of Co 3

   

  (

  PO 4

  )

  2

   

  ( ionic compound )

  : 3 Co × 63 .54 g/mole = 176 .70 grams/mole

  + 2 P × 30 .97 g/mole = 61 .94 grams/mole

  + 8 O × 16 .00 g/mole = 

  128 .00 grams/mole

  _

   total = 

  3 6 6

  . 7 3 g / m o l e

  Certain elements come in the form of molecules. The molar mass of elements, which come in the form of molecules, is based on the formula of the complete molecule.

  MM of O2 (element comes as a molecule): 2 O × 16.00 g/mole = 32.00 g/mole

  PROCEDURE FOR SOLVING MOLE PROBLEMS (COMPOUNDS )

  (1)  Identify the correct form of the [MM = mass(g)/moles] equation, which is required to solve for the unknown variable.

  (2

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  Analytical Chemistry

  • Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
  • 2013(Publication Date)
  • Wiley

    (Publisher)

  Chapter Five STOICHIOMETRIC CALCULATIONS: THE WORKHORSE OF THE ANALYST Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ● How to calculate molarities and moles (key equations: 5.4, 5.5), p. 152 ● How to express analytical results, p. 159 ● How to calculate weight and percent analyted from molarities, volumes, and reaction ratios (key equations: 5.5, 5.17–5.20, 5.25), pp. 152, 169, 171 ● Weight relationships for gravimetric analysis (key equation: 5.28), p. 181 Analytical chemistry deals with measurements of analytes in solids and concentrations in solution, from which we calculate masses. Thus, we prepare solutions of known concentrations that can be used to calibrate instruments or to titrate sample solutions. We calculate the mass of an analyte in a solution from its concentration and the volume. We calculate the mass of product expected from the mass of reactants. All of Stoichiometry deals with the ratios in which chemicals react. these calculations require a knowledge of stoichiometry, that is, the ratios in which chemicals react, from which we apply appropriate conversion factors to arrive at the desired calculated results. In this chapter we review the fundamental concepts of mass, moles, and equivalents; the ways in which analytical results may be expressed for solids and liquids; and the principles of volumetric analysis and how stoichiometric relationships are used in titrations to calculate the mass of analyte. 5.1 Review of the Fundamentals Quantitative analysis is based on a few fundamental atomic and molecular concepts, which we review below. You have undoubtedly been introduced to these in your general chemistry course, but we briefly review them here since they are so fundamental to quantitative calculations. THE BASICS: ATOMIC, MOLECULAR, AND FORMULA WEIGHTS The atomic weight for any element is the weight of a specified number of atoms of that element, and that number is the same from one element to another.

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  SURVIVAL GDE GENERAL CHEM W/ MATH REVIEW

  • Charles Atwood(Author)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Atomic weights are found on a periodic table. *The terms molar mass, molecular weight, and formula weight all apply to the same concept/calculation. Technically, the term molecular weight should be used only with covalent compounds and formula weight applies only to ionic compounds. The more generic term molar mass is used frequently in chemical literature. Sample Exercises Determining Molar Mass 1. What is the molar mass (or formula weight) of calcium phosphate, Ca 3 (PO 4 ) 2 ? The correct answer is: 310.2 g/mol Ca 3 (PO 4 ) 2 Molar mass of Ca 3 (PO 4 ) 2 = (3 x 40.08 g/mol Ca) + (4 x 2 x 16.00 g/mol O) + 2 x 30.97 g/mol P) = 310.18 g /mol Ca 3 (PO 4 ) 2 Module 4 Key Equations & Concepts 1.   ion or molecule, compound, a in atoms of weights atomic mass Molar The molar mass, molecular weight, or formula weight * are calculated by summing the atomic weights of the elements in the compound. Molar mass is the mass in grams of one mole of a substance. 2. One mole = 6.022 x 10 23 particles Avogadro’s relationship converts from the number of moles of a substance to the number of atoms, ions, or molecules of that substance and vice versa. 3. Mass of one atom of an element = mass of an element 1 mole of an element       1 mole of atoms 6.022  10 23 atoms       The mass of one atom, ion, or molecule is used to determine the mass of a few atoms, ions, or molecules of a substance. Notice that the fraction in the first set of parentheses simply is a representation of molar mass. 4. Mole ratio A compounds chemical formula is the ratio of the different types of atom in the compound. The mole ratio is used to convert from mass or moles of a compound to mass or moles of a specific atom in the compound. 34 Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300

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  eBook - ePub

  Understanding General Chemistry

  • Atef Korchef(Author)
  • 2022(Publication Date)
  • CRC Press

    (Publisher)

  weight) percent refers to the ratio of the mass of one element to the total mass of a compound. The mass percent of an element, % (element), is determined bythe following equation:

  % e l e m e n t = n × a t o m i c   m a s s   o f   t h e   e l e m e n t m o l e c u l a r   m a s s   o f   t h e   c o m p o u n d × 100

  where n is the number of atoms of the element in the compound.

  A balanced chemical equation is an equation that represents the correct amounts of reactants and products in a chemical reaction. The number of atoms of each element is the same on both sides of the equation. To balance achemical equation, start by balancing those elements that appear in only one reactant and one product.

  • Example of a balanced chemical equation: 2HgO → 2Hg + 1/2O2

  Stoichiometry is the study of mass relationships that exist between reactants (substances consumed) and products (substances produced) in a chemical reaction. Use coefficients in the balanced equation to determine the relationships between the number of moles of reactants andproducts. Then, calculate the number of moles and the mass of the desired quantity.

  A solution consists of a smaller amount of a substance, the solute, dissolved in a larger amount of another substance, the solvent.

  The concentration of the solution is expressed as the amount of solute dissolved in a given amount of solution. The term most used is Molarity (M), defined as number of moles of solute per liter of solution (M = n/V). A common unit of molarity is M(= mol L−1 ).

  The molality is defined as the number of moles of the solute per kilogram of the solvent. The SI unit of molality is mol kg−1 .

  The solubility designates the maximum mass concentration of the solute in the solvent, at a given temperature. The solution thus obtained is saturated. The solubility depends on temperature and may be expressed in g L−1 , mol L−1 and g 100 g−1

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  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  The equality that relates the substances is the mole-to-mole rela- tionship between glucose and O 2 given by the chemical equation. In this case, the equa- tion tells us that 1 mol C 6 H 12 O 6 3 6 mol O 2 . It is very important to realize that there is no direct conversion between the mass of C 6 H 12 O 6 and the mass of O 2 . We need to con- vert the mass of glucose to moles of glucose, then we use the mole ratio to convert moles of glucose to moles of oxygen, and finally we convert moles of oxygen to mass of oxygen. This sequence, indicating where we use the mole-to-mole equivalence, is shown below 1 mol C 6 H 12 O 6 3 6 mol O 2 1.00 g C 6 H 12 O 6 h mol C 6 H 12 O 6 h mol O 2 h g O 2 Two molar masses are used, once for converting 1.00 g of glucose to moles and again for converting moles of O 2 to grams of O 2 . Figure 3.4 outlines this flow for any stoichiometry problem that relates reactant or product masses. If we know the balanced equation for a reaction and the mass of any reac- tant or product, we can calculate the required or expected mass of any other substance in the equation. Example 3.15 shows how it works. Mass-to-mass calculations using balanced chemical equations Moles of A Moles of B Grams of Substance B Grams of Substance A Molar Mass B Molar Mass A Balanced Chemical Equation Figure 3.4 | The sequence of calculations for solving stoichiometry problems. This sequence applies to all calculations that start with the mass of one substance A and require the mass of a second substance B as the answer. Each box represents a measured or calculated quantity. Each arrow represents one of our chemical tools. 134 Chapter 3 | The Mole and Stoichiometry Portland cement is a mixture of the oxides of calcium, aluminum, and silicon. The raw material for calcium oxide is calcium carbonate, which occurs as the chief component of limestone. When calcium carbonate is strongly heated it decomposes.

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  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  115 This Chapter in Context In 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 impor- tant for future courses in organic chemistry, biochemistry, and almost any other advanced laboratory course in the sciences. Another aspect of stiochiometry is the ratio of elements in compounds. The malachite shown in the opening photograph is composed of copper, oxygen, car- bon, and hydrogen. The ratio of these elements, and the way they are combined gives rise to the beautiful green color or the malachite. 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. 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 3.8, 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. In the later chapters, as we learn new concepts, they will be added to this flowchart to illustrate how many of our chemical ideas are interrelated.

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