Percent Composition

10 Key excerpts on "Percent Composition"

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  Chemistry

  The Molecular Nature of Matter

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

    (Publisher)

  Another way to describe a compound’s makeup is to use the relative the masses of the elements in the compound as a list of percentages by mass, called the compound’s percentage composition. Percentage Composition The percentage by mass of an element in a sample is the number of grams of the element present in 100 g of the sample. In general, a percentage by mass is found by using the following equation. Percentage by mass of elements = mass of element mass of whole sample Ž 100% (3.1) We can determine the percentage composition based on chemical analysis of a substance as shown in the next example. Percentage composition Example 3.7 Calculating a Percentage Composition from Chemical Analysis A sample of a liquid with a mass of 8.657 g was decomposed into its elements and gave 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the percentage composition of this compound? Analysis: Solving problems often requires that we know the meaning of key terms, in this case percentage composition, which is the list of the percentage by mass of the ele- ments in the compound. Assembling the Tools: The tool we need is the percentage by mass equation, Equation 3.1. We are given the mass of each element in the sample, and the total mass, which is also the sum of the masses of the elements. 3.3 | Chemical Formula and Percentage Composition 119 Solution: Using Equation 3.1 for each element gives three equations that we use to compute the needed percentages: For C: 5.217 g 8.657 g Ž 100% = 60.26% C For H: 0.9620 g 8.657 g Ž 100% = 11.11% H For O: 2.478 g 8.657 g Ž 100% = 28.62% O Sum of percentages: 99.99% Is the Answer Reasonable? The “check” is that the percentages must add up to 100%, allowing for small differences caused by rounding. We can also check the individual results by rounding all the numbers to one significant figure to estimate the results.

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  Chemistry 2e

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

    (Publisher)

  But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements. Percent Composition The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s Percent Composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The Percent Composition of this compound could be represented as follows: If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the Percent Composition would be calculated to be 25% H and 75% C: 1 Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447. 3.2 • Determining Empirical and Molecular Formulas 129 EXAMPLE 3.9 Calculation of Percent Composition Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the Percent Composition of this compound? Solution To calculate Percent Composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage: The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

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  Foundations of College Chemistry

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

    (Publisher)

  Percent means parts per 100 parts. Just as each piece of pie is a percent of the whole pie, each element in a compound is a percent of the whole compound. The Percent Composition of a compound is the mass percent of each element in the compound. The molar mass represents the total mass, or 100%, of the compound. Thus, the Percent Composition of water, H 2 O, is 11.19% H and 88.79% O by mass. According to the law of definite composition, the Percent Composition must be the same no matter what size sample is taken. The Percent Composition of a compound can be determined (1) from knowing its formula or (2) from experimental data. LEARNING OBJECTIVE KEY TERM Percent Composition of a compound CHECK YOUR UNDERSTANDING 7.7 Percent Composition ➥ ONLINE LEARNING MODULE Converting Mass to Molecules ➥ 132 CHAPTER 7 • Quantitative Composition of Compounds PROBLEM-SOLVING STRATEGY For Calculating Percent Composition from Formula 1. Calculate the molar mass (Section 7.2). 2. Divide the total mass of each element in the formula by the molar mass and multiply by 100. This gives the Percent Composition: total mass of the element molar mass × 100 = percent of the element ENHANCED EXAMPLE 7.13 Calculate the Percent Composition of sodium chloride (NaCl). SOLUTION Read • Known: NaCl Plan • Use the Problem-Solving Strategy for Calculating Percent Composition from Formula. Calculate • 1. We need to use the atomic masses from the inside front cover of the text to find the molar mass of NaCl. Na Cl 22.99 g + 35.45 g = 58.44 g = molar mass NaCl 2. Calculate the Percent Composition for each element. Na: ( 22.99 g Na 58.44 g ) (100) = 39.34% Na Cl: ( 35.45 g Cl 58.44 g ) (100) = 60.66% Cl 100.00% total NOTE In any two-component system, if the percent of one component is known, the other is automatically defined by difference; that is, if Na is 39.43%, then Cl is 100% − 39.34% = 60.66%.

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  General Chemistry: Atoms First

  • Young, William Vining, Roberta Day, Beatrice Botch(Authors)
  • 2017(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 Unit 8 Stoichiometry 199 8.1b Percent Composition The mole-to-mole relationships in a chemical formula can also be used to determine the Percent Composition of an element in a compound. The Percent Composition of an ele-ment in a compound is the mass of an element present in exactly 100 g of a compound and is calculated using the following equation: % element 5 1 number of atoms of element in formula 21 molar mass of element 2 mass of 1 mol of compound 3 100 % (8.1) For example, the Percent Composition of hydrogen in water, H 2 O , is calculated as follows: % H in H 2 O 5 2 mol H a 1.01 g 1 mol H b 1 mol H 2 O a 18.02 g 1 mol H 2 O b 3 100% 5 11.2% H Notice that Percent Composition is calculated by using the mole-to-mole ratio in the chemical formula of water ( 2 mol H in 1 mol H 2 O ) and converting it to a mass ratio using molar mass. Because water is made up of only two elements and the sum of all the Percent Composition values must equal 100% , the Percent Composition of oxygen in water is % O in H 2 O 5 100% 2 % H 5 100% 2 11.2% 5 88.8% O Use Avogadro’s number and the amount of each element present to determine the number of atoms of each element present in the 2.50-mol sample of acetic acid. 5 .00 mol C 3 6.022 3 10 23 C atoms 1 mol C 5 3.01 3 10 24 C atoms 1 0.0 mol H 3 6.022 3 10 23 H atoms 1 mol H 5 6.02 3 10 24 H atom s 5 .00 mol O 3 6.022 3 10 23 O atoms 1 mol O 5 3.01 3 10 24 O atom s Is your answer reasonable? The sample contains more than one mole of the compound, so it also contains more than one mole of each element and more than Avogadro’s number of atoms of each element in the compound. b Example Problem 8.1.3 (continued) Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 Unit 8 Stoichiometry 200 Example Problem 8.1.4 Calculate Percent Composition from a compound formula.

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  Basic Concepts of Chemistry

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

    (Publisher)

  Percent Composition expresses the mass of each element per 100 mass units of compound. For example, if there is an 82-g quantity of nitrogen present in each 100 g of ammonia (NH 3 ), the Percent Composition is expressed as 82% nitrogen. The mass of one mole of carbon dioxide is 44.01 g, and it is composed of one mole of carbon atoms (12.01 g) and two moles of oxygen atoms (2 * 16.00 = 32.00 g). The Percent Composition is calculated by dividing the total mass of each component ele- ment by the total mass (molar mass) of the compound and then multiplying by 100%. total mass of component element total mass (molar mass) * 100% = Percent Composition In CO 2 , the Percent Composition of C is 12.01 g C 44.01 g CO 2 * 100% = 27.29% C and the Percent Composition of O is 32.00 g O 44.01 g CO 2 * 100% = 72.71% SYNTHESIS Could you go backward with this problem? Could you calculate the mass of limestone that could be made from 5.50 moles of calcium? What conversion factors are needed to solve this problem? Because you want the mass of limestone, you would need its molar mass. Because the problem asks you to relate limestone and calcium, you would also use the mole ratio between them. 5.50 mol Ca * 1 mol CaCO 3 1 mol Ca * 100.1 g CaCO 3 1 mol CaCO 3 = 551 g CaCO 3 Marble is a form of limestone. 164 CHAPTER 5 Quantities in Chemistry What is the Percent Composition of all the elements in borazine (B 3 N 3 H 6 )? PROCEDURE Find the molar mass and convert the total mass of each element to percent of molar mass.

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  Foundations of College Chemistry

  • Morris Hein, Susan Arena, Cary Willard(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  130 CHAPTER 7 Quantitative Composition of Compounds Solitaire/Shutterstock CHAPTER OUTLINE 7.1 The Mole 7.2 Molar Mass of Compounds 7.3 Percent Composition of Compounds 7.4 Calculating Empirical Formulas 7.5 Calculating the Molecular Formula from the Empirical Formula Cereals, cleaning products, and pain remedies all list their ingredients on the package label. The ingredients are listed in order from most to least, but the amounts are rarely given. However, it is precisely these amounts that give products their desired properties and distinguish them from the competition. Understandably, manufacturers carefully regulate the amounts of ingredients to maintain quality and hopefully their custom- ers’ loyalty. The candies in the photo above are made with a sweet center covered in a layer of sucrose or table sugar. They can be measured by counting or weighing. The composition of compounds is an important concept in chemis- try. Determining numerical relationships among the elements in com- pounds and measuring exact quantities of particles are fundamental tasks that chemists routinely perform in their daily work. 7.1 The Mole 131 7.1 The Mole LEARNING OBJECTIVE: Apply the concepts of the mole, molar mass, and Avogadro’s number to solve chemistry problems. The atom is an incredibly tiny object. Its mass is far too small to measure on an ordinary balance. In Chapter 5 (Section 5.6), we learned to compare atoms using a table of atomic mass units. These units are valuable when we compare the masses of individual atoms (mentally), but they have no practical use in the laboratory. The mass in grams for an “average” carbon atom (atomic mass 12.01 u) would be 2.00 × 10 −23 g, which is much too tiny for the best laboratory balance. So how can we confidently measure masses for these very tiny atoms? We increase the number of atoms in a sample until we have an amount large enough to measure on a labora- tory balance.

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  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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  Simple Solutions – Counting Moles...

  • Nigel P. Freestone(Author)
  • 2016(Publication Date)
  • Bentham Science Publishers

    (Publisher)

  64/183.5 x 100 = 34.9 Total 183.5 100%

  Example 1.14: 1.26 g of iron reacts with 0.54 g of oxygen to form rust. What is the percentage composition of each element in the new compound?

  Answer:

  This type of question can be answered using a shortened version of the same calculating frame. Mass of new compound = 1.26 + 0.54 = 1.8 g

  Element	Mass	% composition
  Fe	1.26	1.26 /1.8 x 100 = 70
  O	0.54	0.54/1.8 x 100 = 30
  Total	1.6	100%

  Example 1.15: Decomposition 8.657 g of a liquid sample into its elements gave 5.217g of carbon , 0.9620 g of hydrogen and 2.678 g of oxygen. Determine the percentage composition of the liquid.

  Answer:

  Element	Mass	% composition
  C	5.217	1.26 /1.8 x 100 = 70
  H	0.962	0.962/8.657 x 100 = 11.1
  O	2.478	2.478/8.657x 100 = 28.6
  Total	100%

  Exercise 1.3

  1. What is the percentage composition by mass of silicon and chlorine in SiCl4 ?
  2. Calculate percentage composition of CuSO4
  3. Calculate the mass percentage of hydrogen in aspirin, C9 H8 O4 .
  4. What is the mass of silicon in 10g of clay, Al2 Si2 O5 (OH)4 ?
  5. What is the mass of sulphur in 1 tonne of H2 SO4 ?
  6. Determine the percentage composition of Ca3 (PO4 )2
  7. What is the percentage composition of ammonium sulfate, (NH4 )2 SO4 ?
  8. What is the mass of nitrogen present in 5g of aniline, C6 H5 NH2 ?
  9. 9.03g of Mg combine completely with 3.48g of N to form a compound. What is the percentage composition of this compound?
  10. A 27.0 g sample of a compound contains 7.20 g of C, 2.20 g of hydrogen and 17.6 g of oxygen. Calculate the percentage composition of the compound.

  1.4. Empirical and Molecular Formula

  The empirical formula gives the simplest integer ratio of each element present in a compound. The integers are given as subscripts to right hand side of the chemical symbols. A molecular formula is the same as or a multiple of the empirical formula, i.e. (CH2 O)n. Thus if a compound has an empirical formula of CH2 O its molecular formula could be CH2 O (n= 1), C2 H4 O2 (n= 2), C4 H8 O2 (n = 4) or C10 H20 O10 (n = 10) etc.

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  Chemistry

  • Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
  • 2015(Publication Date)
  • Openstax

    (Publisher)

  Equations • % X = mass X mass compound × 100% • molecular or molar mass ⎛ ⎝ amu or g mol ⎞ ⎠ empirical formula mass ⎛ ⎝ amu or g mol ⎞ ⎠ = n formula units/molecule • (A x B y ) n = A nx B ny • M = mol solute L solution • C 1 V 1 = C 2 V 2 Chapter 3 | Composition of Substances and Solutions 163 • Percent by mass = mass of solute mass of solution × 100 • ppm = mass solute mass solution × 10 6 ppm • ppb = mass solute mass solution × 10 9 ppb Summary 3.1 Formula Mass and the Mole Concept The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be 6.022 × 10 23 , a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H 2 O molecule weighs approximately18 amu and 1 mole of H 2 O molecules weighs approximately 18 g). 3.2 Determining Empirical and Molecular Formulas The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s Percent Composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.

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  Chemistry

  The Molecular Nature of Matter

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

    (Publisher)

  The numerator in Equation 3.1 will be the mass associated with one of the elements in the formula. In the next section we see how this calculated percentage is useful for identifying a substance. Percentage Composition and Chemical Identity We can use Equation 3.1 to determine the percentage composition of any chemical compound from its formula. Nitrogen and oxygen, for example, form all of the following compounds: N 2 O, NO, NO 2 , N 2 O 3 , N 2 O 4 , and N 2 O 5 . To identify an unknown sample of a compound of nitro- gen and oxygen, one might compare the percentage composition found by experiment with the calculated, or theoretical, percentages for each possible formula. A strategy for matching formulas with mass percentages is outlined in the following example. NOTE We use the word theoretical to indicate calculated data rather than experimental. Theoretical data is data derived from highly reliable sources. 128 CHAPTER 3 The Mole and Stoichiometry Solution: As we calculate the theoretical percentage composition for N 2 O 5 , we know that 1 mol of N 2 O 5 must contain 2 mol N and 5 mol O from the mol ratio tool. The corresponding number of grams of N and O are found as follows. 1 mol N 2 O 5 ( 2 mol N __________ 1 mol N 2 O 5 ) ( 14.01 g N ________ 1 mol N ) = 28.02 g N 1 mol N 2 O 5 ( 5 mol O __________ 1 mol N 2 O 5 ) ( 16.00 g O ________ 1 mol O ) = 80.00 g O 1 mole N 2 O 5 = 108.02 g N 2 O 5 Now we can calculate the theoretical percentages for N 2 O 5 . For % N: 28.02 g _______ 108.02 g × 100% = 25.94% N in N 2 O 5 For % O: 80.00 g _______ 108.02 g × 100% = 74.06% O in N 2 O 5 The experimental values of 25.94% N and 74.06% O match the theoretical percentages we just calculated for the formula N 2 O 5 . We are justified in calling the compound N 2 O 5 . Is the Answer Reasonable? You could also calculate the percentage of nitrogen or oxygen in the other compounds, NO, NO 2 , N 2 O 3 , and N 2 O 4 , and compare those values to the experimental values.

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