Chapter 10 Chemical Quantities

Name: Chapter 10 Chemical Quantities
Uploaded: 2017-07-14T09:01:11+00:00
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Description: Chapter 10 Chemical Quantities

Chapter 10 Chemical Quantities
10.1 The Mole: A Measurement of Matter 10.2 Mole-Mass and Mole-Volume Relationships 10.3 Percent Composition and Chemical Formulas Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

What does the percent composition of a compound tell you?
CHEMISTRY & YOU What does the percent composition of a compound tell you? A tag sewn into the seam of a shirt usually tells you what fibers were used to make the cloth and the percent of each. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition of a Compound
How do you calculate the percent composition of a compound? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

The relative amounts of the elements in a compound are expressed as the percent composition or the percent by mass of each element in the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition from Mass Data If you know the relative masses of each element in a compound, you can calculate the percent composition of the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition from Mass Data The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%. % by mass of element = mass of element mass of compound × 100% Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Calculating Percent Composition from Mass Data
Sample Problem 10.9 Calculating Percent Composition from Mass Data When a g sample of a compound containing only magnesium and oxygen is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Analyze List the knowns and the unknowns.
Sample Problem 10.9 Analyze List the knowns and the unknowns. 1 The percent by mass of an element in a compound is the mass of that element divided by the mass of the compound multiplied by 100%. KNOWNS mass of compound = g mass of oxygen = 5.40 g O mass of magnesium = g – 5.40 g O = 8.20 g Mg UNKNOWNS percent by mass of Mg = ?% Mg percent by mass of O = ?% O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Calculate Solve for the unknowns.
Sample Problem 10.9 Calculate Solve for the unknowns. 2 Determine the percent by mass of Mg in the compound. % Mg = mass of Mg mass of compound × 100% = 60.3% Mg = 8.20 g 13.60 g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.9 Calculate Solve for the unknowns. 2 Determine the percent by mass of O in the compound. % O = mass of O mass of compound × 100% 5.40 g = 39.7% O = 13.60 g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Evaluate Does the result make sense?
Sample Problem 10.9 Evaluate Does the result make sense? 3 The percents of the elements add up to 100%. 60.3% % = 100% Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition from the Chemical Formula You can also calculate the percent composition of a compound using its chemical formula. The subscripts in the formula are used to calculate the mass of each element in a mole of that compound. Using the individual masses of the elements and the molar mass, you can calculate the percent by mass of each element. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition from the Chemical Formula You can also calculate the percent composition of a compound using its chemical formula. % by mass of element mass of element in 1 mol compound molar mass of compound × 100% = Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Calculating Percent Composition from a Formula
Sample Problem 10.10 Calculating Percent Composition from a Formula Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Analyze List the knowns and the unknowns.
Sample Problem 10.10 Analyze List the knowns and the unknowns. 1 Calculate the percent by mass of each element by dividing the mass of that element in one mole of the compound by the molar mass of the compound and multiplying by 100%. KNOWNS mass of C in 1 mol C3H8 = 3 mol × 12.0 g/mol = 36.0 g mass of H in 1 mol C3H8 = 8 mol × 1.0 g/mol = 8.0 g molar mass of C3H8 = 36.0 g/mol g/mol = 44.0 g/mol UNKNOWNS percent by mass of C = ?% C percent by mass of H = ?% H Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.10 Calculate Solve for the unknowns. 2 Determine the percent by mass of C in C3H8. % C = mass of C in 1 mol C3H8 molar mass of C3H8 × 100% = 81.8% C = 36.0 g 44.0 g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.10 Calculate Solve for the unknowns. 2 Determine the percent by mass of H in C3H8. % H = mass of H in 1 mol C3H8 molar mass of C3H8 × 100% = 18% H = 8.0 g 44.0 g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.10 Evaluate Does the result make sense? 3 The percents of the elements add up to 100% when the answers are expressed to two significant figures (82% + 18% = 100%). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition as a Conversion Factor You can use percent composition to calculate the number of grams of any element in a specific mass of a compound. To do this, multiply the mass of the compound by a conversion factor based on the percent composition of the element in the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Percent Composition as a Conversion Factor Propane is 81.8% carbon and 18% hydrogen. You can use the following conversion factors to solve for the mass of carbon or hydrogen contained in a specific amount of propane. 81.8 g C 100 g C3H8 and 18 g H Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

CHEMISTRY & YOU What information can you get from the percent composition of a compound? You can use percent composition to determine the mass of an element in a sample of a compound of a given size. You can also determine the empirical formula of the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.11 Calculating the Mass of an Element in a Compound Using Percent Composition Calculate the mass of carbon and the mass of hydrogen in 82.0 g of propane (C3H8). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Analyze List the known and the unknowns.
Sample Problem 10.11 Analyze List the known and the unknowns. 1 Use the conversion factors based on the percent composition of propane to make the following conversions: grams C3H8 → grams C and grams C3H8 → grams H. KNOWN mass of C3H8 = 82.0 g UNKNOWNS mass of carbon = ? g C mass of hydrogen = ? g H Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.11 Calculate Solve for the unknowns. 2 To calculate the mass of C, first write the conversion factor to convert from mass of C3H8 to mass of C. From Sample Problem 10.10, the percent by mass of C in C3H8 is 81.8%. 81.8 g C 100 g C3H8 Multiply the mass of C3H8 by the conversion factor. 81.8 g C 100 g C3H8 82.0 g C3H8 × = 67.1 g C Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.11 Calculate Solve for the unknowns. 2 To calculate the mass of H, first write the conversion factor to convert from mass of C3H8 to mass of H. From Sample Problem 10.10, the percent by mass of H in C3H8 is 18%. 18 g H 100 g C3H8 Multiply the mass of C3H8 by the conversion factor. 18 g H 100 g C3H8 82.0 g C3H8 × = 15 g H Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.11 Evaluate Does the result make sense? 3 The sum of the two masses equals 82 g, the sample size, to two significant figures (67 g C + 15 g H = 82 g C3H8). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

What data can you use to calculate percent composition?
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

What data can you use to calculate percent composition?
You can calculate percent composition if you know the mass of a compound and the masses of the elements contained in the compound, or if you know the chemical formula, the molar mass of the compound, and the atomic weights of the elements contained in the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

How can you calculate the empirical formula of a compound?
Empirical Formulas Empirical Formulas How can you calculate the empirical formula of a compound? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Empirical Formulas The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound. An empirical formula may or may not be the same as a molecular formula. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Empirical Formulas The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound. An empirical formula may or may not be the same as a molecular formula. For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1. Thus, the empirical formula of hydrogen peroxide is HO. The molecular formula, H2O2, has twice the number of atoms as the empirical formula. Notice that the ratio of hydrogen to oxygen is still the same, 1:1. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

For carbon dioxide, the empirical and molecular formulas are the same—
Empirical Formulas For carbon dioxide, the empirical and molecular formulas are the same— CO2. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Empirical Formulas The figure below shows two compounds of carbon and hydrogen having the same empirical formula (CH) but different molecular formulas. Ethyne (C2H2), also called acetylene, is a gas used in welders’ torches. Styrene (C8H8) is used in making polystyrene. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Empirical Formulas The percent composition of a compound can be used to calculate the empirical formula of that compound. The percent composition tells the ratio of masses of the elements in a compound. The ratio of masses can be changed to ratio of moles by using conversion factors based on the molar mass of each element. The mole ratio is then reduced to the lowest whole-number ratio to obtain the empirical formula of the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Determining the Empirical Formula of a Compound
Sample Problem 10.12 Determining the Empirical Formula of a Compound A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Analyze List the knowns and the unknown.
Sample Problem 10.12 Analyze List the knowns and the unknown. 1 The percent composition gives the ratio of the mass of nitrogen atoms to the mass of oxygen atoms in the compound. Change the ratio of masses to a ratio of moles and reduce this ratio to the lowest whole-number ratio. KNOWNS percent by mass of N = 25.9% N percent by mass of O = 74.1% O UNKNOWN empirical formula = N?O? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Calculate Solve for the unknown.
Sample Problem 10.12 Calculate Solve for the unknown. 2 Convert the percent by mass of each element to moles. Percent means “parts per 100,” so g of the compound contains 25.9 g N and 74.1 g O. 25.9 g N × 1 mol N 14.0 g N = 1.85 mol N 74.1 g O × 1 mol O 16.0 g O = 4.63 mol O The mole ratio of N to O is N1.85O4.63. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.12 Calculate Solve for the unknown. 2 Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles. 1.85 mol N 1.85 = 1 mol N 4.63 mol O 1.85 = 2.50 mol O The mole ratio of N to O is N1O2.5. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.12 Calculate Solve for the unknown. 2 Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers. 1 mol N × 2 = 2 mol N 2.5 mol O × 2 = 5 mol O The empirical formula is N2O5. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.12 Evaluate Does the result make sense? 3 The subscripts are whole numbers, and the percent composition of this empirical formula equals the percents given in the original problem. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

You are doing an experiment to try to find the molecular formula of a compound. You discover the percent composition. Can you determine the molecular formula? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

You are doing an experiment to try to find the molecular formula of a compound. You discover the percent composition. Can you determine the molecular formula? You can determine the empirical formula. This might be the same as the molecular formula, or it might not. You would need more data to be sure of the molecular formula. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Comparison of Empirical and Molecular Formulas
Interpret Data Ethyne and benzene have the same empirical formula—CH. Comparison of Empirical and Molecular Formulas Formula (name) Classification of formula Molar mass (g/mol) CH Empirical 13 C2H2 (ethyne) Molecular 26 (2 × 13) C6H6 (benzene) 78 (6 × 13) CH2O (methanol) Empirical and molecular 30 C2H4O2 (ethanoic acid) 60 (2 × 30) C6H12O6 (glucose) 180 (6 × 30) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Interpret Data Methanal, ethanoic acid, and glucose have the same empirical formula—CH2O. Comparison of Empirical and Molecular Formulas Formula (name) Classification of formula Molar mass (g/mol) CH Empirical 13 C2H2 (ethyne) Molecular 26 (2 × 13) C6H6 (benzene) 78 (6 × 13) CH2O (methanal) Empirical and molecular 30 C2H4O2 (ethanoic acid) 60 (2 × 30) C6H12O6 (glucose) 180 (6 × 30) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Interpret Data Notice that the molar masses of the compounds in these two groups are simple whole-number multiples of the molar masses of the empirical formulas, CH and CH2O. Comparison of Empirical and Molecular Formulas Formula (name) Classification of formula Molar mass (g/mol) CH Empirical 13 C2H2 (ethyne) Molecular 26 (2 × 13) C6H6 (benzene) 78 (6 × 13) CH2O (methanal) Empirical and molecular 30 C2H4O2 (ethanoic acid) 60 (2 × 30) C6H12O6 (glucose) 180 (6 × 30) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Molecular Formulas Methanal (formaldehyde), ethanoic acid (acetic acid), and glucose have the same empirical formula—CH2O. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Molecular Formulas The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Molecular Formulas The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula. Once you have determined the empirical formula of a compound, you can determine its molecular formula, if you know the compound’s molar mass. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Molecular Formulas You can calculate the empirical formula mass (efm) of a compound from its empirical formula. This is simply the molar mass of the empirical formula. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Molecular Formulas You can calculate the empirical formula mass (efm) of a compound from its empirical formula. Then you can divide the experimentally determined molar mass by the empirical formula mass. This quotient gives the number of empirical formula units in a molecule of the compound and is the multiplier to convert the empirical formula to the molecular formula. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Finding the Molecular Formula of a Compound
Sample Problem 10.13 Finding the Molecular Formula of a Compound Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Analyze List the knowns and the unknown.
Sample Problem 10.13 Analyze List the knowns and the unknown. 1 Divide the molar mass by the empirical formula mass to obtain a whole number. Multiply the empirical formula subscripts by this value to get the molecular formula. KNOWNS empirical formula = CH4N molar mass = 60.0 g/mol UNKNOWN molecular formula = C?H?N? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.13 Calculate Solve for the unknown. 2 First calculate the empirical formula mass. efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) g/mol = 30.0 g/mol Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem 10.13 Calculate Solve for the unknown. 2 Divide the molar mass by the empirical formula mass. molar mass efm = 60.0 g/mol 30.0 g/mol = 2 Multiply the formula subscripts by this value. (CH4N) × 2 = C2H8N2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

What information, in addition to empirical formula, is necessary to determine the molecular formula of a compound? Molecular formula can be determined if the empirical formula and the molecular mass of a compound are known. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Key Concepts The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%. The percent composition of a compound can be used to calculate the empirical formula of that compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Key Concepts and Key Equations
The molecular formula of a compound is either the same as its experimentally determined formula, or it is a simple whole-number multiple of its empirical formula. % by mass of element mass of element mass of compound = × 100% % by mass of element mass of element in 1 mol compound molar mass of compound × 100% = Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

percent composition: the percent by mass of each element in a compound
Glossary Terms percent composition: the percent by mass of each element in a compound empirical formula: a formula with the lowest whole-number ratio of elements in a compound; the empirical formula of hydrogen peroxide (H2O2) is HO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

The Mole and Quantifying Matter
BIG IDEA The Mole and Quantifying Matter The molecular formula of a compound can be determined by first finding the percent composition of the compound and determining the empirical formula. Using the empirical formula mass and the molar mass of the compound, the molecular formula can be determined. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Chapter 10 Chemical Quantities

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