Textbook Question
(a) By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of a weak acid. What is the molar mass of the acid if it is monoprotic?
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Ch.4 - Reactions in Aqueous Solution
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement140
- Ch.2 - Atoms, Molecules, and Ions192
- Ch.3 - Chemical Reactions and Reaction Stoichiometry172
- Ch.4 - Reactions in Aqueous Solution156
- Ch.5 - Thermochemistry151
- Ch.6 - Electronic Structure of Atoms137
- Ch.7 - Periodic Properties of the Elements127
- Ch.8 - Basic Concepts of Chemical Bonding143
- Ch.9 - Molecular Geometry and Bonding Theories167
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces97
- Ch.12 - Solids and Modern Materials129
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics175
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Acid-Base Equilibria127
- Ch.17 - Additional Aspects of Aqueous Equilibria133
- Ch.18 - Chemistry of the Environment88
- Ch.19 - Chemical Thermodynamics140
- Ch.20 - Electrochemistry137
- Ch.21 - Nuclear Chemistry99
- Ch.22 - Chemistry of the Nonmetals66
- Ch.23 - Transition Metals and Coordination Chemistry69
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry11
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 4, Problem 106b
(b) An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6% C, and 23.5% O by mass. What is its molecular formula?
Verified step by step guidance1
Step 1: Assume that you have 100 g of the compound. This means that you have 5.89 g of Hydrogen (H), 70.6 g of Carbon (C), and 23.5 g of Oxygen (O).
Step 2: Convert these masses into moles using the molar mass of each element. The molar mass of H is approximately 1 g/mol, the molar mass of C is approximately 12 g/mol, and the molar mass of O is approximately 16 g/mol.
Step 3: Divide each mole value by the smallest number of moles calculated. This will give you the mole ratio of H, C, and O in the compound.
Step 4: If the mole ratios are not whole numbers, multiply them by the smallest number that will convert them all to whole numbers. This will give you the empirical formula of the compound.
Step 5: To find the molecular formula, you need to know the molar mass of the compound. Divide the molar mass of the compound by the molar mass of the empirical formula. This will give you a whole number, which you multiply by the subscripts in the empirical formula to get the molecular formula.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Empirical Formula
The empirical formula represents the simplest whole-number ratio of the elements in a compound. To determine the empirical formula from percentage composition, one converts the percentages to moles by dividing by the atomic masses of the elements. This ratio helps in identifying the basic composition of the compound before determining the molecular formula.
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Empirical vs Molecular Formula
Molecular Formula
The molecular formula indicates the actual number of atoms of each element in a molecule of a compound. It can be derived from the empirical formula by multiplying it by a whole number, which is determined by comparing the molar mass of the compound to the molar mass of the empirical formula. This formula provides more specific information about the compound's structure.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To find the molecular formula, one must first calculate the molar mass of the compound using the atomic masses of its constituent elements. This value is essential for determining the relationship between the empirical and molecular formulas.
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Related Practice
Textbook Question
The discovery of hafnium, element number 72, provideda controversial episode in chemistry. G. Urbain, a Frenchchemist, claimed in 1911 to have isolated an elementnumber 72 from a sample of rare earth (elements 58–71)compounds. However, Niels Bohr believed that hafniumwas more likely to be found along with zirconium thanwith the rare earths. D. Coster and G. von Hevesy, workingin Bohr's laboratory in Copenhagen, showed in 1922 thatelement 72 was present in a sample of Norwegian zircon,an ore of zirconium. (The name hafnium comes from theLatin name for Copenhagen, Hafnia). (c) Solid zirconiumdioxide, ZrO2, reacts with chlorine gas in the presenceof carbon. The products of the reaction are ZrCl4 and twogases, CO2 and CO in the ratio 1:2. Write a balanced chemicalequation for the reaction.
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Textbook Question
Suppose you have 3.00 g of powdered zinc metal, 3.00g of powdered silver metal and 500.0 mL of a 0.2 M copper(II) nitrate solution. (b) What is the net ionic equation that describes this reaction?
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Textbook Question
A sample of 8.69 g of Zn(OH)2 is added to 155.0 mL of 0.750 M H2SO4. (c) How many moles of Zn(OH)2, H2SO4, ZnSO4 are present after the reaction is complete?
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Textbook Question
Suppose you have 3.00 g of powdered zinc metal, 3.00g of powdered silver metal and 500.0 mL of a 0.2 M copper(II) nitrate solution. (a) Which metal will react with the copper(II) nitrate solution?
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