Textbook Question
Heating elemental cesium and platinum together for two days at 973 K gives a dark red ionic compound that is 57.67% Cs and 42.33% Pt. (a) What is the empirical formula of the compound?
442
views
Ch.3 - Mass Relationships in Chemical Reactions
McMurry8th EditionChemistryISBN: 9781292336145
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch.1 - Chemical Tools: Experimentation & Measurement143
- Ch.2 - Atoms, Molecules & Ions134
- Ch.3 - Mass Relationships in Chemical Reactions149
- Ch.4 - Reactions in Aqueous Solution157
- Ch.5 - Periodicity & Electronic Structure of Atoms92
- Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory73
- Ch.7 - Covalent Bonding and Electron-Dot Structures47
- Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure51
- Ch.9 - Thermochemistry: Chemical Energy104
- Ch.10 - Gases: Their Properties & Behavior138
- Ch.11 - Liquids & Phase Changes43
- Ch.12 - Solids and Solid-State Materials56
- Ch.13 - Solutions & Their Properties98
- Ch.14 - Chemical Kinetics79
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Aqueous Equilibria: Acids & Bases94
- Ch.17 - Applications of Aqueous Equilibria125
- Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium66
- Ch.19 - Electrochemistry130
- Ch.20 - Nuclear Chemistry73
- Ch.21 - Transition Elements and Coordination Chemistry122
- Ch.22 - The Main Group Elements155
- Ch.23 - Organic and Biological Chemistry43
Chapter 3, Problem 100
Combustion analysis of a 31.472 mg sample of the widely used flame retardant Decabrom gave 1.444 mg of CO2. Is the molecular formula of Decabrom C12Br10 or C12Br10O?
Verified step by step guidance1
Step 1: The first step in solving this problem is to understand what combustion analysis is. Combustion analysis is a method used to determine the elemental composition of a substance. In this case, the combustion of Decabrom produces CO2, which contains the carbon from the original compound.
Step 2: We know that the mass of the CO2 produced is 1.444 mg. We also know that each molecule of CO2 contains one atom of carbon. Therefore, we can use the molar mass of carbon (12.01 g/mol) and the molar mass of CO2 (44.01 g/mol) to find the mass of carbon in the original sample. This can be done using the formula: (mass of CO2) x (molar mass of carbon / molar mass of CO2).
Step 3: Once we have the mass of carbon, we can subtract it from the total mass of the sample to find the mass of bromine. This can be done using the formula: (total mass of sample) - (mass of carbon).
Step 4: Now that we have the mass of carbon and bromine, we can find the number of moles of each element using their respective molar masses. The number of moles is calculated by dividing the mass by the molar mass.
Step 5: Finally, we can compare the ratio of moles of carbon to bromine in the sample to the ratio in the proposed formulas. If the ratio matches the formula C12Br10, then that is the correct formula. If it matches C12Br10O, then that is the correct formula. Remember, the molecular formula represents the actual number of atoms of each element in a molecule of the compound.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Combustion Analysis
Combustion analysis is a technique used to determine the composition of organic compounds by burning a sample in excess oxygen. The products of combustion, typically carbon dioxide (CO2) and water (H2O), are measured to calculate the amounts of carbon and hydrogen in the original sample. This method is essential for deducing the empirical formula of a compound, which can then be used to infer its molecular formula.
Recommended video:
Guided course
01:29
Combustion of Non-Hydrocarbons
Molecular vs. Empirical Formula
The molecular formula of a compound indicates the actual number of atoms of each element in a molecule, while the empirical formula represents the simplest whole-number ratio of these atoms. Understanding the difference is crucial when analyzing combustion data, as the empirical formula derived from combustion analysis may not directly reflect the molecular formula, especially in compounds with multiple elements or complex structures.
Recommended video:
Guided course
02:26Empirical vs Molecular Formula
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions based on the conservation of mass. In combustion analysis, stoichiometric relationships allow us to convert the mass of CO2 produced into moles of carbon, which can then be used to determine the carbon content of the original sample. This quantitative approach is vital for accurately deducing the molecular formula from experimental data.
Recommended video:
Guided course
01:16Stoichiometry Concept
Related Practice
Textbook Question
Disilane, Si2Hx, is analyzed and found to contain 90.28% silicon by mass. What is the value of x?
986
views
Textbook Question
The stimulant amphetamine contains only carbon, hydrogen, and nitrogen. Combustion analysis of a 42.92 mg sample of amphetamine gives 37.187 mg of H2O and 125.75 mg of CO2. If the molar mass of amphetamine is less than 160 g/mol, what is its molecular formula?
660
views
Textbook Question
A certain metal sulfide, MS2, is used extensively as a high-temperature lubricant. If MS2 is 40.06% sulfur by mass, what is the identity of the metal M?
906
views
1
rank