Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (b) NH4NO3(s) → N2O(g) + H2O(g) (c) Zn(OH)2(s) → ZnO(s) + H2O(l) (d) Ag2O(s) → Ag(s) + O2(g)
Ch.3 - Chemical Reactions and Reaction Stoichiometry
Chapter 3, Problem 24
Determine the formula weights of each of the following compounds: (a) Butyric acid, CH3CH2CH2COOH, which is responsible for the rotten smell of spoiled food (b) sodium perborate, NaBO3, a substance used as bleach (c) CF2Cl2, a refrigerant known as Freon (d) NaHCO3, known as baking soda and used in bread and pastry baking (e) isopentyl acetate, CH3CO2C5H11, responsible for the odor of bananas.

1
Identify the elements and their respective counts in the molecular formula of isopentyl acetate, CH3CO2C5H11. The formula can be broken down as: 2 Carbon atoms from the acetyl group (CH3CO), 5 Carbon atoms and 11 Hydrogen atoms from the pentyl group (C5H11), and 2 Oxygen atoms from the acetate group (CO2).
Calculate the total number of each type of atom in the compound: Carbon atoms = 2 (from acetyl) + 5 (from pentyl) = 7, Hydrogen atoms = 3 (from acetyl) + 11 (from pentyl) = 14, Oxygen atoms = 2 (from acetate).
Look up the atomic weights for each element from the periodic table: Carbon (C) approximately 12.01 amu, Hydrogen (H) approximately 1.008 amu, and Oxygen (O) approximately 16.00 amu.
Multiply the atomic weight of each element by the number of atoms of that element in the compound: Carbon contribution = 7 atoms × 12.01 amu, Hydrogen contribution = 14 atoms × 1.008 amu, Oxygen contribution = 2 atoms × 16.00 amu.
Add up all the contributions to get the total formula weight of isopentyl acetate: Total formula weight = Carbon contribution + Hydrogen contribution + Oxygen contribution.

Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
1mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molecular Formula
The molecular formula of a compound indicates the number and types of atoms present in a molecule. For isopentyl acetate, the formula is CH3CO2C5H11, which shows it contains carbon (C), hydrogen (H), and oxygen (O) atoms. Understanding how to interpret molecular formulas is essential for calculating formula weights.
Recommended video:
Guided course
Determining Molecular Formulas
Molar Mass Calculation
Molar mass is the mass of one mole of a substance, calculated by summing the atomic masses of all atoms in the molecular formula. Each element's atomic mass can be found on the periodic table. For isopentyl acetate, you would multiply the number of each type of atom by its atomic mass and then sum these values to find the total formula weight.
Recommended video:
Guided course
Molar Mass Calculation Example
Empirical vs. Molecular Formula
The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms. In the case of isopentyl acetate, understanding the difference helps in recognizing that the molecular formula provides more detailed information about the compound's structure and composition, which is crucial for accurate weight calculations.
Recommended video:
Guided course
Empirical vs Molecular Formula
Related Practice
Textbook Question
337
views
Textbook Question
Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) NaClO3(s) → NaCl(s) + O2(g) (b) NH4OH(aq) → H2O(l) + NH3(g) (c) K(s) + Cl2(g) → KCl(s) (d) C2H5OH(l) + O2(g) → CO2(g) + H2O(l)
335
views
Textbook Question
Calculate the percentage by mass of oxygen in the following compounds: (a) vanillin, C8H8O3
1053
views
Textbook Question
Calculate the percentage by mass of oxygen in the following compounds: (b) isopropyl alcohol, C3H8O
1218
views
1
comments
Textbook Question
Calculate the percentage by mass of oxygen in the following compounds: (c) acetaminophen, C8H9NO2 (d) cyclopropanone, C3H4O
703
views