Problem 47 - Chemistry The Molecular Nature of Matter

Question

Sucrose (table sugar) has the formula \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\) In this compound, what is the (a) atom ratio of \(\mathrm{C}\) to \(\mathrm{H}\) ? (c) atom ratio of \(\mathrm{H}\) to \(\mathrm{O}\) ? (b) mole ratio of \(\mathrm{C}\) to \(\mathrm{O}\) ? (d) mole ratio of \(\mathrm{H}\) to \(\mathrm{O}\) ?

Step-by-Step Solution

Answer

The atom ratio of \(\mathrm{C}\) to \(\mathrm{H}\) is 12:22 or approximately 0.545:1. The atom ratio of \(\mathrm{H}\) to \(\mathrm{O}\) is 22:11 or 2:1. The mole ratio of \(\mathrm{C}\) to \(\mathrm{O}\) is 12:11. The mole ratio of \(\mathrm{H}\) to \(\mathrm{O}\) is 22:11 or 2:1.

Step 1: Identifying the Numbers of Atoms

Look at the chemical formula for sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), and identify the subscript numbers for each element which indicate the number of atoms of each element in one molecule of sucrose.

Step 2: Finding Atom Ratios

Divide the number of atoms of each element by the number of atoms of the second element to find the ratio. For part (a), divide the number of carbon atoms (12) by the number of hydrogen atoms (22). For part (c), divide the number of hydrogen atoms (22) by the number of oxygen atoms (11).

Step 3: Finding Mole Ratios for Carbon to Oxygen

Since the mole ratio is the ratio of moles of one element to moles of another and one molecule of sucrose has the same ratio of moles as it does of atoms for any given element, for part (b), the mole ratio for \(\mathrm{C}\) to \(\mathrm{O}\) is the same as the atom ratio, so you would divide the number of carbon atoms (12) by the number of oxygen atoms (11).

Step 4: Finding Mole Ratios for Hydrogen to Oxygen

Similarly, for part (d), the mole ratio for \(\mathrm{H}\) to \(\mathrm{O}\) is the same as the atom ratio. Divide the number of hydrogen atoms (22) by the number of oxygen atoms (11).

Key Concepts

Chemical Formula

A chemical formula, such as the one for sucrose \( \mathrm{C}_{12} \mathrm{H}_{22}\mathrm{O}_{11} \), provides a wealth of information in a simple and compact form. It tells us the types of atoms that make up a compound as well as the precise number of each type of atom.

When looking at sucrose’s chemical formula, we can see there are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. This formula is fundamental for further calculations, such as determining atom and mole ratios, which are crucial for understanding molecular composition and stoichiometry.

Atom Ratio

The atom ratio in a compound provides insight into the proportion of one type of atom relative to another. In our sucrose example, to find the atom ratio of carbon to hydrogen (part a), divide the number of carbon atoms by the number of hydrogen atoms which gives us \( \frac{12}{22} \), or simply \( \frac{6}{11} \) when reduced.

Similarly, for the ratio of hydrogen to oxygen (part c), the process involves dividing 22 hydrogen atoms by 11 oxygen atoms resulting in a ratio of 2:1. Such ratios are fundamental to understanding the qualitative composition of the molecule. They also play a role in stoichiometry when balancing equations or predicting the outcome of a reaction.

Molecular Composition

Molecular composition refers to the elements that make up a molecule and their specific arrangement. In the case of sucrose, the molecule is composed of carbon, hydrogen, and oxygen atoms arranged in a specific pattern that gives sucrose its unique properties.

The molecular composition gives chemists and students alike a blueprint of the molecule, which can be used to understand its behavior, reactivity, and how it interacts with other substances. The arrangement and proportions of these atoms are also fundamental in the study of isomerism where molecules have the same molecular formula but different structures.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships, or ratios, between the reactants and products in a chemical reaction. It’s grounded in the conservation of mass and the concept of moles.

For instance, in our sucrose example, we use stoichiometry to deduce that the mole ratio (part b) is the same as the atom ratio because mole reflects the amount of substance. Hence, for carbon to oxygen, it is \( \frac{12}{11} \). Likewise, for hydrogen to oxygen (part d), the mole ratio is 2:1. Understanding these ratios is crucial when predicting how much of a reactant is needed to produce a desired amount of product in a chemical reaction. It also helps in calculating yields and determining reaction rates.