Chapter 20: Problem 23

The molar mass of cellulose is approximately \(6.0 \times 10^{5} \mathrm{~g} / \mathrm{mol}\), and the molar mass of a soluble starch is on the order of \(4.0 \times 10^{3} \mathrm{~g} / \mathrm{mol}\). The monomer unit in both of these molecules has the empirical formula \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{5}\). The units are about \(5.0 \times 10^{-10} \mathrm{~m}\) long. About how many units occur in each molecule, and how long are the molecules of cellulose and starch as a result?

Short Answer

Expert verified

Cellulose has about 3704 units and is about \(1.85 \times 10^{-6} \mathrm{m}\) long. Starch has about 25 units and is about \(1.25 \times 10^{-8} \mathrm{m}\) long.

Step by step solution

Understand the problem

Determine the number of monomer units in each molecule by dividing the molar mass of each polymer by the molar mass of a single monomer unit. Then, calculate the length of each molecule by multiplying the number of monomer units by the length of one monomer unit.

Find the molar mass of a single monomer unit

First, calculate the molar mass of one monomer unit, \(\text{C}_6\text{H}_{10}\text{O}_5\). Using the atomic masses (approximate): \C: 12 \mathrm{g/mol}\, \H: 1 \mathrm{g/mol}\, \O: 16 \mathrm{g/mol}\, we calculate: \[ 6(12) + 10(1) + 5(16) = 72 + 10 + 80 = 162 \mathrm{g/mol} \]

Calculate the number of monomer units

- For cellulose: \[ N_{cellulose} = \frac{6.0 \times 10^5 \mathrm{g/mol}}{162 \mathrm{g/mol}} \approx 3704 \] \- For starch: \[ N_{starch} = \frac{4.0 \times 10^3 \mathrm{g/mol}}{162 \mathrm{g/mol}} \approx 24.7 \approx 25 \]

Calculate the length of each molecule

- For cellulose: \[ L_{cellulose} = N_{cellulose} \times 5.0 \times 10^{-10} \mathrm{m} \approx 3704 \times 5.0 \times 10^{-10} \mathrm{m} \approx 1.85 \times 10^{-6} \mathrm{m} \] \- For starch: \[ L_{starch} = N_{starch} \times 5.0 \times 10^{-10} \mathrm{m} \approx 25 \times 5.0 \times 10^{-10} \mathrm{m} \approx 1.25 \times 10^{-8} \mathrm{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

polymer chemistry

Polymer chemistry focuses on the study of polymers, which are large molecules made up of repeating subunits called monomers.
Polymers can be natural, like cellulose and starch, or synthetic.
In this exercise, we are examining natural polymers: cellulose and starch.
Both have the same monomer unit but differ greatly in their molar masses.
Because of their large sizes, they have unique properties that make them useful in various applications.
Understanding their structure helps in tailoring these properties for industrial or practical uses.

molar mass

Molar mass is the mass of one mole of a substance in grams per mole (g/mol).
In this exercise, we need to find the molar mass of a monomer unit and use it to calculate the number of such units in polymers.
The molar mass of cellulose is approximately \(6.0 \times 10^{5} \mathrm{~g} / \mathrm{mol}\), while that of soluble starch is around \(4.0 \times 10^{3} \mathrm{~g} / \mathrm{mol}\).
Calculating the molar masses accurately allows us to determine the number of monomer units in these polymers.
This makes molar mass a foundational concept for understanding polymer structures.

monomer units

Monomer units are the repeating building blocks of polymers.
In our example, both cellulose and starch have the same monomer unit with an empirical formula \(\text{C}_6\text{H}_{10}\text{O}_5\)
This monomer unit forms the basis of their structures.
To determine the number of monomer units in a polymer, divide the polymer's molar mass by the molar mass of a single monomer unit.
For cellulose: \(\frac{6.0 \times 10^5 \mathrm{g/mol}}{162 \mathrm{g/mol}} \approx 3704\).
For starch: \(\frac{4.0 \times 10^3 \mathrm{g/mol}}{162 \mathrm{g/mol}} \approx 25\).
This reveals how many times the monomer repeats in the chains.

empirical formula

The empirical formula of a compound expresses the simplest whole-number ratio of the atoms in that molecule.
For our monomer unit, the empirical formula is \(\text{C}_6\text{H}_{10}\text{O}_5\).
This informs us about the relative quantities of carbon, hydrogen, and oxygen atoms in each monomer.
Knowing the empirical formula helps in calculating the molar mass of the monomer.
For example: \(6(12) + 10(1) + 5(16) = 72 + 10 + 80 = 162 \mathrm{g/mol}\).
Accurate empirical formulas are crucial for various calculations in chemistry, including determining the composition of polymers.

molecular length calculations

To find the length of a polymer molecule, we multiply the number of monomer units by the length of each monomer unit.
For cellulose, this is \(3704 \times 5.0 \times 10^{-10} \mathrm{m} = 1.85 \times 10^{-6} \mathrm{m}\).
For starch, this is \(25 \times 5.0 \times 10^{-10} \mathrm{m} = 1.25 \times 10^{-8} \mathrm{m}\).
Such calculations are important for understanding the physical dimensions of polymers, impacting their application and functionality.
Longer chains often mean stronger and more durable materials, essential for various industries and products.

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Short Answer

Step by step solution

Understand the problem

Find the molar mass of a single monomer unit

Calculate the number of monomer units

Calculate the length of each molecule

Key Concepts

polymer chemistry

molar mass

monomer units

empirical formula

molecular length calculations

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