Problem 44 - Introductory Chemistry Atoms First

Question

How many years are there in 1 mole of seconds? Use conversion factors, and show which units cancel.

Step-by-Step Solution

Answer

There are approximately \(1.91 \times 10^{16}\) years in one mole of seconds.

Step 1: Determine the conversion factors required

To convert from seconds to years, we'll need to go through some intermediate units: minutes, hours, and days. The necessary conversion factors are: - 1 minute = 60 seconds - 1 hour = 60 minutes - 1 day = 24 hours - 1 year = 365.25 days (accounting for leap years)

Step 2: Write the given amount of time in terms of a mole of seconds

A mole is simply a very large number used in chemistry, defined as Avogadro's number, which is approximately \(6.022 \times 10^{23}\) units in one mole. So, we are given that the time is 1 mole of seconds, or \(t = 6.022 \times 10^{23}\) seconds.

Step 3: Convert seconds to minutes

Now, let's start converting seconds to minutes using the conversion factors. $$ t \,(\text{minutes}) = 6.022 \times 10^{23}\ \text{seconds} \times \frac{1 \, \text{minute}}{60\, \text{seconds}} $$

Step 4: Convert minutes to hours

Next, we'll convert minutes to hours using the conversion factors. $$ t \,(\text{hours}) = 6.022 \times 10^{23}\ \text{seconds} \times \frac{1 \, \text{minute}}{60\, \text{seconds}} \times \frac{1\, \text{hour}}{60\,\text{minutes}} $$

Step 5: Convert hours to days

Continuing with the conversion process, we will now convert hours to days. $$ t \,(\text{days}) = 6.022 \times 10^{23}\ \text{seconds} \times \frac{1 \, \text{minute}}{60\, \text{seconds}} \times \frac{1\, \text{hour}}{60\,\text{minutes}} \times \frac{1\, \text{day}}{24\, \text{hours}} $$

Step 6: Convert days to years

Finally, we'll convert days to years, taking into account leap years by using an average of 365.25 days per year. $$ t \,(\text{years}) = 6.022 \times 10^{23}\ \text{seconds} \times \frac{1 \, \text{minute}}{60\, \text{seconds}} \times \frac{1\, \text{hour}}{60\,\text{minutes}} \times \frac{1\, \text{day}}{24\, \text{hours}} \times \frac{1 \,\text{year}}{365.25\,\text{days}} $$

Step 7: Calculate and simplify

Now that all the conversion factors are in place, calculate the final expression to find the number of years. \[ t \,(\text{years}) = 6.022 \times 10^{23}\ \text{seconds} \times \frac{1 \, \text{minute}}{60\, \text{seconds}} \times \frac{1\, \text{hour}}{60\,\text{minutes}} \times \frac{1\, \text{day}}{24\, \text{hours}} \times \frac{1 \,\text{year}}{365.25\,\text{days}} \] Notice that the units of seconds, minutes, hours, and days all cancel out, leaving us only with years. \[ t \, (years) = 6.022 \times 10^{23} \times \frac{1}{(60)(60)(24)(365.25)} \, \text{years} \] Using a calculator, we find: \[ t \,(\text{years}) \approx 1.91 \times 10^{16} \, \text{years} \] So there are approximately \(1.91 \times 10^{16}\) years in one mole of seconds.

Key Concepts

Avogadro's Number

Avogadro's Number is a fundamental constant in chemistry, representing a very large quantity. Named after the scientist Amedeo Avogadro, this number is approximately \(6.022 \times 10^{23}\), meaning it corresponds to the number of atoms, molecules, or other elementary entities found in one mole of a substance.
Knowing Avogadro's Number helps scientists measure and compare large quantities of tiny particles. For example, in the given exercise, 1 mole of seconds is a representation of \(6.022 \times 10^{23}\) seconds, which underscores how substantial and abstract such a number is.
The concept is central to converting between atoms or molecules and grams, often seen in stoichiometry calculations in chemistry. Grasping this idea is key to understanding chemistry at the molecular level. It's not just a large number—it's the bridge between the atomic universe and the macroscopic world.

Unit Conversion

Unit conversion involves changing one unit of measurement to another without changing the quantity's value. In the exercise, the task was to find the number of years in one mole of seconds by converting through intermediate time units such as minutes, hours, and days.
In unit conversion:

• Use conversion factors, which are ratios expressing how many of one unit correspond to another.
• Set up your equations so that unwanted units cancel out, leaving the desired units.
• Be mindful of precision and accuracy, especially when dealing with small or large numbers.

Practicing unit conversions is crucial for students to move seamlessly between units in various scientific computations. Effective unit conversion aids in understanding different scales and contexts in scientific work.

Dimensional Analysis

Dimensional Analysis is a method used to convert one kind of unit into another by employing conversion factors strategically. This systematic approach incorporates the principle that the dimensions (units) of both sides of any equation must be the same.
In the given solution:

• We systematically converted seconds into years.
• We achieved this by ensuring each intermediate step involved only cancelling units that were unnecessary for the desired output.

Here is how it works: - Start with the known quantity (1 mole of seconds). - Use conversion factors which will cancel previous units gradually leading to the required units. - Understanding the sequential process helps validate the solution for errors and confirm the final unit matches the required outcome.
It's a powerful method for ensuring clarity and accuracy when solving complex problems involving different units.

Scientific Notation

Scientific Notation is a way of expressing very large or very small numbers compactly, utilizing powers of ten. It is used extensively in science to make calculations more manageable and to simplify data interpretation.
In this problem, we expressed the incredibly large number of seconds in a mole as \(6.022 \times 10^{23}\) using scientific notation. This notation makes:

• Complex numbers easier to read and comprehend.
• Mathematical operations, like multiplication and division, simpler as it involves basic exponent arithmetic.

Scientific notation is essential for handling extremely high or low values, which are frequent in scientific and engineering fields. By reducing room for error in arithmetic and providing a standard form, it assists in maintaining consistency across data handling and reporting in scientific research.