Question

It has been estimated that each square meter of the earth's surface supports \(1 \times 10^{7} \mathrm{~g}\) of air above it. If air is \(20 \%\) oxygen \(\left(\mathrm{O}_{2}\right.\), molecular weight \(\left.=32 \mathrm{~g} / \mathrm{mole}\right)\) by weight, approximately how many moles of \(\mathrm{O}_{2}\) are there above each square meter of the earth?

Short answer

There are approximately \(6.25 \times 10^{4}\) moles of oxygen above each square meter of the earth's surface.

Step-by-step solution

Determine the Weight of Oxygen above Each Square Meter

Find the percentage of oxygen in the air, 20%, and multiply it by the weight of the air above each square meter (1 x 10^7 g) to get the weight of oxygen. Weight of oxygen = (Weight of air) x (Percentage of oxygen in the air) = (1 x 10^7 g) x 0.20

Calculate the Weight of Oxygen above Each Square Meter

Compute the weight of oxygen by multiplying as mentioned in the previous step. Weight of oxygen = (1 x 10^7 g) x 0.20 = 2 x 10^6 g

Find the Moles of Oxygen Above Each Square Meter

Now we know the weight of oxygen above each square meter. To find the number of moles, we use the molecular weight of oxygen, which is given to be 32 g/mole. Moles of oxygen = (Weight of oxygen) / (Molecular weight of oxygen) = (2 x 10^6 g) / (32 g/mole)

Calculate the Moles of Oxygen Above Each Square Meter

Perform the division to find the moles of oxygen above each square meter: Moles of oxygen = (2 x 10^6 g) / (32 g/mole) = 6.25 x 10^4 moles So, there are approximately \(6.25 \times 10^{4}\) moles of oxygen above each square meter of the earth's surface.

Key concepts

Atmospheric Chemistry

In the field of atmospheric chemistry, we study the components and processes happening within the Earth's atmosphere. One of the significant elements of our air is oxygen (\(\mathrm{O}_{2}\)), which plays crucial roles not only in sustaining life but also in chemical reactions that occur in our atmosphere.

A key aspect here is understanding the "composition of air," which essentially is a mix of gases with nitrogen comprising about 78% and oxygen accounting for roughly 20% by volume.

Atmospheric chemists are particularly interested in the concentration of oxygen because it affects:

• Respiration in living creatures
• Combustion processes
• Various biochemical reactions

Understanding how much oxygen is present can be pivotal for studies concerning climate change, pollution, and the overall health of our planet.

Molecular Weight

Molecular weight, also known as molecular mass, refers to the mass of a molecule. It is calculated by summing the atomic weights of all atoms present in a specific molecule.
The molecular weight of a substance is usually expressed in atomic mass units (amu) or grams per mole (g/mol).

In the case of oxygen gas (\(\mathrm{O}_2\)), its molecular weight is determined by adding the atomic weights of two oxygen atoms. Since each oxygen atom has an atomic weight of 16 g/mol, the molecular weight of \(\mathrm{O}_2\)turns out to be:

• 16 g/mol + 16 g/mol = 32 g/mol

Molecular weight helps us understand and calculate how molecules react and combine in chemical processes. It serves as a conversion bridge between the mass of a substance and the number of molecules or moles, which is essential for accurately interpreting chemical reactions and quantities.

Percentage by Weight

The concept of "percentage by weight" allows chemists to express the concentration of a component within a mixture based on weight rather than volume. It's a simple yet powerful way to denote proportions, crucial when dealing with non-liquid phases like gases or solids.

It is calculated by the formula:\[\text{Percentage by weight} = \left( \frac{\text{weight of the component}}{\text{total weight of the mixture}} \right) \times 100%\]In our example, oxygen comprises 20% of the air by weight. This means that for any given weight of air, 20% of it is oxygen. For air weighing \(1 \times 10^{7} \mathrm{~g}\) above each square meter of Earth, the weight of oxygen is:

• \(0.20 \times 1 \times 10^{7} \mathrm{~g} = 2 \times 10^6 \mathrm{~g}\)

This percentage allows us to easily compute the weight of oxygen or any other component in the air, facilitating further analyses or reactions.

Chemical Calculations

Chemical calculations are essential for converting the information between different physical quantities, like weight and moles. These calculations allow scientists and students to understand and predict the outcomes of chemical reactions.

This process often involves:

• Determining the weight of a component in a mixture
• Converting weight to moles using molecular weight
• Applying stoichiometric ratios to solve problems

Let's apply these steps to determine the moles of oxygen in our exercise:
First, we have calculated the weight of oxygen above each square meter, which is \(2 \times 10^6 \mathrm{g} \).

To find out how many moles of oxygen this corresponds to, we use the molecular weight of oxygen, which is \(32 \mathrm{g/mol} \). The number of moles is calculated as follows:\[\text{Moles of oxygen} = \frac{2 \times 10^6 \mathrm{g}}{32 \mathrm{g/mol}} = 6.25 \times 10^4 \ ext{moles}\]Such calculations illustrate how physicochemical quantities are interconnected, ensuring that results are precise, scalable, and applicable to various scientific investigations.