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Air is a mixture of many gases. However, in calculating its molar mass we need consider only the three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is made up of 78.08 percent nitrogen, 20.95 percent oxygen, and 0.97 percent argon, what is the molar mass of air?

Short Answer

Expert verified
The molar mass of air is 28.96 g/mol.

Step by step solution

01

Understand the Components of Air

Air consists of nitrogen (N₂), oxygen (O₂), and argon (Ar). These components make up the air in different percentage by moles: 78.08% nitrogen, 20.95% oxygen, and 0.97% argon.
02

Identify Molar Masses of Individual Gases

To calculate the molar mass of each gas: - Nitrogen (N₂) has a molar mass of 28.02 g/mol. - Oxygen (O₂) has a molar mass of 32.00 g/mol. - Argon (Ar) has a molar mass of 39.95 g/mol.
03

Calculate Molar Mass Contribution of Each Gas

To find the molar mass contribution of each gas, multiply the molar mass by its mole fraction (percentage divided by 100):\[\text{Nitrogen contribution} = 28.02 \times \frac{78.08}{100} = 21.87 \, \text{g/mol}\]\[\text{Oxygen contribution} = 32.00 \times \frac{20.95}{100} = 6.70 \, \text{g/mol}\]\[\text{Argon contribution} = 39.95 \times \frac{0.97}{100} = 0.39 \, \text{g/mol}\]
04

Sum the Contributions to Find Total Molar Mass

Add up the contributions of all the gases to find the total molar mass of air:\[\text{Total molar mass of air} = 21.87 + 6.70 + 0.39 = 28.96 \, \text{g/mol}\]

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Key Concepts

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

Understanding Nitrogen in Air
Nitrogen is a key component of Earth's atmosphere, making up approximately 78% by volume. This high percentage plays a crucial role in determining the molar mass of air. Nitrogen in the atmosphere exists primarily in the form of diatomic nitrogen gas, represented by the chemical formula \( \text{N}_2 \). Each nitrogen molecule is composed of two nitrogen atoms that are bound together. This bond is very strong, which makes nitrogen less reactive under normal conditions.

The molar mass of nitrogen, \( \text{N}_2 \), is calculated by adding the atomic masses of the two nitrogen atoms, each approximately 14.01 g/mol. Thus, the molar mass of nitrogen gas is 28.02 g/mol. In the context of atmospheric air, nitrogen's contribution to the molar mass is determined by multiplying its molar mass with its mole fraction: \[ \text{Nitrogen contribution} = 28.02 \, \text{g/mol} \times \frac{78.08}{100} = 21.87 \, \text{g/mol} \] This indicates that nitrogen has the largest impact on the total molar mass of air due to its abundance.
Role of Oxygen in Molar Mass of Air
Oxygen is the second most abundant component in Earth's atmosphere, making up about 21% by volume. This abundance significantly influences the molar mass of the air we breathe. Oxygen is present as a diatomic molecule, \( \text{O}_2 \), meaning each molecule of oxygen contains two oxygen atoms. This molecular structure is crucial for many life processes, including respiration.

The molar mass of an \( \text{O}_2 \) molecule is the sum of the masses of two oxygen atoms, which adds up to 32.00 g/mol. To calculate oxygen's contribution to the molar mass of air, multiply by its mole fraction: \[ \text{Oxygen contribution} = 32.00 \, \text{g/mol} \times \frac{20.95}{100} = 6.70 \, \text{g/mol} \] Oxygen contributes significantly to the total molar mass of air despite its lower percentage compared to nitrogen, due to its higher individual molar mass.
Inclusion of Argon in Air Composition
Argon is an inert gas that comprises about 0.97% of the atmosphere. Though present in much smaller quantities than nitrogen and oxygen, argon's role cannot be overlooked when calculating the molar mass of air. Argon atoms exist as monatomic species, meaning they are not bonded to other atoms. This characteristic makes argon highly stable and chemically non-reactive under most conditions.

The molar mass of argon is 39.95 g/mol, which is quite high compared to nitrogen and oxygen. Although its percentage in air is small, the heavy molar mass of argon means it still contributes to the overall molar mass of air. To see its impact, multiply its molar mass by its relative presence: \[ \text{Argon contribution} = 39.95 \, \text{g/mol} \times \frac{0.97}{100} = 0.39 \, \text{g/mol} \] While argon contributes less to the total molar mass due to low abundance, the difference is not negligible because of its heavy atomic mass.

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Most popular questions from this chapter

When \(0.273 \mathrm{~g}\) of \(\mathrm{Mg}\) is heated strongly in a nitrogen \(\left(\mathrm{N}_{2}\right)\) atmosphere, a chemical reaction occurs. The product of the reaction weighs \(0.378 \mathrm{~g}\). Calculate the empirical formula of the compound containing \(\mathrm{Mg}\) and \(\mathrm{N}\). Name the compound.

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) \(2 \mathrm{NaHCO}_{3} \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O},(\mathrm{b}) \mathrm{NH}_{3}+\mathrm{HCl} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}\) (c) \(2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

What mole ratio of molecular chlorine \(\left(\mathrm{Cl}_{2}\right)\) to molecular oxygen \(\left(\mathrm{O}_{2}\right)\) would result from the breakup of the compound \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) into its constituent elements?

A certain metal oxide has the formula MO where \(\mathrm{M}\) denotes the metal. A \(39.46-\mathrm{g}\) sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, \(31.70 \mathrm{~g}\) of the metal is left over. If \(\mathrm{O}\) has an atomic mass of 16.00 amu, calculate the atomic mass of \(\mathrm{M}\) and identify the element.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

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