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How many \(\mathrm{O}_{2}\) molecules are there in \(1.00 \mathrm{~g}\) of \(\mathrm{O}_{2}\) molecules?

Short Answer

Expert verified
There are approximately \(1.88 \times 10^{22}\) \(\mathrm{O}_2\) molecules in 1.00 gram of \(\mathrm{O}_2\).

Step by step solution

01

Calculate the molecular weight of \(\mathrm{O}_2\)

First, let's determine the molecular weight (also called molar mass) of an \(\mathrm{O}_2\) molecule. Oxygen has an atomic weight of approximately 16.00 grams/mol. Since \(\mathrm{O}_2\) consists of two oxygen atoms, we can calculate its molecular weight as follows: Molecular weight of \(\mathrm{O}_2 = 2 \times \mathrm{Atomic~weight~of~Oxygen} = 2 \times 16.00 \mathrm{g/mol} = 32.00 \mathrm{g/mol}\)
02

Convert grams of \(\mathrm{O}_2\) to moles

Now, we will convert the grams of \(\mathrm{O}_2\) to moles using the molecular weight we calculated in the previous step. To convert grams to moles, we use the formula: Moles = (grams of substance) / (molecular weight of substance) Following this formula, we will find the moles of \(\mathrm{O}_2\): Moles of \(\mathrm{O}_2\) = \(\frac{1.00 \mathrm{~g}}{32.00 \mathrm{~g/mol}} = 0.03125 \mathrm{~mol}\)
03

Use Avogadro's number to find the number of molecules of \(\mathrm{O}_2\)

Lastly, we will use Avogadro's number (approximately \(6.022 \times 10^{23}\)) to determine the number of \(\mathrm{O}_2\) molecules in the given moles. We can do this by multiplying the moles of \(\mathrm{O}_2\) by Avogadro's number: Number of \(\mathrm{O}_2\) molecules = (moles of \(\mathrm{O}_2\)) × (Avogadro's number) Number of \(\mathrm{O}_2\) molecules = \(0.03125 \mathrm{~mol} \times 6.022 \times 10^{23} \mathrm{molecules/mol} = 1.88 \times 10^{22} \mathrm{molecules}\) Therefore, there are approximately \(1.88 \times 10^{22}\) \(\mathrm{O}_2\) molecules in 1 gram of \(\mathrm{O}_2\).

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Key Concepts

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

Molecular Weight Calculation
Understanding how to calculate molecular weight is essential for various chemical calculations. Each atom has an atomic weight usually expressed in grams per mole (g/mol). Molecular weight, also known as molar mass, is the sum of the atomic weights of all the atoms in a molecule. For instance, an oxygen (\text{O}) atom has an atomic weight of approximately 16.00 g/mol. A molecule of oxygen (\text{O}_2) consists of two oxygen atoms, hence the molecular weight of \text{O}_2 is the sum of the atomic weights of two oxygen atoms which can be calculated as follows:
Molecular weight of \text{O}_2 = 2 \( \times \) Atomic weight of Oxygen = 2 \( \times \) 16.00 g/mol = 32.00 g/mol.
This calculation is vital when transitioning from grams to moles and vice versa, as it acts as a conversion factor in stoichiometry.
Avogadro's Number
Avogadro's number plays a pivotal role in chemistry when dealing with atoms, ions, or molecules on a macroscopic scale. This constant, approximately \(6.022 \times 10^{23}\), represents the number of particles in one mole of a substance. It's named after the Italian scientist Amedeo Avogadro and is a fundamental component in the mole concept. Whether you're counting out molecules of oxygen, carbon, or any other element or compound, Avogadro's number is the bridge connecting the microscopic world of atoms to the quantities we can measure and observe in the laboratory.
Molecules to Moles Conversion
The path from molecules to moles is guided by Avogadro's number. To perform a conversion, you must understand the reciprocal nature of this constant: while one mole of any substance contains \(6.022 \times 10^{23}\) entities, the number of moles corresponding to a given number of molecules is determined by dividing the number of molecules by Avogadro's number.
For instance, if you have \(1.88 \times 10^{22}\) molecules of \text{O}_2, the number of moles can be calculated by dividing this by Avogadro's number, effectively reversing the multiplication process used to find the number of molecules from moles.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's based on the conservation of mass and the mole concept and involves calculations that use the balanced chemical equation to understand these relationships.
Molecular weight calculations, Avogadro's number, and the mole concept are cornerstones of stoichiometric calculations. They allow chemists to predict the quantities of substances consumed and produced in a reaction, facilitating the optimization of the reactants needed and predicting the yield of the products.

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Most popular questions from this chapter

Butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), used as the fuel in disposable lighters, reacts with oxygen \(\left(\mathrm{O}_{2}\right)\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Suppose \(10.00 \mathrm{~g}\) butane is combined with \(10.00 \mathrm{~g} \mathrm{O}_{2}\). (a) Write a balanced chemical equation for the combustion of butane. (b) Which reactant is limiting? (c) What is the theoretical yield of each product for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) How many additional grams of the limiting reactant are required to run this reaction in a balanced fashion?

Consider the unbalanced chemical equation \(\mathrm{Al}_{2} \mathrm{~S}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{~S}\) If \(56.0 \mathrm{~g}\) of aluminum sulfide reacts with \(48.2 \mathrm{~g}\) of water, (a) Which is the excess reactant? (b) What mass in grams of the excess reactant remains after the reaction is complete?

Consider the unbalanced chemical equation \(\mathrm{CaCN}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3}+\mathrm{NH}_{3}\) If you began the reaction with \(5.65 \mathrm{~g}\) of \(\mathrm{CaCN}_{2}\) and \(12.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{NH}_{3}\) would be produced if the reaction had an \(86.0 \%\) yield?

Acetaminophen, used in many over-the-counter pain relievers, has a molecular formula of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} .\) Calculate the mass percentage of each element in acetaminophen.

In a chemical reaction, the mass of the products is equal to the mass of the reactants consumed. Are the moles of product equal to the moles of reactant consumed? Explain your answer.

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