Question

How many moles of oxygen atoms are contained in each compound? a. 2.50 mol of KMnO \(_{4}\) b. 45.9 mol of \(\mathrm{CO}_{2}\) c. \(1.25 \times 10^{-2}\) mol of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Short answer

There are 10.0 moles of oxygen atoms in 2.50 moles of KMnO$_4$, 91.8 moles of oxygen atoms in 45.9 moles of CO$_2$, and \(1.125 \times 10^{-1}\) moles of oxygen atoms in \(1.25 \times 10^{-2}\) moles of CuSO$_4 \cdot 5 \mathrm{H}_{2}\mathrm{O}$.

Step-by-step solution

Identify the number of oxygen atoms in one mole of the compound

In the given compound KMnO4, there are 4 oxygen atoms present. Now we need to calculate the moles of oxygen atoms, given that we have 2.50 moles of KMnO4.

Calculate the moles of oxygen atoms

Using the mole-to-mole ratio, we can calculate the moles of oxygen atoms: Moles of oxygen atoms = (Moles of compound) × (Number of oxygen atoms in one mole of the compound) Moles of oxygen atoms = (2.50 moles) × (4) = 10.0 moles So, there are 10.0 moles of oxygen atoms in 2.50 moles of KMnO4. #b: Moles of oxygen atoms in CO2#

Identify the number of oxygen atoms in one mole of the compound

In the given compound CO2, there are 2 oxygen atoms present. Now we need to calculate the moles of oxygen atoms, given that we have 45.9 moles of CO2.

Calculate the moles of oxygen atoms

Using the mole-to-mole ratio, we can calculate the moles of oxygen atoms: Moles of oxygen atoms = (Moles of compound) × (Number of oxygen atoms in one mole of the compound) Moles of oxygen atoms = (45.9 moles) × (2) = 91.8 moles So, there are 91.8 moles of oxygen atoms in 45.9 moles of CO2. #c: Moles of oxygen atoms in CuSO4·5H2O#

Identify the number of oxygen atoms in one mole of the compound

In the given compound CuSO4·5H2O, there are 4 oxygen atoms in CuSO4 and 5 oxygen atoms in 5H2O, making a total of 9 oxygen atoms. Now we need to calculate the moles of oxygen atoms, given that we have \(1.25 \times 10^{-2}\) moles of CuSO4·5H2O.

Calculate the moles of oxygen atoms

Using the mole-to-mole ratio, we can calculate the moles of oxygen atoms: Moles of oxygen atoms = (Moles of compound) × (Number of oxygen atoms in one mole of the compound) Moles of oxygen atoms = \((1.25 \times 10^{-2} moles)\) × (9) = \(1.125 \times 10^{-1}\) moles So, there are \(1.125 \times 10^{-1}\) moles of oxygen atoms in \(1.25 \times 10^{-2}\) moles of CuSO4·5H2O.

Key concepts

Molar Mass Calculation

Understanding molar mass is crucial when you're dealing with chemical compounds and reactions. Molar mass is the weight of one mole (or 6.022 × 10²³ particles) of any chemical species, be it an element or a molecule. The unit for molar mass is grams per mole (g/mol).

To calculate the molar mass, look at the chemical formula and add up the atomic masses of all the atoms present - you can find these atomic masses on the periodic table. For example, in water (H₂O), you'd calculate the molar mass by adding twice the atomic mass of hydrogen (each about 1 g/mol) to the atomic mass of oxygen (about 16 g/mol), giving you approximately 18 g/mol for the molar mass of water.

Stoichiometry

Stoichiometry is the heart of chemical equations. It involves the calculation of reactants and products in chemical reactions. It's like a recipe that tells you how much of each ingredient you need and how much of the product you'll get.

The key principle in stoichiometry is the conservation of mass. This principle states that in a chemical reaction, the total mass of the reactants equals the total mass of the products. To perform stoichiometric calculations, you need a balanced chemical equation and knowledge of the mole concept, which allows you to use molar ratios to convert between amounts of different substances. The steps you've seen in the moles of oxygen atoms problem are based on stoichiometric calculations - using the mole ratio from the chemical formula to find the amount of a particular atom in a given amount of a compound.

Chemical Formulas

A chemical formula is a shorthand method to describe the composition of a chemical compound. It shows the elements present and the number of each atom in a molecule. For instance:

Empirical vs. Molecular Formulas

The empirical formula is the simplest ratio of the elements in a compound (e.g., CH for benzene), while the molecular formula is the exact number of each type of atom in a molecule (e.g., C₆H₆ for benzene).

Complex Formulas

Some formulas, like for hydrates (compounds with water molecules) which are often written with a dot (as in CuSO₄·5H₂O), include both the compound and the number of water molecules associated with it.

By knowing the chemical formula, one can deduce a lot about the substance - such as the relative masses of the elements present, the structure of the compound, and the stoichiometric relationships that govern the compound's reactions.