Question

Verify the conservation of mass law using the molar masses of reactants and products for each substance in the following balanced equations: (a) \(\mathrm{P}_{4}(\mathrm{~s})+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{P}_{2} \mathrm{O}_{3}(\mathrm{~s})\) (b) \(\mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{P}_{2} \mathrm{O}_{5}(\mathrm{~s})\)

Short answer

In both equations, the total mass of reactants equals the total mass of products, confirming the conservation of mass.

Step-by-step solution

Identify Reactants and Products

For each equation, identify the reactants and products. For (a), the reactants are \(\mathrm{P}_{4} (\mathrm{s})\) and \(3 \mathrm{O}_{2} (g)\), and the product is \(2 \mathrm{P}_{2} \mathrm{O}_{3} (\mathrm{s})\). For (b), the reactants are \(\mathrm{P}_{4} (\mathrm{s})\) and \(5 \mathrm{O}_{2} (g)\), and the product is \(2 \mathrm{P}_{2} \mathrm{O}_{5} (\mathrm{s})\).

Calculate Molar Masses

Calculate the molar masses of each reactant and product. - The molar mass of \(\mathrm{P}_{4}\) is \(4\times \text{molar mass of}\, \mathrm{P} = 4 \times 30.97 = 123.88 \text{ g/mol}\). - The molar mass of \(\mathrm{O}_2\) is \(2\times \text{molar mass of}\, \mathrm{O} = 2\times 16.00 = 32.00 \text{ g/mol}\).For (a):- The molar mass of \(\mathrm{P}_{2} \mathrm{O}_{3}\) is \(2\times 30.97 + 3\times 16.00 = 61.94 + 48.00 = 109.94 \text{ g/mol}\).For (b):- The molar mass of \(\mathrm{P}_{2} \mathrm{O}_{5}\) is \(2\times 30.97 + 5\times 16.00 = 61.94 + 80.00 = 141.94 \text{ g/mol}\).

Calculate Total Mass of Reactants

Calculate the total mass of the reactants for each equation. For (a):- Mass of \(\mathrm{P}_{4}\) is \(1 \times 123.88 \text{ g/mol} = 123.88 \text{ g/mol}\).- Mass of \(3 \mathrm{O}_2\) is \(3 \times 32.00 \text{ g/mol} = 96.00 \text{ g/mol}\).- Total mass of reactants = \(123.88 + 96.00 = 219.88 \text{ g/mol}\).For (b):- Mass of \(\mathrm{P}_{4}\) is \(1 \times 123.88 \text{ g/mol} = 123.88 \text{ g/mol}\).- Mass of \(5 \mathrm{O}_2\) is \(5 \times 32.00 \text{ g/mol} = 160.00 \text{ g/mol}\).- Total mass of reactants = \(123.88 + 160.00 = 283.88 \text{ g/mol}\).

Calculate Total Mass of Products

Calculate the total mass of the products for each equation. For (a):- Mass of \(2 \mathrm{P}_{2} \mathrm{O}_{3}\) is \(2 \times 109.94 \text{ g/mol} = 219.88 \text{ g/mol}\).- Total mass of products = \(219.88 \text{ g/mol}\).For (b):- Mass of \(2 \mathrm{P}_{2} \mathrm{O}_{5}\) is \(2 \times 141.94 \text{ g/mol} = 283.88 \text{ g/mol}\).- Total mass of products = \(283.88 \text{ g/mol}\).

Verify Conservation of Mass

Compare the total mass of reactants and products to verify the conservation of mass. For (a):- Total mass of reactants = \(219.88 \text{ g/mol}\), total mass of products = \(219.88 \text{ g/mol}\). Mass is conserved.For (b):- Total mass of reactants = \(283.88 \text{ g/mol}\), total mass of products = \(283.88 \text{ g/mol}\). Mass is conserved.

Key concepts

Molar Mass

The concept of molar mass is fundamental in chemistry. It refers to the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). Molar mass allows chemists to convert between the number of moles of a substance and its mass. For example, when calculating the molar mass of a molecule like \( \mathrm{P}_4 \), we consider the molar mass of a single phosphorus atom, approximately 30.97 g/mol, and multiply it by the number of phosphorus atoms in the molecule, which, in this case, is 4.

To find the molar mass of compounds such as \( \mathrm{P}_{2} \mathrm{O}_{3} \) or \( \mathrm{P}_{2} \mathrm{O}_{5} \), we follow a similar process:

• For \( \mathrm{P}_{2} \mathrm{O}_{3} \), add up the masses: \(2 \times 30.97 + 3 \times 16.00 = 109.94 \text{ g/mol}\).
• For \( \mathrm{P}_{2} \mathrm{O}_{5} \), it's \(2 \times 30.97 + 5 \times 16.00 = 141.94 \text{ g/mol}\).

This calculation helps us determine how much the substance weighs when there is one mole of it, which is crucial for stoichiometric calculations in chemical reactions.

Balanced Chemical Equations

Balanced chemical equations are essential in chemistry. They show how reactants transform into products while conserving atoms. This ensures that the same number of each type of atom appears on both sides of the equation. Balancing is vital because it reflects the law of conservation of mass; matter is neither created nor destroyed in chemical reactions.

Here's how balancing works in our example:

• For \( \mathrm{P}_{4} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{P}_{2} \mathrm{O}_{3} \):
  • The reactants contain 4 phosphorus atoms and 6 oxygen atoms.
  • The products also contain 4 phosphorus atoms and 6 oxygen atoms through 2 molecules of \( \mathrm{P}_{2} \mathrm{O}_{3} \).
• For \( \mathrm{P}_{4} + 5 \mathrm{O}_{2} \rightarrow 2 \mathrm{P}_{2} \mathrm{O}_{5} \):
  • The reactants have 4 phosphorus atoms and 10 oxygen atoms.
  • The products equate this with 4 phosphorus atoms and 10 oxygen atoms in 2 molecules of \( \mathrm{P}_{2} \mathrm{O}_{5} \).

Balancing ensures the "equation" remains true for the transformation, keeping a logical, consistent foundation for conducting experiments and calculations.

Chemical Reactions

Chemical reactions represent the processes where reactants convert into products, reflecting changes at the molecular level. These changes often involve the making or breaking of bonds. Understanding how reactions occur can help predict the products formed and the energy changes associated with the reactions.

For reactions like those involving phosphorus and oxygen we looked at:

• Reactants like \( \mathrm{P}_{4} \) and \( \mathrm{O}_{2} \) start on the left, meaning they are consumed.
• Products \( \mathrm{P}_{2} \mathrm{O}_{3} \) and \( \mathrm{P}_{2} \mathrm{O}_{5} \) appear on the right, signifying new substances formed.

Reactions are governed by various conditions such as temperature, pressure, and the presence of catalysts, influencing the rate and extent of reaction.
Chemical reactions follow principal laws and theories, such as the law of conservation of mass, which asserts that the mass of reactants equals the mass of products. This can be demonstrated quantitatively with balanced equations outlining total masses for both sides as shown in the provided exercises.