A sample of sodium sulfite (Na2SO3) has a mass of 2.25 g
The correct answer and explanation is :
To solve the problem where you are given a sample of sodium sulfite (Na₂SO₃) with a mass of 2.25 g, we need to determine the number of moles of sodium sulfite in the sample. This requires applying the concepts of molar mass and stoichiometry.
Step 1: Find the Molar Mass of Sodium Sulfite (Na₂SO₃)
The molar mass of a compound is the sum of the atomic masses of all the elements in that compound. We can find the molar mass of sodium sulfite by adding up the atomic masses of sodium (Na), sulfur (S), and oxygen (O). From the periodic table:
- Atomic mass of sodium (Na) = 22.99 g/mol
- Atomic mass of sulfur (S) = 32.07 g/mol
- Atomic mass of oxygen (O) = 16.00 g/mol
Now, let’s calculate the molar mass of Na₂SO₃:
[
\text{Molar mass of Na₂SO₃} = (2 \times 22.99 \, \text{g/mol}) + (1 \times 32.07 \, \text{g/mol}) + (3 \times 16.00 \, \text{g/mol})
]
[
\text{Molar mass of Na₂SO₃} = 45.98 \, \text{g/mol} + 32.07 \, \text{g/mol} + 48.00 \, \text{g/mol} = 126.05 \, \text{g/mol}
]
Step 2: Calculate the Number of Moles of Sodium Sulfite
To find the number of moles, we use the formula:
[
\text{Number of moles} = \frac{\text{mass of sample}}{\text{molar mass}}
]
Given that the mass of the sodium sulfite sample is 2.25 g, we can now calculate the number of moles:
[
\text{Number of moles of Na₂SO₃} = \frac{2.25 \, \text{g}}{126.05 \, \text{g/mol}} = 0.01783 \, \text{mol}
]
Step 3: Final Answer
The number of moles of sodium sulfite (Na₂SO₃) in the sample is approximately 0.01783 moles.
Explanation
This calculation is based on the principle of converting the mass of a substance into the number of moles using its molar mass. The molar mass tells us how much 1 mole of a substance weighs in grams, so dividing the mass of the sample by the molar mass gives us the number of moles. The result, 0.01783 moles, represents the amount of sodium sulfite in the 2.25 g sample. Understanding this concept is important in chemistry because it allows for the conversion between mass and moles, which is essential for stoichiometric calculations, determining reactant quantities, and working with chemical equations.