A sample of sodium sulfite (Na2SO3) has a mass of 2.25 g.
a. How many Na+ ions are present?
b. How many SO3²? ions are present?
c. What is the mass in grams of one formula unit of Na2SO3?
The correct answer and explanation is :
Let’s solve each part of the question step by step.
Given:
- Mass of sodium sulfite (Na₂SO₃) = 2.25 g
Step 1: Molar mass of Na₂SO₃
First, we need the molar mass of Na₂SO₃. To calculate this, we add the atomic masses of the elements involved:
- Atomic mass of Na = 22.99 g/mol
- Atomic mass of S = 32.07 g/mol
- Atomic mass of O = 16.00 g/mol
The formula for sodium sulfite is Na₂SO₃, which contains:
- 2 Na atoms
- 1 S atom
- 3 O atoms
So, the molar mass of Na₂SO₃ is:
[
\text{Molar mass of Na₂SO₃} = (2 \times 22.99) + (1 \times 32.07) + (3 \times 16.00)
]
[
\text{Molar mass of Na₂SO₃} = 45.98 + 32.07 + 48.00 = 126.05 \, \text{g/mol}
]
a. How many Na⁺ ions are present?
- First, we calculate the number of moles of Na₂SO₃ in the 2.25 g sample.
[
\text{Moles of Na₂SO₃} = \frac{\text{mass of Na₂SO₃}}{\text{molar mass of Na₂SO₃}} = \frac{2.25 \, \text{g}}{126.05 \, \text{g/mol}} = 0.0178 \, \text{mol}
]
- From the formula Na₂SO₃, each formula unit contains 2 Na⁺ ions. Therefore, the number of Na⁺ ions is:
[
\text{Number of Na⁺ ions} = 2 \times \text{moles of Na₂SO₃} \times \text{Avogadro’s number}
]
[
\text{Number of Na⁺ ions} = 2 \times 0.0178 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol}
]
[
\text{Number of Na⁺ ions} = 2.14 \times 10^{22} \, \text{Na⁺ ions}
]
b. How many SO₃²⁻ ions are present?
- From the formula Na₂SO₃, each formula unit contains 1 SO₃²⁻ ion. Therefore, the number of SO₃²⁻ ions is the same as the number of Na₂SO₃ units:
[
\text{Number of SO₃²⁻ ions} = \text{moles of Na₂SO₃} \times \text{Avogadro’s number}
]
[
\text{Number of SO₃²⁻ ions} = 0.0178 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol}
]
[
\text{Number of SO₃²⁻ ions} = 1.07 \times 10^{22} \, \text{SO₃²⁻ ions}
]
c. What is the mass in grams of one formula unit of Na₂SO₃?
To find the mass of one formula unit of Na₂SO₃, we need to use the molar mass. The molar mass of Na₂SO₃ is 126.05 g/mol, and this corresponds to 1 mole of Na₂SO₃, which contains Avogadro’s number of formula units.
The mass of one formula unit is:
[
\text{Mass of one formula unit} = \frac{\text{Molar mass of Na₂SO₃}}{\text{Avogadro’s number}} = \frac{126.05 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{formula units/mol}}
]
[
\text{Mass of one formula unit} = 2.09 \times 10^{-22} \, \text{g}
]
Summary of answers:
- a. The number of Na⁺ ions present is ( 2.14 \times 10^{22} ) ions.
- b. The number of SO₃²⁻ ions present is ( 1.07 \times 10^{22} ) ions.
- c. The mass of one formula unit of Na₂SO₃ is ( 2.09 \times 10^{-22} \, \text{g} ).
This problem involved converting the mass of Na₂SO₃ to moles, then using stoichiometry to find the number of ions, and finally determining the mass of a single formula unit based on the molar mass and Avogadro’s number.