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Sagot :
Solution:
Step 1: Calculate the molar mass of sodium lauryl sulfate (NaC₁₂H₂₅SO₄).
molar mass = (22.99 g/mol × 1) + (12.01 g/mol × 12) + (1.008 g/mol × 25) + (32.07 g/mol × 1) + (16.00 g/mol × 4)
molar mass = 288.38 g/mol
Step 2: Calculate the percentage composition of sodium lauryl sulfate (NaC₁₂H₂₅SO₄).
For mass percent of Na
[tex]\text{mass percent of Na} = \frac{\text{22.99 g/mol × 1}}{\text{288.38 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Na = 7.97\%}}[/tex]
For mass percent of C
[tex]\text{mass percent of C} = \frac{\text{12.01 g/mol × 12}}{\text{288.38 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of C = 49.98\%}}[/tex]
For mass percent of H
[tex]\text{mass percent of H} = \frac{\text{1.008 g/mol × 25}}{\text{288.38 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of H = 8.74\%}}[/tex]
For mass percent of S
[tex]\text{mass percent of S} = \frac{\text{32.07 g/mol × 1}}{\text{288.38 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of S = 11.12\%}}[/tex]
For mass percent of O
[tex]\text{mass percent of O} = \frac{\text{16.00 g/mol × 4}}{\text{288.38 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of O = 22.19\%}}[/tex]
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