Final answer:
The mass percent of 47.2g C₁₂H₂₂O₁₁ in 508g of H₂O is approximately 4.69%.
Step-by-step explanation:
To calculate the mass percent of C₁₂H₂₂O₁₁ in 508g of H₂O, we need to determine the mass of C₁₂H₂₂O₁₁ in the solution. First, let's calculate the molar mass of C₁₂H₂₂O₁₁: Molar mass of C = 12.01 g/mol. Molar mass of H = 1.008 g/mol. Molar mass of O = 16.00 g/mol. Summing up the molar masses, we get: Molar mass of C₁₂H₂₂O₁₁ = 12.01 * 12 + 1.008 * 22 + 16.00 * 11 = 342.34 g/mol
Now, we can calculate the mass percent: Mass percent = (mass of C₁₂H₂₂O₁₁ / mass of H₂O) * 100. Mass of C₁₂H₂₂O₁₁ = (mass percent / 100) * mass of H₂O. Mass of C₁₂H₂₂O₁₁ = (47.2 / 100) * 508g = 239.36g. Therefore, the mass percent of 47.2g C₁₂H₂₂O₁₁ in 508g of H₂O is approximately 4.69%.