Answer:
The empirical formula is C₆H₁₀O₇.
Step-by-step explanation:
1. Calculate the masses of C, H, and O from the masses given.
Mass of C = 2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568 g C
Mass of H = 0.6955 g H₂O × (2.016 H/18.02 g H₂O) = 0.077 81 g H
Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O
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2. Convert these masses to moles.
Moles C = 0.5568 × 1/12.01 = 0.046 36
Moles H = 0.077 81 × 1/1.008 = 0.077 19
Moles O = 0.8654 × 1/16.00 = 0.054 09
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3. Find the molar ratios.
Moles C = 0.046 36/0.046 36 = 1
Moles H = 0.077 19/0.046 36 = 1.665
Moles O = 0.054 09/0.046 36 = 1.167
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4. Multiply the ratios by a number to make them close to integers
C = 1 × 6 = 6
H = 1.665 × 6 = 9.991
O = 1.167 × 6 = 7.001
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5. Round the ratios to integers
C:H:O =6:10:7
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6. Write the empirical formula
The empirical formula is C₆H₁₀O₇.
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7. Calculate the empirical formula mass
C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00
C₆H₁₀O₇ = 72.01 + 10.08+ 112.0
C₆H₁₀O₇ = 194.09
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8. Divide the molecular mass by the empirical formula mass.
MM/EFM = 194.14/194.09 = 1.000 ≈ 1
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9. Determine the molecular formula
MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇
