Answer:
[tex]\boxed{\text{C$_{3}$H$_{6}$O}}[/tex]
Explanation:
Let's call the ethyl butyrate X.
1. Calculate the mass of each element in 2.78 mg of X.
(a) Mass of C
[tex]\text{Mass of C} = \text{6.32 mg } \text{CO}_{2}\times \dfrac{\text{12.01 mg C}}{\text{44.01 mg }\text{CO}_{2}}= \text{1.725 mg C}[/tex]
(b) Mass of H
[tex]\text{Mass of H} = \text{2.58 mg }\text{H$_{2}$O}\times \dfrac{\text{2.016 mg H}}{\text{18.02 mg } \text{{H$_{2}$O}}} = \text{0.2886 mg H}[/tex]
(d) Mass of O
Mass of O = 2.78 - 1.725 - 0.2886 = 0.7667 g
2. Calculate the moles of each element
[tex]\text{Moles of C = 1.725 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{0.1436 mmol C}\\\\\text{Moles of H = 0.2886 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{0.2863 mmol H}\\\\\text{Moles of O = 0.7767 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{0.04792 mmol O}[/tex]
3. Calculate the molar ratios
Divide all moles by the smallest number of moles.
[tex]\text{C: } \dfrac{0.1436}{0.04972}= 2.997\\\\\text{H: } \dfrac{0.2863}{0.04972} = 5.976\\\\\text{Fe: } \dfrac{0.04972}{0.04972} = 1[/tex]
4. Round the ratios to the nearest integer
C:H:O = 3:6:1
5. Write the empirical formula [tex]\text{The empirical formula is } \boxed{\textbf{C$_{3}$H$_{6}$O}}[/tex]
The empirical formula of 2.78 mg of ethyl butyrate which produces 6.32 mg of CO₂ and 2.58 mg of H₂O is C₃H₆O
Mass of CO₂ = 6.32 mg
Molar mass of CO₂ = 44 g/mol
Molar mass of C = 12 g/mol
Mass of C = 12/44 × 6.32
Mass of H₂O = 2.58 mg
Molar mass of H₂O = 18 g/mol
Molar mass of H₂ = 1 × 2 = 2 g/mol
Mass of H = 2/18 × 2.58
Mass of C = 1.72 mg
Mass of H = 0.29 mg
Mass of compound = 2.78 mg
Mass of O = (Mass of compound ) – (mass of C + mass of H)
Mass of O = 2.78 – (1.72 + 0.29)
Mass of C = 1.72 mg
Mass of H = 0.29 mg
Mass of O = 0.77 mg
Divide by their molar mass
C = 1.72 / 12 = 0.143
H = 0.29 / 1 = 0.29
O = 0.77 / 16 = 0.048
Divide by the smallest
C = 0.143 / 0.048 = 3
H = 0.29 / 0.048 = 6
O = 0.048 / 0.048 = 1
Therefore, the empirical formula of the compound is C₃H₆O
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