Answer: <em>C₃H₆O</em>
Explanation:
1) First, calculate the chemical composition, which you do using balance mass using the data given.
2) Data:
i) mass of ethyl butyrate: 2.22 mg
ii) mass of CO₂: 5.06 mg
iii) mass of H₂O: 2.06 mg
3) Solution
i) All the carbon in CO₂ produced was present in the ehtyl butyrate sample.
ii) The molar mass of CO₂ is 44.01 g/mol
iii) Use proportionality:
12.01 g C / 44.01 g CO₂ = x / 5.06 mg C ⇒ x = 1.38 mg C
iv) All the H in H₂O was present in the original sample of ethyl butyrate
v) The molar mass of H₂O is 18.015 g/mol
vi) Use proportionality
2.016 g H / 18.015 g H₂O = x / 2.06 mg H ⇒ x = 0.23 mg H
vii) Mass balance:
mass of O in the sample = mass of the sample - mass of H - mass of C
mass of O = 2.22 mg - 1.38 mg - 0.23 mg = 0.61 mg O.
viii) Composition:
C = 1.38 mg
H = 0.23 mg
O = 0.61 mg
Which in terms of composition is the same that:
C = 1.38 g
H = 0.23 g
O = 0.61 g
ix) Divide each by the molar atomic mass of the corresponding element, to obtain the number of moles:
C = 1.38 g / 12.01 g/mol = 0.115 mol
H = 0.23 g / 1.008 g/mol = 0.228 mol
O = 0.61 g / 16 g/mol = 0.0381 mol
x) Divide each by the least number of moles to obtain mole proportion
C = 0.115 / 0.0381 = 3.02 ≈ 3
H = 0.228 / 0.0381 = 1.98 ≈ 5.98 ≈ 6
O = 0.0381 / 0.0381 = 1
xi) Therefore the empirical formula searched is <em>C₃H₆O</em>.
