multiple choice 1 point\na compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102…

multiple choice 1 point\na compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102 g/mol. what is the molecular formula for this compound\nc₂h₁₀o₃\nc₅h₅o₃\nch₃o₃\nc₅h₁₀o₂\nclear my selection\nprevious

multiple choice 1 point\na compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102 g/mol. what is the molecular formula for this compound\nc₂h₁₀o₃\nc₅h₅o₃\nch₃o₃\nc₅h₁₀o₂\nclear my selection\nprevious

Answer

Answer:

D. (C_5H_{10}O_2)

Explanation:

Step1: Assume 100g of the compound

So, (m(C)=58.8g), (m(H) = 9.8g), (m(O)=31.4g)

Step2: Calculate the number of moles of each element

  • For (C): (n(C)=\frac{m(C)}{M(C)}=\frac{58.8g}{12g/mol}=4.9mol\approx5mol)
  • For (H): (n(H)=\frac{m(H)}{M(H)}=\frac{9.8g}{1g/mol}=9.8mol\approx10mol)
  • For (O): (n(O)=\frac{m(O)}{M(O)}=\frac{31.4g}{16g/mol}=1.96mol\approx2mol) The empirical formula is (C_5H_{10}O_2)

Step3: Calculate the molar mass of the empirical formula

(M_{empirical}=(5\times12 + 10\times1+2\times16)g/mol=(60 + 10+32)g/mol = 102g/mol) Since the molar mass of the compound ((102g/mol)) is equal to the molar mass of the empirical formula ((C_5H_{10}O_2)), the molecular formula is the same as the empirical formula.