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Answer:
C₃H₅O₂
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O
Explanation:
The reaction can be expressed as:
CₓHₓOₓ + nO₂ → CO₂ + H₂O
Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:
All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:
Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:
0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O
In order to determine the empirical formula, we calculate the moles of each component:
- mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
- mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
- mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O
Then we divide those values by the lowest one:
0.0236 mol C ÷ 0.0141 = 1.67
0.0354 mol H ÷ 0.0141 = 2.51
0.0141 mol O ÷ 0.0141 = 1
If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.
- The reaction is:
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O
Answer:
a. alkyne
b. alkane
c. alkyne
d. alkene
Explanation:
The general formula for each class of compound is given below
Alkane:
Alkene:
Alkyne: (assuming single multiple bonds)
Now let us classify according to the above formulas:
a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane
c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
d. It has hydrogen atoms two times of carbon atoms hence, it's alkene