Question

Complete combustion of a 0.70-mol sample of a hydrocarbon, CxHy, gives 2.80 mol of CO2 and 3.50 mol of H2O. The molecular formula of the original hydrocarbon is:
a. C2H5
b. C4H10
c. CH3
d. C5H5
e. C4H5​

Answer 1

Answer:

Explanation:

B. C4H10

Answer 2

The molecular formula of the original hydrocarbon is C4H10. Option b is correct.

The chemical equation for combustion reaction can be represented as:

$\mathbf{C_xH_y + O_2 \to xCO_2 + \dfrac{y}{2} H_2O}$

From the above reaction;

• moles of CxHy = 0.70 mol
• moles of CO2 = 2.80 mol
• moles of H2O = 3.50 mol

Using the conditions in the given question.

since 0.70 mol of CxHy gives 2.80 mol of CO2 and 3.50 mol of H2O

∴

number of carbon (x) $\mathbf{=\dfrac{2.80}{0.70}}$

x = 4

Similarly, the number of hydrogen $\mathbf{\dfrac{y}{2}= \dfrac{3.50}{0.7}}$

y = 10

The balanced equation can now be expressed as:

$\mathbf{C_4H_{10} + \dfrac{13}{2}O_2 \to 4CO_2 + \dfrac{10}{2} H_2O}$

Therefore, we can conclude that the molecular formula of the original hydrocarbon is C4H10.

Learn more about combustion reaction here:

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