Question

An unidentified organic compound X, containing C, H, and O was subjected to combustion analysis. When 228.4 mg of pure X was burned 627.4 mg of CO2 and 171.2 mg of H2O were obtained. Determine the simplest formula of the compound X. 1. C5H8O2 2. C6H10O2 3. C10H8O2 4. C6H8O 5. C2H4O

Answer 1

Answer:

The formula would be: C₆H₈O

Explanation:

The mass of carbon dioxide formed = 627.4 mg

The percentage of carbon in the sample would be:

% carbon = $\frac{12Xmass of carbondioxideX100}{44Xmassofsample}=\frac{12X627.4X100}{44X228.4}=74.92$

Where

44 is the molar mass of carbon dioxide.

The percentage of hydrogen in the sample would be:

% Hydrogen = $\frac{2Xmassofwater}{18Xmassofsample}=\frac{2X171.2X100}{228.4X18}=8.33$

Rest will be oxygen = 100 -[74.92+8.33]=16.75

Let us calculate the moles of each of the elements present:

moles of carbon = $\frac{mass}{atomicmass}=\frac{74.92}{12}=6.24$

moles of hydrogen = $\frac{mass}{atomicmass}=\frac{8.33}{1}=8.33$

moles of oxygen = $\frac{massofoxygen}{atomicmass}=\frac{16.75}{16}=1.05$

The mole ration of carbon, hydrogen and oxygen = C:H:O = 6 : 8 : 1

so the formula would be:

C₆H₈O

Answer 2

Answer:

Option D. C₆H₈O

Explanation:

We can calculate this just using a relation ratio between Hydrogen and Carbon. You may wondering why? well, because the products of the combustion of X, gives CO2 and H2O, and data obtained involves those two elements, so, we can have a ratio of those elements to get the correct formula.

The previous answer was ok, but here is another method

Let's just say that X has the general formula CₓHₙOₐ

When this compound has the combustion we have the following:

CₓHₙOₐ + O₂(excess) ---------> xCO₂ + n/2H₂O

Let's calculate the moles of CO₂  and H₂O:

moles CO₂ = 0.627 / 44 = 0.01425 moles

moles H₂O = 0.1712 / 18 = 9.51x10⁻³ moles

the moles of CO₂ are the same moles of Carbon, and the moles of Hydrogen will be the twice of moles of H₂O according to the equation I wrote so:

moles of H = 9.51x10⁻³ * 2 = 1.90x10⁻² moles

Now, the ratio of Hydrogen and Carbon would be:

H/C = 1.90x10⁻² / 0.01425 = 1.333

So, all we have to do now, is look into the options and see which one has the same ratio of 1.3333:

C5H8O ----> H/C = 8/5 = 1.6

C6H10O2 ----> H/C = 10/6 = 1.67

C10H8O2 ----> H/C = 8/10 = 0.8

C2H4O ----> H/C = 4/2 = 2

C6H8O ----> H/C = 8/6 = 1.3333

Therefore the formula for compound X is C6H8O. This is an alternate method of resolution.