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Step-by-Step Solution
Step 1: Understand the given data
The problem provides the mass percentage ratios of carbon to hydrogen (C : H) and
carbon to oxygen (C : O) in a saturated acyclic organic compound X:
Mass ratio of C to H = 4 : 1
Mass ratio of C to O = 3 : 4
We are asked to find the number of moles of O_2 required for the complete
combustion of two moles of this organic compound.
Step 2: Express the compound as C_xH_yO_z
Let the organic compound be C_xH_yO_z . Suppose it contains:
x moles of carbon (C)
y moles of hydrogen (H)
z moles of oxygen (O)
Step 3: Convert mass ratios into mole ratios
The mass of carbon is 12x , the mass of hydrogen is 1 \times y ,
and the mass of oxygen is 16z . The ratios are given by:
3.1 Ratio of C to H
Mass ratio = \frac{W_C}{W_H} = \frac{4}{1} . In terms of moles:
\frac{x / \text{(number of moles of C)}}{y / \text{(number of moles of H)}}
= \frac{W_C / 12}{W_H / 1} = \frac{4}{1}.
Simplifying:
\frac{x/12}{y/1} = \frac{4}{1}
\quad \Longrightarrow \quad
\frac{x}{y} = \frac{4}{1} \times \frac{12}{1} \times \frac{1}{12}.
However, an easier way can be:
\frac{W_C}{12} : \frac{W_H}{1} = \frac{4}{1}.
Let W_C = 4k and W_H = k for some constant k . Then
\text{moles of C} = \frac{4k}{12}, \quad
\text{moles of H} = \frac{k}{1}.
So
x = \frac{4k}{12} = \frac{k}{3}, \quad
y = \frac{k}{1} = k.
Hence
\frac{x}{y} = \frac{1}{3}.
Thus we get
x : y = 1 : 3.
3.2 Ratio of C to O
Mass ratio = \frac{W_C}{W_O} = \frac{3}{4} . Similarly:
\frac{x/ \text{(moles of C)}}{z/ \text{(moles of O)}} = \frac{W_C/12}{W_O/16} = \frac{3}{4}.
This means:
\frac{x/12}{z/16} = \frac{3}{4}
\quad \Longrightarrow \quad
\frac{x}{z} = \frac{3}{4} \times \frac{16}{12} = 1.
Hence
x = z.
Step 4: Determine the empirical formula
From the above steps, we have:
x : y = 1 : 3
x : z = 1 : 1
Thus if x = 1 , then y = 3 and z = 1 . The empirical formula
is therefore CH_3O .
Step 5: Find the molecular formula based on "saturated acyclic compound"
A saturated acyclic compound containing C, H, and O with the empirical formula
CH_3O could be a multiple of CH_3O . One common saturated (single-bonded)
possibility is C_2H_6O_2 . This matches the condition of “saturated acyclic”
and also fits well with the given ratios. We assume the simplest such multiple
to be C_2H_6O_2 .
Step 6: Write the balanced combustion reaction
The complete combustion of C_2H_6O_2 is:
C_2H_6O_2 + \frac{5}{2} O_2 \;\rightarrow\; 2 CO_2 + 3 H_2O.
For each 1 mole of C_2H_6O_2 ,
\frac{5}{2}
moles of O_2 are required.
Step 7: Calculate the O_2 required for 2 moles of C_2H_6O_2
If 1 mole of C_2H_6O_2 needs
\frac{5}{2}
moles of O_2 , then 2 moles of C_2H_6O_2 would require:
\left( \frac{5}{2} \right) \times 2 = 5
Therefore, the moles of oxygen gas required for complete combustion of
two moles of the compound X is 5.
