Since the volumes of steam produced were in the ratio \(5: 4\), hydrogen atoms present must be in the ratio \(5: 4\). Don't forget that they must have the same number of carbon atoms. \(C _{4} H _{9} OH\) and \(C _{3} H _{7} CHO\) conform to these.
METHOD 2
One can write equation for the combustions and then find each co-efficient in term of \(n\).or the alkanol:
\(C _{ n } H _{2 n +1} OH +a O _{2} \rightarrow b CO _{2}+c H _{2} O\): \(n=b, H : 2 n+2=2 c, O : 2 a+1=2 b+ c\)
\(b=\) is already equal to " \(n\) ". let's find \(c\) and \(a\) also in term of \(n\).
\(2 n+2=2 c, c=\frac{2 n+2}{2}\) \(=\frac{2(n+1)}{2}=n+1\) \(2 a+1=2 b+c\) \(2 a=2 b+c-1\) \(=2 n+n+1-1 \Rightarrow 3 n\) Then, \(a=3 n / 2\)
The complete combustion equation for the alkanol can then be written as:
$$
\begin{aligned} _{n} H _{2 n}+{ }_{1} OH +\frac{3 n}{2} O _{2} \rightarrow \\
n CO _{2}+(n+1) H _{2} O
\end{aligned}
$$
It is now the turn of the alkanal. Care must be taken here if they must contain the same number of carbon atoms and ' \(n\) ' in the two equations must have the same meaning. The alkyl group in the alkanal must be one carbon less than that of the alkanol. This is compensated by the carbon atom in the carbonyl group in the alkanal.
Let us write it!
\(C _{n-1} H _{2(n-1)} CHO +x O _{2} \rightarrow y CO +z H _{2} O\)
\(C : n-1+1=y, n=y\)
\(H : 2 n=2 z, n=z\)
O: \(2 x+1=2 y+2 z\)
Substituting the values of \(y\) and \(z\) in the last equation; we
can get \(x\)
\(2 x+1=2 n+2 n\)
\(2 x=4 n-1\),
\(x=\frac{4 n-1}{2}\)
Rewriting the equation, we have:
\(C _{n-1} H _{2(n-1)+1} CHO +\left(\frac{4 n-1}{2}\right) O _{2} \rightarrow CO _{2}+ zH _{2} O\)
It is said that steam in the alkanol and the alkanal were in the ration \(5: 4\), therefore
\(\frac{n+1}{n}=\frac{5}{4}\)
\(4 n+4=5 n, n=4\)
substituting \(n=4\) in the formular of the two compounds, we have
\(C _{4} H _{9} OH\) and \(C _{3} H _{7}- CHO\)