Question

an alcohol is 54.52 % c and 9.170 % h by mass. the rest is oxygen. what is the empirical formula of the alcohol? enter the elements in the order c, h, and then o.

Explanation:

Step1: Calculate the mass - percentage of oxygen

The sum of mass - percentages of all elements in a compound is 100%. Given the mass - percentage of C is 54.52% and of H is 9.170%. So the mass - percentage of O is $100-(54.52 + 9.170)=36.31\%$.

Step2: Assume a 100 - g sample

If we assume a 100 - g sample of the alcohol, then the masses of C, H, and O are:
Mass of C: $m_C = 54.52$ g, mass of H: $m_H=9.170$ g, mass of O: $m_O = 36.31$ g.

Step3: Calculate the number of moles of each element

The molar mass of C is $M_C=12.01$ g/mol, the molar mass of H is $M_H = 1.008$ g/mol, and the molar mass of O is $M_O=16.00$ g/mol.
The number of moles of C: $n_C=\frac{m_C}{M_C}=\frac{54.52}{12.01}\approx4.54$ mol.
The number of moles of H: $n_H=\frac{m_H}{M_H}=\frac{9.170}{1.008}\approx9.10$ mol.
The number of moles of O: $n_O=\frac{m_O}{M_O}=\frac{36.31}{16.00}\approx2.27$ mol.

Step4: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (in this case, $n_O = 2.27$ mol).
For C: $\frac{n_C}{n_O}=\frac{4.54}{2.27}=2$.
For H: $\frac{n_H}{n_O}=\frac{9.10}{2.27}\approx4$.
For O: $\frac{n_O}{n_O}=1$.

Answer:

$C_2H_4O$