Question

an alcohol is 58.80 % c and 9.889 % 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 C and H is $58.80\%+9.889\% = 68.689\%$. So the mass - percentage of O is $100\% - 68.689\%=31.311\%$.

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: $m_C = 58.80\ g$; Mass of H: $m_H=9.889\ g$; Mass of O: $m_O = 31.311\ g$.

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: $n_C=\frac{m_C}{M_C}=\frac{58.80\ g}{12.01\ g/mol}\approx4.90\ mol$.
The number of moles of H: $n_H=\frac{m_H}{M_H}=\frac{9.889\ g}{1.008\ g/mol}\approx9.81\ mol$.
The number of moles of O: $n_O=\frac{m_O}{M_O}=\frac{31.311\ g}{16.00\ g/mol}\approx1.96\ mol$.

Step4: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles ($n_O = 1.96\ mol$).
For C: $\frac{n_C}{n_O}=\frac{4.90\ mol}{1.96\ mol}\approx2.5$.
For H: $\frac{n_H}{n_O}=\frac{9.81\ mol}{1.96\ mol}\approx5$.
For O: $\frac{n_O}{n_O}=1$.
Multiply each ratio by 2 to get whole - numbers. The mole - ratio of C : H : O is 5 : 10 : 2.

Answer:

$C_5H_{10}O_2$