# Question 96c21

Jun 3, 2015

The compound's empirical formula is ${C}_{2} {H}_{6} {O}_{3}$.

The first thing you need to figure out is how much oxygen your compound contains. Since it only contains carbon, hydrogen, and oxygen, you can use the total amss of the compound and the masses of carbon and hydrogen to figure that out.

${m}_{\text{total" = m_"carbon" + m_"hydrogen" + m_"oxygen}}$

${m}_{\text{oxygen" = m_"total" - m_"carbon" - m_"hydrogen}}$

${m}_{\text{oxygen" = 50 - 24.35 - 4.05 = "21.6 g}}$

Now you need to determine how many moles of each element your sample contains. You do that by dividing each element's mass by its molar mass

$\text{For C": (24.35cancel("g"))/(12.01cancel("g")/"mol") = "2.03 moles}$

$\text{For H": (4.05cancel("g"))/(1.01cancel("g")/"mol") = "4.01 moles}$

$\text{For O": (21.6cancel("g"))/(16.0cancel("g")/"mol") = "1.35 moles}$

Next, determine what the mole ratios are between your elements by dividing each of those three numbers by the smallest one

$\text{For C": (2.03cancel("moles"))/(1.35cancel("moles")) = "1.50}$

$\text{For H": (4.01cancel("moles"))/(1.35cancel("moles")) = "2.97} \cong 3.0$

"For O": (1.35cancel("moles"))/(1.35cancel("moles")) = 1.0#

This means that you get

${C}_{1.5} {H}_{3} {O}_{1}$

Since you cannot have fractional subscripts in the empirical formula, multiply all the subscripts by 2 to get

${C}_{3} {H}_{6} {O}_{2}$ $\to$ your compound's empirical formula.