# Question a1833

Jun 3, 2015

Your acid's molecular formula is ${C}_{6} {H}_{10} {O}_{4}$.

Since you're dealing with an organic acid, and since the only products formed when it undergoes combustion are carbon dioxide and water, you know that it contains three elements: carbon, hydrogen, and oxygen.

The carbon dioxide will contain all the carbon the inital sample contained. At the same time, the water will contain all the hydrogen, which means that you can use the amounts produced to get the mass of carbon and hydrogen in the sample.

Since you know the total mass of the sample, this will allow you to calculate the mass of oxygen as well.

Determine the percent composition of carbon in $C {O}_{2}$ and of hydrogen in ${H}_{2} O$ by using molar masses

(12.0cancel("g/mol"))/(44.0cancel("g/mol")) * 100 = "27.27% $\to$ carbon in $C {O}_{2}$

(2 * 1.01cancel("g/mol"))/(18.02cancel("g/mol")) * 100 = "11.21% $\to$ hydrogen in ${H}_{2} O$

This means that, for every 100 g of $C {O}_{2}$, you get 27.27 g of carbon; likewise, for every 100 g of water, you get 11.21 g of hydrogen.

Therefore, the initial sample contained

1.81cancel("g"CO_2) * "27.27 g C"/(100cancel("g"CO_2)) = "0.4936 g C"

and

0.616cancel("g"H_2O) * "11.21 g H"/(100cancel("g"H_2O)) = "0.0691 g H"

The mass of oxygen in the initial sample will thus be

${m}_{\text{sample" = m_"O" + m_"H" + m_"C}}$

${m}_{\text{O" = 1 - 0.4936 - 0.0691 = "0.4373 g O}}$

Determine how many moles of each element you get in that much mass

$\text{For C": (0.4936cancel("g"))/(12.0cancel("g")/"mol") = "0.04113 moles C}$

$\text{For H": (0.0691cancel("g"))/(1.01cancel("g")/"mol") = "0.6842 moles H}$

$\text{For O": (0.4373cancel("g"))/(16.0cancel("g")/"mol") = "0.02733 moles O}$

Divide all these numbers by the smallest one to get the mole ratios of the elements within the compound

$\text{For C": (0.04113cancel("moles"))/(0.02733cancel("moles")) = "1.5}$

$\text{For H": (0.6842cancel("moles"))/(0.02733cancel("moles")) = "2.5}$

"For O": (0.02733cancel("moles"))/(0.02733cancel("moles")) = 1#

You'll get

${C}_{1.5} {H}_{2.5} O$

Since the empirical formula cannot contain fractional subscripts, multiply all the subscripts by 2 to get

${C}_{3} {H}_{5} {O}_{2}$

To get the molecular formula, use the acid's molar mass and the molar mass of its empirical formula

${\left({C}_{3} {H}_{5} {O}_{2}\right)}_{\textcolor{b l u e}{\text{n}}} = 150$

$\left(3 \cdot 12.0 + 5 \cdot 1.01 + 2 \cdot 16.0\right) \cdot \textcolor{b l u e}{\text{n}} = 150$

$\textcolor{b l u e}{\text{n}} = \frac{150}{73.05} = 2.05 \cong 2$

Thus, the acid's molecular formula will be

${C}_{6} {H}_{10} {O}_{4}$