# Question 7f838

Jun 3, 2015

Your compound's molecular formula is ${C}_{4} {H}_{8} {O}_{2}$.

This one is a little complex because there are many calculations to do before being able to determine the empirical and molecular formulas of the compound.

So, you know that your compound contains carbon, hydrogen, and oxygen. Moreover, you know that a 0.392-g sample occupies a volume of $100 {\text{cm}}^{3}$ at STP - Standard Temperature and Pressure.

At STP, one mole of any ideal gas occupies exactly 22.4 L - this is known as the molar volume of a gas at STP. This means that your sample contains

$n = \frac{V}{V} _ \text{molar" = (0.1cancel("L"))/(22.4cancel("L")) = "0.004464 moles}$

This implies that your compound's molar mass will be

${M}_{M} = \frac{m}{n} = \text{0.392 g"/"0.004464 moles" = "87.8 g/mol}$

Moving on. You know that this compound undergoes combustion and that 0.32 g of water are produced. You'll use this value to determine how much hydrogen the initial sample contained.

Start by determining the percent composition of hydrogen in water

((2 * 1.00794)cancel("g/mol"))/(18.02cancel("g/mol")) * 100 = "11.19%"

This means that, for every 100 g of water, you get 11.19 g of hydrogen. Therefore, for 0.32 g, you get

0.32cancel("g"H_2O) * "11.19 g H"/(100cancel("g"H_2O)) = "0.03581 g H"

Next, determine how much carbon your initial sample contained. Since the combustion reaction produces ${\text{400 cm}}^{3}$ of $C {O}_{2}$, you can determine how many moles of $C {O}_{2}$ you get at STP

n_(CO_2) = (0.4cancel("L"))/(22.4cancel("L")) = "0.01786 moles" $C {O}_{2}$

Since you get 1 mole of carbon for every 1 mole of $C {O}_{2}$, you have

0.01786cancel("moles"CO_2) * "1 mole C"/(1cancel("mole"CO_2)) = "0.01786 moles C"

Use carbon's molar mass to determine how many grams would contain this many moles

0.01786cancel("moles") * "12.0 g"/(1cancel("mol")) = "0.2143 g C"

You can now figure out how much oxygen your initial sample had by

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

${m}_{\text{O" = 0.392 - 0.2143 - 0.03581 = "0.1419 g O}}$

It's all downhill from here. Calculate how many moles of each element your sample contained by using their molar masses

$\text{For C": (0.2143cancel("g"))/(12.0cancel("g")/"mol") = "0.01786 moles C}$

$\text{For H": (0.03581cancel("g"))/(1.01cancel("g")/"mol") = "0.03546 moles H}$

$\text{For O": (0.1419cancel("g"))/(16.0cancel("g")/"mol") = "0.008869 moles O}$

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

"For C": (0.01786cancel("moles"))/(0.008869cancel("moles")) = 2.1 ~= 2

"For H": (0.03546cancel("moles"))/(0.008869cancel("moles")) = 4

"For O": (0.008869cancel("moles"))/(0.008869cancel("moles")) = 1

You compound's empirical formula will be

${C}_{2} {H}_{4} O$

To get its molecular formula, you have to use the molar mass of the empirical formula and the molar mass of the compound.

(C_2H_4O)_color(blue)("n") = "87.8 g/mol"#

$\left(12 \cdot 2 + 1.10 \cdot 4 + 16\right) \cancel{\text{g/mol") * color(blue)("n") = 87.8cancel("g/mol}}$

$\textcolor{b l u e}{\text{n}} = \frac{87.8}{44.04} = 1.99 \cong 2$

Thus, your compound's molecular formula will be

${C}_{4} {H}_{8} {O}_{2}$

SIDE NOTE You get the same result by using the current definition of STP conditions, at which the volume occupied by an ideal gas is equal to 22.7 L, not 22.4 L.

However, I assume that your problem intended for you to use the old value, so I did the calculations using 22.4 L.