# An unknown compound has the formula C_xH_yO_z. You burn 0.1523 g of the compound and isolate 0.3718 g of CO_2 and 0.1522 g of H_2O. What is the empirical formula of the compound?

## If the molar mass is 72.1 g/mol, what is the molecular formula?

Jul 20, 2016

Empirical and molecular formulae are both ${C}_{4} {H}_{8} O$

#### Explanation:

First, work out the mass of each element that was present in the original compound. Carbon is always present as $C {O}_{2}$ in the ratio (12.011 g / 44.0098 g), and hydrogen is always present as ${H}_{2} O$ in the ratio (2.0158 g / 18.0152 g).

So you need to work out the mass of carbon in 0.3718 g of $C {O}_{2}$ and the mass of hydrogen in 0.1522 g of water..

Carbon: 0.3718 x (12.011 / 44.0098) = 0.10147 g
Hydrogen: 0.1522 x (2.0158 / 18.0152) = 0.01703 g

You can determine the mass of oxygen by difference. 0.1523 - 0.10147 - 0.01703 = 0.0338 g
.
Next, convert each of these to numbers of moles:

Carbon: 0.10147 / 12.011 = 0.00845 mol
Hydrogen: 0.01703 / 1.0079 = 0.01689 mol
Oxygen: 0.0338 / 15.994 = 0.002113 mol
Remember that you're working out moles of hydrogen atoms here, not moles of molecular hydrogen gas. So use the atomic weight of hydrogen 1.0079.
.
Next, divide each number of moles by the lowest value, so that you get as close as possible to whole numbers.

Carbon: 0.00845 / 0.002113 = 4
Hydrogen: 0.001689 / 0.002113 = 8
Oxygen: 0.002113/ 0.002113 = 1

So the emprical formula is ${C}_{4} {H}_{8} O$

The molar mass is 72.1 g/mol - work out the mass of ${C}_{4} {H}_{8} O$ - this is (12 x 4) + (8 x 1) + 16 = 72. Therefore the molecular formula is the same as the empirical..

Jul 20, 2016

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

#### Explanation:

Balanced equation of cmbustion reaction

${C}_{x} {H}_{y} {O}_{z} + \left(x + \frac{y}{4} - \frac{z}{2}\right) {O}_{2} \to x C {O}_{2} + \frac{y}{2} {H}_{2} O$

$\text{No.of moles of "C_xH_yO_z " reacted}$

$= \text{its mass"/"Its molar mass"=(0.1523g)/(72.1g/"mol")=0.1523/72.1"mol}$

$\text{No.of moles of "CO_2" formed}$

="its mass"/"Its molar mass"=(0.3718g)/(44g/"mol")=0.3718/44mol

$\text{No.of moles of "H_2O" formed}$

="its mass"/"Its molar mass"=(0.1522g)/(18g/"mol")=0.1522/18mol

By the balanced equation 1 mole of the compound produces x moles $C {O}_{2}$ and $\frac{y}{2} \text{ moles } {H}_{2} O$

Hence $\frac{0.1523}{72.1} \text{mol}$ compound will produce $\frac{0.1523}{72.1} \cdot x \text{ mol } C {O}_{2}$

$\therefore \frac{0.1523}{72.1} \cdot x = \frac{0.3718}{44}$

$\implies x = \frac{0.3718}{44} \cdot \frac{72.1}{0.1523} \approx 4$

Again $\frac{0.1523}{72.1} \text{mol}$
compound will produce $\frac{0.1523}{72.1} \cdot \frac{y}{2} \text{ mol } {H}_{2} O$

So
$\frac{0.1523}{72.1} \cdot \frac{y}{2} = \frac{0.1522}{18}$

$\implies y = \frac{0.1522}{18} \cdot \frac{72.1 \cdot 2}{0.1533} \approx 8$

Finally

$\text{Molar mass of } {C}_{x} {H}_{y} {O}_{z} = 72.1$

$\implies 12 x + y + 16 z = 72.1$

$\implies 12 \cdot 4 + 8 + 16 z = 72.1$

$z = 1$

Hence both empirical and molecular formula of the compound is${C}_{4} {H}_{8} O$