A compound containing carbon, hydrogen, and oxygen. Combustion analysis of a $4.30 g$ sample of butyric acid produced $8.59 g$ ${CO}_{2}$ and $3.52 g H_{2} O$ . Find the empirical formula.

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Solution

Empirical formula:
The empirical formula represents the simplest whole-number ratio of various atoms in a compound
Hence here $n = 2$ and the empirical formula is $C_{2} H_{4} O$ .
Step 1: Finding the mass of carbon in ${CO}_{2}$
The mass of ${CO}_{2}$ in the problem is $8.59 g$
Convert it to moles by dividing the mass in grams by the molar mass of ${CO}_{2}$
${CO}_{2} = \frac{8.59}{44} = 0.193 moles$
${CO}_{2}$ contains one mole of $C$ and two moles of $O$ .
As a result, the mass of $C$ in ${CO}_{2}$ is
$1 mole {CO}_{2} = \frac{0.193 moles {CO}_{2} \times 1 mole C}{1 mole {CO}_{2}} = 0.193 moles of carbon$
Convert moles of $C$ to mass presently ( multiple the moles with molecular mass of $C$ )
$0.193$ moles of $C$ multiplied by $12.01$ Equals $2.34 g$ of $C$
Step 2: Finding the mass of $H$ in a given quantity of $H_{2} O$ .
$\frac{3.52 g H_{2_{}} O}{18 g H_{2} O} x \frac{2 moles H}{1 mole H_{2_{}} O} x 1 g H = 0.394 g H$
Step 3: Finding the mass of $O$ in a given quantity
The overall mass of the compound is $4.3000 g$ .
$O$ mass = $compound mass - (C + H) mass$
$O$ moles = $0.0977 g$
Step 4: Calculate the ratio
Examine the mass of $C + H + O = 4.3000 g$ :
$C moles = 0.1952 H moles = 0.3911 O moles = 0.0977$
To calculate the ratio, divide by the smallest number.
$C : 2.00 H : 4.00 O : 1.00$
$C_{2} H_{4} O$ is the empirical formula.

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A compound containing carbon, hydrogen, and oxygen. Combustion analysis of a 4.30gsample of butyric acid produced 8.59g CO2 and3.52gH2O. Find the empirical formula.

A compound containing carbon, hydrogen, and oxygen. Combustion analysis of a $4.30 g$ sample of butyric acid produced $8.59 g$ ${CO}_{2}$ and $3.52 g H_{2} O$ . Find the empirical formula.