Question

Find the empirical formula of the compound which contains 33.18 percent of carbon 4.60 percent of hydrogen 29.49 percent of Oxygen and 32.72 percentage of chlorine respectively. ​

Answer 1

The empirical formula of a compound is, $C_3H_5O_2Cl$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 33.18 g

Mass of H = 4.60 g

Mass of O = 29.49 g

Mass of Cl = 32.72 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{33.18g}{12g/mole}=2.765moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.60g}{1g/mole}=4.60moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{29.49g}{16g/mole}=1.843moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{32.72g}{35.5g/mole}=0.9217moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2.765}{0.9217}=2.99\approx 3$

For H = $\frac{4.60}{0.9217}=4.99\approx 5$

For O = $\frac{1.843}{0.9217}=1.99\approx 2$

For Cl = $\frac{0.9217}{0.9217}=1$

The ratio of C : H : O : Cl = 3 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_5O_2Cl_1=C_3H_5O_2Cl$

Therefore, the empirical of the compound is, $C_3H_5O_2Cl$

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Answer 2

empirical formula C3H5O2Cl