Question

Determine the empirical formulas for compounds with the following percent compositions:

Answer 1

Answer: a)  $CS_2$

b) $CH_2O$

Explanation:

a) 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 = 15.8g

Mass of S= 84.2 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{15.8g}{12g/mole}=1.32moles$

Moles of S=$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{84.2g}{32g/mole}=2.63moles$

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

For C = $\frac{1.32}{1.32}=1$

For S =$\frac{2.63}{1.32}=2$

The ratio of C  : S = 1:2

Hence the empirical formula is $CS_2$

b) Mass of C= 40 g

Mass of H= 6.7 g

Mass of O = 53.3 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.7g}{1g/mole}=6.7moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles$

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

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{6.7}{3.33}=2$

For O =$\frac{3.33}{3.33}=1$

The ratio of C : H: O= 1 :2: 1

Hence the empirical formula is $CH_2O$