Question

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

Answer 1

Answer: The empirical formula is $CH_2$.

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles$

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

For C =$\frac{0.142}{0.142}=1$

For H =$\frac{0.287}{0.142}=2$

The ratio of C: H = 1: 2

Hence the empirical formula is $CH_2$.