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`.

Step-by-step 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}}= (1.71g)/(12g/mole)=0.142moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (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 =
`(0.142)/(0.142)=1`

For H =
`(0.287)/(0.142)=2`

The ratio of C: H = 1: 2

Hence the empirical formula is
`CH_2`.