Question

Determine the number of atoms of each element in the empirical formula of a compound with the following composition:

68.75 percent C, 10.90 percent H, 20.35 percent O.

Answer 1

Answer: The empirical formula is
`C_9H_(17)O_2`

Step-by-step explanation:

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= 68.75 g

Mass of H = 10.90 g

Mass of O = 20.35 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (68.75g)/(12g/mole)=5.73moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (10.90g)/(1g/mole)=10.90moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (20.35g)/(16g/mole)=1.27moles`

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

For C =
`(5.73)/(1.27)=4.5`

For H =
`(10.90)/(1.27)=8.6`

For O =
`(1.27)/(1.27)=1`

The ratio of C : H: O= 4.5: 8.6: 1

converting them into whole number ratio by multiplying by 2

`2* C_(4.5)H_(8.6)O=C_9H_(17)O_2`

Hence the empirical formula is
`C_9H_(17)O_2`