Question

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

Answer 1

To determine the empirical formula of the compound, we assume a basis of 100 g of this compound. We calculate as follows:

C = 68.75 g
H = 10.90 g
O = 20.35 g

We convert these mass to moles,
C = 68.75 g / 12.01 g/mol = 5.72 mol
H = 10.90 g / 1.01 g/mol = 10.79 mol
O = 20.35 g / 16 g/mol = 1.27 mol

C = 5.72 mol / 1.27 mol = 5
H = 10.79 mol / 1.27 mol = 8
O = 1.27 mol / 1.27 mol = 1

C5H8O

Answer 2

Answer: The empirical formula is $C_9H_{17}O_2$

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 =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{68.75g}{12g/mole}=5.73moles$

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

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

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

For C = $\frac{5.73}{1.27}=4.5$

For H = $\frac{10.90}{1.27}=8.6$

For O =$\frac{1.27}{1.27}=1$

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

converting them into whole number ratio by multiplying by 2

$2\times C_{4.5}H_{8.6}O=C_9H_{17}O_2$

Hence the empirical formula is $C_9H_{17}O_2$