Question

A compound is found to contain 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. Determine the simplest formula.

Answer 1

Answer : The simplest formula of a compound is,
`C_2H_4O`

Solution :

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 = 54.5 g

Mass of H = 9.1 g

Mass of O = 36.4 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (54.5g)/(12g/mole)=4.54moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (9.1g)/(1g/mole)=9.1moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (36.4g)/(16g/mole)=2.28moles`

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

For C =
`(4.54)/(2.28)=1.99\approx 2`

For H =
`(9.1)/(2.28)=3.99\approx 4`

For O =
`(2.28)/(2.28)=1`

The ratio of C : H : O = 2 : 4 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`C_2H_4O_1=C_2H_4O`

Therefore, the simplest formula of the compound is,
`C_2H_4O`

Answer 2

The simplest formula is calculated as follows

find the moles of each element = 5 composition/molar mass of element

C= 54.5 / 12 = 4.542 moles
H= 9.1/1= 9.1 moles
O = 36.4 /16= 2.275 moles

find the mole ratio by diving each mole with the smallest mole ( 2.275 mole)

C = 4.542/ 2.275 = 2
H= 9.1/ 2.275= 4
O= 2.275/2.275= 1

the simple formula is therefore = C2H4O