Question

Suppose you analyze the composition of an unnamed compound. Your analysis shows that

the compound is 51.30% oxygen, 42.20% carbon, and 6.50% hydrogen by mass. What can
you conclude about the compound?

Answer 1

Answer : The compound is formaldehyde
`CH_2O`.

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 = 42.20 g

Mass of H = 6.50 g

Mass of O = 51.30 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}}= (42.20g)/(12g/mole)=3.52moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (6.50g)/(1g/mole)=6.50moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (51.30g)/(16g/mole)=3.21moles`

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

For C =
`(3.52)/(3.21)=1.09\approx 1`

For H =
`(6.50)/(3.21)=2.02\approx 2`

For O =
`(3.21)/(3.21)=1`

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

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

The Empirical formula =
`C_1H_2O_1=CH_2O`

Hence, from this we conclude that the compound is formaldehyde.

Answer 2

First from the data given we are able to determine the empirical formula of the product. To do that we use the following algorithm:

1) Divide each percentage at the atomic mass of each element:

O 51.3 / 16 = 3.21

C 42.2 / 12 = 3.52

H 6.5 / 1 = 6.5

2) Now we divide the result to the lowest number, which is 3.21:

O 3.21 /3.21 = 1

C 3.52 /3.21 = 1.1 ≈ 1

H 6.5/3.21 = 2

So the empirical formula of the unknown compounds is:

CH₂O

We can conclude that the compound contains carbon, hydrogen and oxygen in the ratio given by the empirical formula:

C : H : O = 1 : 2 : 1