Question

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

Answer 1

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

Answer 2

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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.20g}{12g/mole}=3.52moles$

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

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

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

For C = $\frac{3.52}{3.21}=1.09\approx 1$

For H = $\frac{6.50}{3.21}=2.02\approx 2$

For O = $\frac{3.21}{3.21}=1$

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.