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 : The compound is formaldehyde [tex]CH_2O[/tex].
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 = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.20g}{12g/mole}=3.52moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.50g}{1g/mole}=6.50moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{51.30g}{16g/mole}=3.21moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.52}{3.21}=1.09\approx 1[/tex]
For H = [tex]\frac{6.50}{3.21}=2.02\approx 2[/tex]
For O = [tex]\frac{3.21}{3.21}=1[/tex]
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 = [tex]C_1H_2O_1=CH_2O[/tex]
Hence, from this we conclude that the compound is formaldehyde.
