58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
Answer : The empirical and molecular formula of the compound is,
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 = 58.8 g
Mass of H = 9.8 g
Mass of O = 31.4 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 =
Moles of H =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For O =
The ratio of C : H : O = 2.5 : 5 : 1
To make the ratio in a whole number we are multiplying ratio by 2, we get:
The ratio of C : H : O = 5 : 10 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula =
The empirical formula weight = 5(12) + 10(1) + 2(16) = 102 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
Molecular formula =
Therefore, the molecular formula of the compound is,
(1 mol)(102 g/mol) = 102 g
Determine the amount of C, H, and O in mole.
C = (102 g)(0.588)(1 mol/12 g) = 4.998 mols C
H = (102 g)(0.098)(1 mol/1 g) = 9.996 mols H
O = (102 g)(0.314)(1 mol/16 g) = 2 mols O
The empirical formula of the substance is C5H10O2. The molar mass of the empirical formula is 102. This means that this is also its molecular formula.
Assuiming a 100 g sample,
58.8g C / (1 mole/12g) = 4.9 moles C
9.8gH / (1 mole/1g) = 9.8 moles H
31.4 gO / (1mole/16g) = 1.96 moles O
4.9 moles C/1.96 = 2.5moles C x 2 = 5moles C
9.8 moles H/1.96 = 5 moles H x 2 = 10 moles H
1.96 moles O/1.96 = 1 mole O x 2= moles O