A compound contains 38.7% by mass of C, 9.7% by mass of hydrogen and the remainder is oxygen (51.6% by mass).
Find the empirical formula. In this calculation C=12, H=1 and O=16.
The first step is to convert the percentages into moles of atoms. If you consider 100g of the compound, there will be 38.7g C, 9.7g H and 51.6g O. To convert these into moles, divide the mass by the molar mass.
|Mass present (g)||38.7||9.7||51.6|
|Moles of atoms||(38.7/12)||(9.7/1)||(51.6/16)|
|Moles of atoms||3.225||9.7||3.225|
|Divide by smallest||3.225/3.225||9.7/3.225||3.225/3.225|
|Ratio of atoms||1.00||3.007||1.00|
So empirical formula = CH3O
If the molar mass of the compound = 62.0g, find the molecular formula.
To do this, find the molar mass of the empirical formula. CH2O = 12 + 1x3 + 16 = 31g
The number of times moles mass of compound is bigger than the mass of the empirical formula is found by dividing the molar mass of the compound by the molar mass of the empirical formula.
Answer = 62.0/31 = 2.
So the molecular formula = (CH3O) x 2 = C2H6O2