A compound on analysis as 87:8% c,12:2%h determine its empirical formular&molecular formular given mass=56gmol

Answer 1

Take a 100 g sample which gives you
87.8g C
12.2 g H
Convert to mols = g/atomic mass

87.8/12 = approx 7.3
12.2/1 = 12.2

Determine the ratio of the two elements to each other with the lowest number being no less than 1.00. The easy to do that is to divide both numbers by the smaller number.
7.3/7.3 = 1
12.2/7.3 = 1.67

Those are not whole numbers so multiply them by 2 and see if that works.
1 x 2 = 2
1.67 x 2 = 3.34. Doesn't work. Multiply by 3
1 x 3 = 3
1.67 x 3 = 5.01. Close so round to whole numbers; therefore, the empirical formula is C3H5. If you didn't make a typo with the 56 molar mass, I would say the molecular formula is C3H5.