In 100 g of oxalic acid, there is: 26,7 g C, 2,2 g H and 71,1 g O.
n = m/M
nC= 26,7/12 = 2,225 mol
nH = 2,2/1,01 = 2,18 mol
nO = 71,1/16 = 4,44 mol
To find the empirical formula we note how many moles of each element we have. Then divide by the smallest number to get the ratios of each element and round off to nearest whole number.
That is C= 2.25 H= 2.18 O= 4.44. Smallest number is 2.18 therefore C= 2.225/2.18 = 1, H=2.18/2.18= 1, O =4.44/2.18 = 2
The empirical formula is CHO2. The molar mass of oxalic acid using the empirical formula is 45 g·mol^−1 . However the question gives the molar mass as 90 g·mol^−1 . Therefore the actual number of moles of each element must be double what it is in the empirical formula (90/45 = 2). The molecular formula is therefore C2H2O4.