Step-by-step explanation:
49.31 + 43.79 = 93.10
finding the percentage of hydrogen in the compound hence 100-93.10 =6.9
therefore 6.9 percent of hydrogen is present in the compound
assuming that the compound is in 100 g so the elements present would be 49.31g of C , 43.79g of O and 6.9g of H
no. of moles of C = given mass÷molar mass
= 49.31 ÷ 12
=4.11 mol
no of moles of O = givem mass ÷molar mass
=43.79÷ 16
=2.73 mol
no. of moles of H =6.9÷1
=6.9 mol
dividing the no. of moles of each element by the least count of the elemnt to get the ratio of the simple whole numbr
therefore
so by dividing we get 1:1:2 of C:O:H
the empirical formula is CH2O
empirical formula mass of CH2O IS
= (12)+(2×1)+(16)
= 12+2+16
=30 G/MOL
n= molar mass of the compound ÷ empirical formula mass of the compound
so, n= 146.1 ÷30
n=5
molecular formula = n×empirical formula
molecular formula= 5×CH2O
molecular formula=C5H10O5 (ribose is the compound)