A compound is made of 49.31% C, 43.79% 0 and the rest hydrogen. The molar mass is 146.1g/mol, what are the empirical and molecular formulas?
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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)