Asked by Macey
                what is the empirical formula of a compound that is 62.0% C, 10.4% H, and 27.5% O by mass?
            
            
        Answers
                    Answered by
            DrBob222
            
    Take a 100 g sample. This will give you
62.0 g C
10.4 g H
27.5 g O
Now determine moles of each.
62.0/atomic mass C = ??
10.4/atomic mass H = ??
27.5/atomic mass O = ??
Now you want to determine the mole ratios of these three to each other in small whole numbers. The easy way to do this is to divide the smallest number by itself; obviously, that gives 1.000 for that element. Then divide all the other values by that same small number and round to whole numbers. That should provide the empirical formula.
    
62.0 g C
10.4 g H
27.5 g O
Now determine moles of each.
62.0/atomic mass C = ??
10.4/atomic mass H = ??
27.5/atomic mass O = ??
Now you want to determine the mole ratios of these three to each other in small whole numbers. The easy way to do this is to divide the smallest number by itself; obviously, that gives 1.000 for that element. Then divide all the other values by that same small number and round to whole numbers. That should provide the empirical formula.
                    Answered by
            me
            
    C3H6O
    
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