Acompound is found to contain 59.94 % carbon , 13.44 % hydrogen , and 26.62 % oxygen by mass.

(1) the empirical formula for this compound is .
(2) the molecular weight for this compound is 60.11 amu. the molecular formula for this compound is

Answer : (1) The empirical formula of a compound is,

(2) The molecular formula of a compound is,

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 59.94 g

Mass of H = 13.44 g

Mass of O = 26.62 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =

Moles of H =

Moles of O =

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

For O =

The ratio of C : H : O = 3 : 8 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =

The empirical formula weight = 3(12) + 8(1) + 1(16) = 60 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

Molecular formula =

Therefore, the molecular of the compound is,

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i believe the answer is true.