The percent composition by mass of an unknown compound with a molecular mass of 60.052 amu is 40.002% C, 6.7135% H, and 53.284% O. Determine the compound's empirical and molecular formulas. Empirical formula:

The empirical formula and molecular formula of the compound with a molecular mass of 60.052 amu containing 40.002% C, 6.7135% H, and 53.284% O are:

1. The empirical formula of the compound is CH₂O

2. The molecular formula of the compound is C₂H₄O₂

1. Determination of the empirical formula of the compound.

C = 40.002%

H =  6.7135%

O = 53.284%

Divide by their molar mass

C = 40.002 / 12 = 3.3335

H = 6.7135 / 1 = 6.7135

O = 53.284 / 16 = 3.33025

Divide by the smallest

C = 3.3335 / 3.33025 = 1

H = 6.7135 / 3.33025 = 2

O = 3.33025 / 3.33025 = 1

Thus, the empirical formula of the compound is CH₂O

2. Determination of the molecular formula of the compound

Empirical formula = CH₂O

Molar mass = 60.052 amu

Molecular formula = Empirical × n = molar mass

[CH₂O]n = 60.052

[12 + (2×1) + 16]n = 60.052

30n = 60.052

Divide both side by 30

n = 60.052 / 30

n = 2

Molecular formula = [CH₂O]n

Molecular formula = [CH₂O]₂

Therefore, the molecular formula of the compound is C₂H₄O₂

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Empirical formula and molecular formula are and

Explanation:

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= 40.002 g

Mass of H= 6.7135 g

Mass of O = 53.284 g

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= 1 : 2 : 1

Hence the empirical formula is

The empirical weight of = 1(12)+2(1)+1(16)= 30 g.

The molecular weight = 60.052 g/mole

Now we have to calculate the molecular formula.

The molecular formula will be=

Thus empirical formula and molecular formula  are and

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