Question

The percent composition by mass of a compound is 76.0% c, 12.8% h, and 11.2% o. the molar mass of this compound is 284.5 g/mol. what is the molecular formula of the compound?

Answer 1

You have a few steps to solve this one. First, we'll find the molar mass by percentage of each element in the molecule. Then, we'll divide each of those relative masses by the atomic mass of each element. The number of times the mass divides into the relative mass is the number of atoms of that element in the molecule:

C: 284.5 x .76 = 216.22
H: 284.5 x .128= 36.416
O: 284.5 x .112 = 31.864.

Now we divide out each element's atomic mass (from the periodic table). it's okay if they're approximated from the decimal answer.
C: 216.22 ÷ 12.011 ≈ 18
H: 36.416 ÷ 1.008 ≈36
O: 31.864 ÷ 15.999 ≈ 2

Therefore, the molecular formula is C18H36O2. 

The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O

Answer 2

Taking into accoun the definition of molecular and empirical formula, the molecular formula is: (C₉H₁₈O₁)₂= C₁₈H₃₆O₂

• Definitions

In first place, the empirical formula is the simplest expression to represent a chemical compound. It indicates the elements that are present and the minimum proportion in whole numbers that there are between their atoms.

On the other side, the molecular formula is the chemical formula that indicates the number and type of different atoms present in the molecule. The molecular formula is the actual number of atoms that make up a molecule.

Finally, percent by mass is a way of representing the concentration of an element in a compound or a component in a mixture.

• Calculations

Through the percentage composition of a compound, its molecular formula can be known as follows:

• Assuming you have 100 grams of the compound, the mass percentages will indicate the mass of each element present in those 100 grams. In this case: 76 g of C, 12.8 g of H and 11.2 g of O are present in 100 grams of the compound.

• Using the atomic mass of each element, it is possible to transform its mass into number of moles. In this case:

C: 76 g× $\frac{1 mole}{12.01 g}$= 6.34 moles of C

H: 12.8 g× $\frac{1 mole}{1.01 g}$= 12.7 moles of H

O: 11.2 g×$\frac{1 mole}{16 g}$= 0.7 moles of O

• The molar quantity of each element is then divided by the smallest molar quantity. In this case:

C: $\frac{6.34 moles}{0.7 moles}$= 9

H: $\frac{12.7 moles}{0.7 moles}$= 18

O: $\frac{0.7 moles}{0.7 moles}$= 1

The empirical formula of the compound will therefore be C₉H₁₈O₁.

The molecular formula can coincide with the empirical formula or be a multiple of it. To find the molecular formula, the empirical formula must be multiplied by an integer "n", which will be the quotient of dividing the molecular mass by the mass of the empirical formula.

In this case, the molar mass of the empirical formula is 142.27 g/mol. Then, the integer "n" is calculated as:

284.5÷142.27= 2

Then, the molecular formula is: (C₉H₁₈O₁)₂= C₁₈H₃₆O₂

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