Question

The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4% O, and its molar mass is 102 g/mol. Calculate the empirical and molecular formulas for this compound.

Answer 1

Let our basis for the calculation be 1 mol of the substance.
      (1 mol)(102 g/mol) = 102 g
Determine the amount of C, H, and O in mole.
          C = (102 g)(0.588)(1 mol/12 g) = 4.998 mols C
          H = (102 g)(0.098)(1 mol/1 g) = 9.996 mols H
          O = (102 g)(0.314)(1 mol/16 g) = 2 mols O

The empirical formula of the substance is C5H10O2. The molar mass of the empirical formula is 102. This means that this is also its molecular formula.

Answer 2

Answer : The empirical and molecular formula of the compound is, $C_5H_{10}O_2$

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 = 58.8 g

Mass of H = 9.8  g

Mass of O = 31.4 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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{58.8g}{12g/mole}=4.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.8g}{1g/mole}=9.8moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{31.4g}{16g/mole}=1.9moles$

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

For C = $\frac{4.9}{1.9}=2.5$

For H = $\frac{9.8}{1.9}=5.16\approx 5$

For O = $\frac{1.9}{1.9}=1$

The ratio of C : H : O = 2.5 : 5 : 1

To make the ratio in a whole number we are multiplying ratio by 2, we get:

The ratio of C : H : O = 5 : 10 : 2

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

The Empirical formula = $C_5H_{10}O_2$

The empirical formula weight = 5(12) + 10(1) + 2(16) = 102 gram/eq

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

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{102}{102}=1$

Molecular formula = $(C_5H_{10}O_2)_n=(C_5H_{10}O_2)_1=C_5H_{10}O_2$

Therefore, the molecular formula of the compound is, $C_5H_{10}O_2$