Question

6. Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; OF: 9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 486g

Answer 1

Answer : The empirical formula of a compound is, $C_6H_{10}S_2O_1$  and the molecular of the compound is, $C_{18}H_{30}S_6O_3$

Solution : Given,

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 = 44.4 g

Mass of H = 6.21 g

Mass of S = 39.5 g

Mass of O = 9.86 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of S = 32 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{44.4g}{12g/mole}=3.7moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.21g}{1g/mole}=6.21moles$

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{39.5g}{32g/mole}=1.23moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{9.86g}{16g/mole}=0.62moles$

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

For C = $\frac{3.7}{0.62}=5.96\approx 6$

For H = $\frac{6.21}{0.62}=10.01\approx 10$

For S = $\frac{1.23}{0.62}=1.98\approx 2$

For O = $\frac{0.62}{0.62}=1$

The ratio of C : H : S : O = 6 : 10 : 2 : 1

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

The Empirical formula = $C_6H_{10}S_2O_1$

The empirical formula weight = 6(12) + 10(1) + 2(32) + 1(16) = 162 gram/eq

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

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{486}{162}=3$

Molecular formula = $(C_6H_{10}S_2O_1)_n=(C_6H_{10}S_2O_1)_3=C_{18}H_{30}S_6O_3$

Therefore, the molecular of the compound is, $C_{18}H_{30}S_6O_3$