Question

An organic substance containing carbon, hydrogen and oxygen gave the percentage composition as c= 40.687%, h=5.085% and o=54.228% .the vapor density of compound is 59.calculate the molecular formula of the compound.

Answer 1

Answer : The molecular formula of the com[pound is, $C_4H_6O_4$

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 = 40.687 g

Mass of H = 5.085 g

Mass of O = 54.228 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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.687g}{12g/mole}=3.39moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.085g}{1g/mole}=5.085moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{54.228g}{16g/mole}=3.39moles$

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

For C = $\frac{3.39}{3.39}=1$

For H = $\frac{5.085}{3.39}=1.5$

For O = $\frac{3.39}{3.39}=1$

To make the whole number, multiplying the ratio by 2.

The ratio of C : H : O = 2 : 3 : 2

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

The Empirical formula = $C_2H_3O_2$

The empirical formula weight = 12(2) + 3(1) + 2(16) = 59 gram/eq

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}\\\\n=\frac{2\times \text{Vapor density}}{\text{Empirical formula weight}}=\frac{2\times 59}{59}=2$

Molecular formula = $(C_2H_3O_2)_n=(C_2H_3O_2)_2=C_4H_6O_4$

Therefore, the molecular formula of the com[pound is, $C_4H_6O_4$

Answer 2

Answer:

also pls answer this

Explanation:

An organic substance containing C; H and oxygen gave the following percentage composition.

C=52.17%

H=13.04%

O=34.78%

The molecular wt of the compound is 4.6.Calculate the molecular formula of the compound.