Chapter 10, Problem 16STP

Interpretation Introduction

Interpretation:

The molecular formula of the compound needs to be identified.

Concept introduction:

The simplest positive integer ratio of the atoms present in the chemical compound is known as empirical formula. Molecular formula tells about the actual number of atoms of elements present in the given 1 molecule of the compound.

Answer to Problem 16STP

The molecular formula of the compound is C₆H₁₂O₄

Explanation of Solution

Given information

Total given mass of the unknown molecule is = 25 g

Mass of carbon in the given molecule is = 12.16 gram

Mass of hydrogen in the given molecule is = 2.027 gram

Mass of Oxygen in the given molecule is = 10.81 gram

The Calculation of % composition of carbon in the molecule is shown below:

$\text{\%Composition = }\frac{\text{Given mass}}{\text{Total mass}}\times 100$

$\begin{array}{c}\text{\% Composition of carbon = }\frac{\text{12}\text{.16}}{\text{25}}\text{×100}\\ \text{= 48}\text{.64\%}\end{array}$

The Calculation of % composition of hydrogen in the molecule is shown below:

$\begin{array}{l}\text{\% Composition of hydrogen = }\frac{2.027}{25}\times 100\\ =8.108\%\end{array}$

The Calculation of % composition of oxygen in the molecule is shown below:

$\begin{array}{l}\text{\% Composition of oxygen = }\frac{10.81}{25}\times 100\\ =43.24\%\end{array}$

The Calculation of Empirical formula is shown below:

$\begin{array}{l}\text{Fraction of carbon atoms =}\frac{\text{\% Composition}}{\text{Molar mass}}\\ \text{Fraction of carbon atoms}=\frac{48.64}{12}\\ =4.053\end{array}$

$\begin{array}{l}\text{Fraction of Hydrogen atoms = }\frac{\text{\% Composition}}{\text{Molar mass}}\\ \text{Fraction of Hydrogen atoms = }\frac{\text{8}\text{.108}}{\text{1}\text{.008}}\\ \text{= 8}\text{.044}\end{array}$

$\begin{array}{l}\text{Fraction of Oxygen atoms = }\frac{\text{\% Composition}}{\text{Molar mass}}\\ \text{Fraction of Oxygen atoms}=\frac{43.24}{16}\\ =2.7025\end{array}$

The calculation of ratio of number of atoms by dividing the each moles of smallest moles is shown below:

$\begin{array}{c}\text{ratio}\text{of}\text{C}\text{=}\frac{\text{4}\text{.053}}{\text{2}\text{.7025}}\\ \text{=}\text{1}\text{.5}\\ \text{ratio}\text{of}\text{H}\text{=}\frac{\text{8}\text{.044}}{\text{2}\text{.7025}}\\ \text{=}\text{3}\\ \text{ratio}\text{of}\text{O}\text{=}\frac{\text{2}\text{.7025}}{\text{2}\text{.7025}}\\ \text{= 1}\end{array}$

Therefore; the ratio is

1.5:3:1

3:6:2

The table of empirical formula is shown below:

Element	%	fraction	ratio	Whole no.
C	48.64	4.053	4.053/2.7025 = 1.5	3
H	8.108	8.044	8.044/2.7025 = 3	6
O	43.24	2.7025	2.7025/2.7025 = 1	2

Hence the empirical formula is = C₃H₆O₂

The calculation of molecular formula is shown below:

Molecular formula = (Empirical formula)_n

$\begin{array}{c}\text{n = }\frac{\text{molar mass}}{\text{empirical mass}}\\ \text{=}\frac{\text{148}}{\text{74}}\\ \text{= 2}\end{array}$

Hence, the molecular formula = (C₃H₆O₂)₂

= C₆H₁₂O₄

Conclusion

The molecular formula of the compound is C₆H₁₂O₄