Chapter 6, Problem 59A

Interpretation Introduction

Interpretation:

The empirical formula of the compound needs to be determined.

Concept Introduction:

The simplest positive integer ratio of the atoms present in the chemical compound is known as empirical formula.

Answer to Problem 59A

The empirical formula of the compound is CuSH₁₀O₉

Explanation of Solution

Given information:

Cu = 25.45 %

S = 12.84 %

H = 4.036 %

O = 57.67 %

Conversion of percentage to grams with 100 g sample:

Mass of Cu = 25.45 g

Mass of S = 12.84 g

Mass of H = 4.036 g

Mass of O = 57.67 g

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Cu}\text{=}\frac{\text{mass}\text{of}\text{Cu}}{\text{molar}\text{mass}\text{of}\text{Cu}}\\ \text{=}\frac{\text{25}\text{.45}\text{g}}{\text{63}\text{.546}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.4005}\text{mol}\\ \text{moles}\text{of}\text{S}\text{=}\frac{\text{mass}\text{of}\text{S}}{\text{molar}\text{mass}\text{of}\text{S}}\\ \text{=}\frac{\text{12}\text{.84}\text{g}}{\text{32}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.4012}\text{mol}\\ \text{moles}\text{of}\text{H}\text{=}\frac{\text{mass}\text{of}\text{H}}{\text{molar}\text{mass}\text{of}\text{H}}\\ \text{=}\frac{\text{4}\text{.036}\text{g}}{\text{1}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{4}\text{.036}\text{mol}\\ \text{moles}\text{of}\text{O}\text{=}\frac{\text{mass}\text{of}\text{O}}{\text{molar}\text{mass}\text{of}\text{O}}\\ \text{=}\frac{\text{57}\text{.67}\text{g}}{\text{16}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{3}\text{.604}\text{mol}\end{array}$

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

  $\begin{array}{c}\text{ratio}\text{of}\text{Cu}\text{=}\frac{\text{0}\text{.4005}}{\text{0}\text{.4005}}\\ \text{=}\text{1}\\ \text{ratio}\text{of}\text{S}\text{=}\frac{\text{0}\text{.4012}}{\text{0}\text{.4005}}\\ \text{=}\text{1}\\ \text{ratio}\text{of}\text{H}\text{=}\frac{\text{4}\text{.036}}{\text{0}\text{.4005}}\\ \text{=}\text{10}\\ \text{ratio}\text{of}\text{O}\text{=}\frac{\text{3}\text{.60}}{\text{0}\text{.4005}}\\ \text{=}\text{9}\end{array}$

The table of empirical formula is shown below:

Element	%	moles	ratio	Whole no.
Cu	25.45	0.4005	0.400/0.400	1
S	12.84	0.400	0.400/0.400	1
H	4.036	4.036	4.036/0.400	10
O	57.67	3.60	3.60/0.400	9

Conclusion

The empirical formula of the compound is CuSH₁₀O₉