Chapter 3, Problem 18STP

Interpretation Introduction

Interpretation:

The compound needs to be determined. If mass of sample is 125 g and there is $\text{16}\text{.7}\text{g}$ hydrogen, $\text{75}\text{.0}\text{g}$ carbon, and $\text{33}\text{.0}\text{g}$ oxygen.

Concept introduction:

The mass percentage is defined as the percentage of component present in it. The mass percentage of the element is the mass of element divided by the total mass of the compound and multiplied by 100.

$\text{Mass percentage=}\frac{\text{mass}\text{of}\text{element}}{\text{mass}\text{of}\text{total}\text{compound}}\text{×100}$

Answer to Problem 18STP

Correct answer: Option (e), isopropanol

Explanation of Solution

Reason for correct option: The given table:

Percentage by mass of carbon, hydrogen, and oxygen in selected compounds
Compound	%H	%C	%O
Carbonic acid	$3.2$	$19.4$	$77.4$
Acetic acid	$6.7$	$40.0$	$53.3$
Methanol	$12.5$	$37.5$	$40.0$
Methanol	$6.7$	$40.0$	$53.3$
Isopropanol	$13.3$	$60.0$	$26.7$

For the percentage by mass calculation- The mass of sample (compound) $\text{= 125}\text{.0}\text{g}$

Mass of hydrogen $\text{=}\text{16}\text{.7}\text{g}$

The percentage by mass of hydrogen $\text{=}\frac{\text{mass}\text{of}\text{hydrogen}}{\text{mass}\text{of}\text{compound}}\text{×100}$

For the percentage by mass of hydrogen $\text{=}\frac{\text{16}\text{.7}\text{g}}{\text{125}\text{.0}\text{g}}\text{×100}\text{=}\text{13}\text{.3}\text{\%}$

Mass of carbon $\text{= 75}\text{.0}\text{g}$

The percentage by mass of carbon $\text{=}\frac{\text{mass}\text{of}\text{carbon}}{\text{mass}\text{of}\text{compound}}\text{×100}$

For the percentage by mass of carbon $\text{=}\frac{\text{75}\text{.0}\text{g}}{\text{125}\text{.0}\text{g}}\text{×100}\text{=}60.0\text{\%}$

Mass of oxygen $\text{=}\text{33}\text{.0}\text{g}$

The percentage by mass of oxygen $\text{=}\frac{\text{mass}\text{of}\text{oxygen}}{\text{mass}\text{of}\text{compound}}\text{×100}$

For the percentage by mass of oxygen $\text{=}\frac{\text{33}\text{.0}\text{g}}{\text{125}\text{.0}\text{g}}\text{×100}\text{=}\text{26}\text{.7}\text{\%}$

From the above calculation the value of percentage by mass of hydrogen, carbon and oxygen is matched with the isopropanol from the given table.