Chapter 10, Problem 4STP

Interpretation Introduction

Interpretation:

The grams of formaldehyde should be predicted.

Concept introduction:

The representation of atoms of a compound in simple whole number ratio is known as empirical formula.

Chemical formula or molecular formula represents the type of atoms and number of atoms present in a molecule using numerical subscripts and atomic symbol.

The rules for determining empirical formula is given as follows:

• Determine the mass of elements in given compound.
• Calculate number of moles using molar mass of each compound or element.
• Divide each number of moles by smallest number of mole value calculated in second step.
• Round the value calculated in step 3 to nearest whole number.

Answer to Problem 4STP

The grams of formaldehyde are 60.06 g.

Explanation of Solution

Assuming the percentage is given by mass that means 40.0 g, 6.7 g, and 53.3 g of carbon, hydrogen, and oxygen is present in butanoic acid.

Mass of carbon = 40.0 g

Mass of hydrogen = 6.7 g

Mass of oxygen = 53.3 g

Now, calculating the moles of carbon, hydrogen, and oxygen using their molar mass as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{40.0\text{g}}{12.0107\text{g/mol}}\\ {\text{n}}_{\text{C}}=3.33\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{6.7\text{g}}{\text{1}\text{.008}\text{g/mol}}\\ {\text{n}}_{\text{H}}=6.64\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{\text{53}\text{.3}\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=3.33\text{mol}\end{array}$

1. Now, dividing each mole value by smallest mole value as follows:

$\begin{array}{c}{\text{n}}_{\text{C}}=\frac{3.33}{3.33}\\ =1\\ {\text{n}}_{\text{H}}=\frac{6.64}{3.33}\\ =2\\ {\text{n}}_{\text{O}}=\frac{3.33}{3.33}\\ =1\end{array}$

2. Thus, the empirical formula and molecular formula for formaldehyde is ${\text{CH}}_{\text{2}}\text{O}$.

Now, the molar mass of formaldehyde as follows:

$\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{formaldehyde}=1\times \text{C}+2\times \text{H}+1\times \text{O}\\ =\left(1\times 12.0107+2\times 1.008+1\times 15.999\right)\text{g/mol}\\ =30.0257\text{g/mol}\end{array}$

Now, the grams of formaldehyde in 2.000 mol are calculated as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{formaldehyde}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ \text{Given}\text{mass}=\text{Moles}\text{of}\text{formaldehyde}\times \text{Molar}\text{mass}\\ =2.000\text{mol}\times 30.0257\text{g/mol}\\ \text{Given}\text{mass}=60.06\text{g/mol}\end{array}$

Thus, there are 60.06 g of formaldehyde in 2.000 moles of formaldehyde.