Chapter 10, Problem 161A

Interpretation Introduction

Interpretation:

The empirical formula for the given compound 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 161A

The empirical formula for given compound is ${\text{BaCO}}_3$.

Explanation of Solution

Assuming the percentage is given by mass that means the elements are present in that much amount.

Mass of carbon = 6.09 g

Mass of barium = 69.58 g

Mass of oxygen = 24.32 g

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

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{6.09\text{g}}{\text{12}\text{.0107}\text{g/mol}}\\ {\text{n}}_{\text{C}}=0.51\text{mol}\\ \text{Moles}\text{of}\text{barium}=\frac{69.58\text{g}}{\text{137}\text{.327}\text{g/mol}}\\ {\text{n}}_{\text{Ba}}=0.51\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{24.32\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=1.52\text{mol}\end{array}$

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

$\begin{array}{c}{\text{n}}_{\text{c}}=\frac{0.51}{0.51}\\ =1\\ {\text{n}}_{\text{Ba}}=\frac{0.51}{0.51}\\ =1\\ {\text{n}}_{\text{O}}=\frac{1.52}{0.51}\\ =3\end{array}$

2. Thus, the empirical formula of given compound is ${\text{BaCO}}_3$.