Question

Question:A sample of soda-lime glass for tableware is analyzed and found to contain the following percentages by mass of oxides: ${\mathbf{SiO}}_{\mathbf{2}},\mathbf{72}.\mathbf{4}\%;{\mathbf{Na}}_{\mathbf{2}}\mathbf{O},\mathbf{18}.\mathbf{1}\%;\mathbf{CaO},\mathbf{8}.\mathbf{1}\%;{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}},\mathbf{1}.\mathbf{0}\%;\mathbf{MgO},\mathbf{0}.\mathbf{2}\%;\mathbf{BaO},\mathbf{0}.\mathbf{2}\%.$ (The elements are not actually present as binary oxides, but this is the way compositions are usually given.) Calculate the chemical amounts of Si, Na, Ca, Al, Mg, and Ba atoms per mole of O atoms in this sample.

Short answer

Answer

$S{i}_{0.42}N{a}_{0.10}C{a}_{0.05}A{l}_{0.0068}M{g}_{0.0017}B{a}_{0.00045}{O}_{1.00}$

Step-by-step solution

Chemical amounts of each element concerning oxygen

The chemical amounts of each element with respect to oxygen can be calculated by using the below steps:

1. Consider the 100 g of sample.
2. The mass of each oxide with respect to 100 g is written.
3. Then calculate the number of moles of each compound by dividing by their molar mass.
4. The total number of moles of oxygen should be calculated.
5. Divide the number of moles of each constituent element by the total number of moles of oxygen.

Calculation

If 100 g of the sample that is soda-lime glass is taken, then the amount of each oxide is as shown below:

$$\begin{gathered} Si{O}_2--- 72.4 g \\ N{a}_{2}O---18.1 g \\ CaO--- 8.1 g \\ A{l}_2{O}_3---1.0 g \\ MgO--- 0.2 g \\ BaO---0.2 g \end{gathered}$$

The number of moles of each oxide is obtained by dividing its mass by its respective molar mass:

$$\begin{gathered} Si{O}_2--- \frac{72.4 g}{60.08 g/mol}=1.21 mol \\ N{a}_{2}O---\frac{18.1 g}{61.98 g/mol}=0.292 mol \\ CaO--- \frac{8.1 g}{56.08 g/mol}=0.144 mol \\ A{l}_2{O}_3---\frac{1.0 g}{101.96 g/mol}=0.00980 \\ MgO--- \frac{0.2 g}{40.30 g/mol}=0.00496 mol \\ BaO---\frac{0.2 g}{153.3 g/mol}=0.00130 mol \end{gathered}$$

The total number of moles of Oxygen in the sample:

$$\begin{gathered} n_{O_2}=\left[\left(2\times 1.21\right)+0.292+0.144+\left(3\times 0.00980\right)+0.00496+0.00130\right]mol \\ =2.87206 mol \end{gathered}$$

Chemical amounts of each element: