Question

Question:A sample of Portland cement is analyzed and found to contain the following percentages by mass of oxides: $\mathbf{CaO},\mathbf{64}.\mathbf{3}\%;{\mathbf{SiO}}_{\mathbf{2}},\mathbf{21}.\mathbf{2}\%;{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}},\mathbf{5}.\mathbf{9}\%;{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}},\mathbf{2}.\mathbf{9}\%;\mathbf{MgO},\mathbf{2}.\mathbf{5}\%;{\mathbf{SO}}_{\mathbf{3}},\mathbf{1}.\mathbf{8}\%;{\mathbf{Na}}_{\mathbf{2}}\mathbf{O},\mathbf{1}.\mathbf{4}\%.$ Calculate the chemical amounts of Ca, Si, Al, Fe, Mg, S, and Na atoms per mole of O atoms in this sample.

Short answer

Answer

.$C{a}_{0.514}S{i}_{0.158}A{l}_{0.052}F{e}_{0.016}M{g}_{0.028}{S}_{0.010}N{a}_{0.020}{O}_{1.00}$

Step-by-step solution

Chemical amounts of each element

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 with their molar mass.
4. The total number of moles of oxygen should be calculated.
5. Divide the number of moles of each constituent element with the total number of moles of oxygen.

Explanation

If 100 g of Portland cement is taken, then the mass of each oxide is as shown below:

$$\begin{gathered} CaO---64.3 g \\ Si{O}_2---21.2 g \\ A{l}_2{O}_3---5.9 g \\ F{e}_2{O}_3---2.9 g \\ MgO---2.5 g \\ S{O}_3---1.8 g \\ N{a}_{2}O---1.4 g \end{gathered}$$

The chemical amounts of oxides can be obtained by dividing with their respective molar mass.

$$\begin{gathered} CaO---\frac{64.3 g}{56.08 g/mol}=1.15 mol \\ Si{O}_2---\frac{21.2 g}{60.09}=0.353 mol \\ A{l}_2{O}_3---\frac{5.9 g}{101.96 g/mol}=0.0578 mol \\ F{e}_2{O}_3---\frac{2.9 g}{159.69 g/mol}=0.0181 mol \\ MgO---\frac{2.5 g}{40.30 g/mol}=0.0620 mol \\ S{O}_3---\frac{1.8 g}{80.07 g/mol}=0.0224 mol \\ N{a}_{2}O---\frac{1.4 g}{61.98 g/mol}=0.0225 mo \end{gathered}$$

The total number of moles of oxygen:

$$\begin{gathered} n_{O_2}=\left[1.15 +\left(2\times 0.353\right)+\left(3\times 0.0578\right)+\left(3\times 0.0181\right)+0.0620+\left(3\times 0.0224\right)+0.0225\right] mol \\ =2.2354 mol \end{gathered}$$

Then,

$$\begin{gathered} n_{Ca}=\frac{1.15 mol}{2.2354 mol} \\ =0.514 \\ {n}_{Si}=\frac{0.353 mol}{2.2354 mol} \\ =0.158 \\ {n}_{Al}=\frac{0.0578\times 2 mol}{2.2354 mol} \\ =0.052 \\ {n}_{Fe}=\frac{0.0181\times 2 mol}{2.2354 mol} \\ =0.016 \\ {n}_{Mg}=\frac{0.0620 mol}{2.2354 mol} \\ =0.028 \\ {n}_{Fe}=\frac{0.0224 mol}{2.2354 mol} \\ =0.010 \end{gathered}$$