Question

An analysis of pyrex glass showed $12.9\% {\text{ }}{{\text{B}}_2}{{\text{O}}_3}$ , $2.2\% {\text{ A}}{{\text{l}}_2}{{\text{O}}_3}$ , $3.8\% {\text{ N}}{{\text{a}}_2}{\text{O}}$ , $0.4\% {\text{ }}{{\text{K}}_2}{\text{O}}$ and the remaining is ${\text{Si}}{{\text{O}}_2}$ . What is the ratio of silicon to boron atoms?

Answer 1

To solve this question, it is required to have knowledge about pyrex glass. We shall calculate the percentage of silicon present in glass. Then, we shall calculate moles of boron and silicon present from the mole concept (formula given). Then, we shall find the ratio of silicon to boron.

Formula used: ${\text{moles = }}\dfrac{{{\text{mass of compound}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
As we know that, pyrex glass is made of something called borosilicate glass. It is very resistant to thermal shock. It is used for laboratory glassware and kitchenware. Now, we shall find out the percentage of ${\text{Si}}{{\text{O}}_2}$ present in pyrex glass. As we know that, ${{\text{B}}_2}{{\text{O}}_3}$ is present at $12.9\%$ , ${\text{A}}{{\text{l}}_2}{{\text{O}}_3}$ is present at $2.2\%$ , ${\text{N}}{{\text{a}}_2}{\text{O}}$ is present at $3.8\%$ and ${{\text{K}}_2}{\text{O}}$ is present at $0.4\%$ . This means that ${\text{Si}}{{\text{O}}_2}$ will be present at $100 - \left( {12.9 + 2.2 + 3.8 + 0.4} \right){\text{ = }}80.7\%$ . Now, if we assume the weight of the glass compound to be 100g, this means that the weight of ${\text{Si}}{{\text{O}}_2}$ will be $80.7{\text{ g}}$ and the weight of ${{\text{B}}_2}{{\text{O}}_3}$ will be $12.9{\text{ g}}$.
Now, we shall convert them into the number of moles present:
${\text{moles of }}{{\text{B}}_2}{{\text{O}}_3}{\text{ = }}\dfrac{{{\text{12}}{\text{.9}}}}{{70}}{\text{ = 0}}{\text{.184}}$
As there are 2 atoms of boron in each molecule of ${{\text{B}}_2}{{\text{O}}_3}$ . This means that the number of moles of boron is twice the number of moles of ${{\text{B}}_2}{{\text{O}}_3}$. So:
${\text{moles of Boron = 0}}{\text{.184}} \times {\text{2 = 0}}{\text{.368}}$
For ${\text{Si}}{{\text{O}}_2}$ :
${\text{moles of Si}}{{\text{O}}_2}{\text{ = }}\dfrac{{{\text{80}}{\text{.7}}}}{{60}} = 1.34$
So, the moles of silicon will be $1.34{\text{ moles}}$ .
Now, we know that the number of moles of an element will be proportional to the number of atoms present. Thus, the ratio of silicon atoms to boron atom will be:
$\Rightarrow \dfrac{{{\text{number of Silicon atoms}}}}{{{\text{number of Boron atoms}}}} = \dfrac{{1.345}}{{0.368}}$
$\therefore$ The ratio of the atoms of silicon to the atoms of boron will be $\dfrac{{1.345}}{{0.368}}$ .

Note: In this type of question, we have used the number of moles instead of the weight of the elements because the number of moles is proportional to the number of atoms. Weight of the compounds are not proportional to the number of atoms present in them as each compound has different number of atoms in the same weight.