Question 8 of 8: Composition of a barium-borate glass-ceramic in weight percent

Question 8: Composition of a barium-borate glass-ceramic in weight percent (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A two-constituent batch specified on a molar basis, with atomic weights taken from the periodic table as the question directs:

Find. The composition of the nucleated batch in weight percent — reported both on the oxide basis and, as the question’s reference to “each component element” requires, on the elemental basis.

Approach. Take a convenient basis of 100 moles of batch, which turns the given mole percentages directly into moles; convert each constituent to mass through its formula weight; then divide by the total mass, first grouping the atoms by oxide and then by element.

1. Build the formula weights from the atomic weights. Working up from the elements, $$M_{\text{BaO}} = 137.33 + 16.00 = 153.33 \qquad M_{\text{B}_2\text{O}_3} = 2(10.81) + 3(16.00) = 69.62$$ $$M_{\text{BaO}\cdot 4\text{B}_2\text{O}_3} = 153.33 + 4(69.62) = 153.33 + 278.48 = 431.81\ \text{g/mol}$$ $$M_{\text{TiO}_2} = 47.87 + 2(16.00) = 79.87\ \text{g/mol}$$ Note that the base glass is treated as a single molecular unit of formula BaB₈O₁₃ — one barium, eight borons and thirteen oxygens.
2. Choose a basis and convert mole percent to moles. Taking 100 mol of batch, the 10 mole % of nucleating agent means $$n_{\text{TiO}_2} = 10\ \text{mol}, \qquad n_{\text{glass}} = 90\ \text{mol}$$ A basis of 100 moles is the whole trick of this question: it makes the mole percentages into moles with no arithmetic at all.
3. Convert moles to masses. $$m_{\text{glass}} = 90(431.81) = 38\,862.9\ \text{g}, \qquad m_{\text{TiO}_2} = 10(79.87) = 798.7\ \text{g}$$ $$m_{\text{total}} = 38\,862.9 + 798.7 = 39\,661.6\ \text{g per 100 mol of batch}$$
4. Report the two batch constituents in weight percent. Dividing each mass by the total, $$\boxed{\ 97.99\ \text{wt}\%\ \text{BaO}\cdot 4\text{B}_2\text{O}_3 \quad\text{and}\quad 2.01\ \text{wt}\%\ \text{TiO}_2\ }$$ The contrast with the molar specification is striking and is the point of the question: 10 mole % of the nucleating agent is only 2 wt % of the batch, because a mole of the base glass is more than five times heavier than a mole of titania.
5. Break the glass down into its oxides. Each mole of base glass carries one mole of BaO and four of B₂O₃, so $$w_{\text{BaO}} = \frac{90(153.33)}{39\,661.6}\times 100 = \frac{13\,799.7}{39\,661.6}\times 100 = 34.79\ \%$$ $$w_{\text{B}_2\text{O}_3} = \frac{90(4)(69.62)}{39\,661.6}\times 100 = \frac{25\,063.2}{39\,661.6}\times 100 = 63.19\ \%$$ $$\boxed{\ 34.79\ \%\ \text{BaO} \ +\ 63.19\ \%\ \text{B}_2\text{O}_3 \ +\ 2.01\ \%\ \text{TiO}_2 = 100.00\ \%\ }$$ This is the form in which a glass technologist would normally quote a batch.
6. Break the batch down into its elements. The question asks for the atomic weights of each component element, so the elemental composition is also required. Counting atoms per 100 mol of batch: 90 mol Ba, 90(8) = 720 mol B, 10 mol Ti, and for oxygen $$n_{\text{O}} = 90(1 + 12) + 10(2) = 1170 + 20 = 1190\ \text{mol}$$ Multiplying each by its atomic weight gives 12 359.7 g Ba, 7783.2 g B, 478.7 g Ti and 19 040 g O, which sum to 39 661.6 g — the same total as Step 3, and the check that no atom has been lost.
7. Express the elemental masses as weight percentages. Dividing through by the same total, $$\boxed{\ 31.16\ \%\ \text{Ba} \ +\ 19.62\ \%\ \text{B} \ +\ 1.21\ \%\ \text{Ti} \ +\ 48.01\ \%\ \text{O} = 100.00\ \%\ }$$ Oxygen is the largest single component by weight despite being the lightest element present, simply because there is so much of it: nearly twelve moles of oxygen for every mole of everything else combined. That is a general feature of oxide glasses and is worth remembering as a sanity check on any batch calculation.