Question 6 of 8: Composition of a barium-borate glass-ceramic in weight percent
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2017 — 16-Mec-B8 Engineering Materials. Three hours, open book; any non-communicating calculator permitted. Eight problems, all of equal value; any FIVE constitute a complete paper, so each problem is worth 20 marks. Candidates are urged to submit a clear statement of any assumptions made. All eight problems are solved below, because the set as a whole is the study resource.
Groover, Fundamentals of Modern Manufacturing, 7th ed. — shaping and consolidation of polymer-matrix composites.
Note on this sitting. Two of them are worth flagging because the fresh data changes the answer, not merely the arithmetic. Question 8 raises the copper content of the incumbent alloy, which makes it denser, and as a result both candidate alloys now meet the customer’s weight request so the selection turns on strength parity rather than on who clears the bar. Question 4 changes the aeroplane from a twelve-seat business jet to a hundred-seat commercial transport and the stretch from 25 to 30 per cent, which sharpens rather than softens the damage-tolerance objection. Every number below has been re-worked from this paper's own data.
Question 6: Composition of a barium-borate glass-ceramic in weight percent (20 marks)
Given. A batch of 7.5 mole per cent TiO2 nucleating agent in 92.5 mole per cent of the barium-borate glass BaO·4B2O3, with atomic weights read from the periodic table.
Given data
Quantity
Value
Glass former
BaO·4B2O3
Nucleating agent
TiO2, 7.5 mole%
Atomic weight of Ba
137.33
Atomic weight of B
10.81
Atomic weight of O
16.00
Atomic weight of Ti
47.87
Find. The composition of the resulting glass-ceramic in weight percent. The question asks for the composition “in weight percent of each component element,” so the elemental split is the required answer; it is reached through the oxide basis, which is how a batch is actually weighed out, and both are reported.
Approach. Take a convenient basis of 100 moles of batch, convert each constituent from moles to mass through its formula weight, and express each mass as a percentage of the total. The whole calculation is one mole-to-mass conversion carried out three ways.
Build the formula weights from the atomic weights. Adding the constituent atoms, $$M_{\text{BaO}} = 137.33 + 16.00 = 153.33$$ $$M_{\text{B}_{2}\text{O}_{3}} = 2(10.81) + 3(16.00) = 69.62$$ so the glass itself, one BaO with four B2O3, weighs $$M_{g} = 153.33 + 4(69.62) = \boxed{431.81\ \text{g/mol}}$$ and the nucleating agent weighs $M_{\text{TiO}_{2}} = 47.87 + 2(16.00) = 79.87$ g/mol. The glass unit is 5.4 times heavier than the titania unit, and that single ratio governs everything that follows.
Choose a basis and convert moles to mass. Taking 100 mol of batch, of which 7.5 mol is TiO2 and the remaining 92.5 mol is glass: $$m_{g} = 92.5(431.81) = 39\,942.43\ \text{g}$$ $$m_{t} = 7.5(79.87) = 599.03\ \text{g}$$ and the batch total is $m_{tot} = 40\,541.45$ g. Every percentage below is a fraction of this one number.
Report the two batch constituents in weight percent. Dividing through, $$\text{wt\% glass} = \frac{39942.43}{40541.45}\times100, \; \text{wt\% TiO}_{2} = \frac{599.03}{40541.45}\times100$$ which gives $$\boxed{98.52\ \text{wt\% BaO}\cdot4\text{B}_{2}\text{O}_{3} \; \text{and} \; 1.48\ \text{wt\% TiO}_{2}}$$ The contrast with the 7.5 mole per cent charged is the point of the question: a noticeable molar addition is a very small weight addition, because the heavy component is the one present in the majority.
Break the glass into its own oxides. Each of the 93 mol of glass carries one mole of BaO and four of B2O3, so $$\text{wt\% BaO} = \frac{92.5(153.33)}{40541.45}\times100 = 34.98\%$$ $$\text{wt\% B}_{2}\text{O}_{3} = \frac{92.5\times4\times69.62}{40541.45}\times100 = 63.54\%$$ Together with the 1.48 per cent TiO2 these sum to 100.00 per cent, which is the arithmetic check.
Composition of the glass-ceramic on an oxide basis. Boron oxide dominates the mass even though barium is by far the heaviest element present, because there are four B2O3 units for every BaO.
The oxide statement above is what a batch house would weigh out. The question, however, asks for the weight percent of each component element, so the same 40 541.45 g is now split element by element. This is the answer the paper is actually after.
Count the atoms of each element in 100 mol of batch. The glass contributes 92.5 mol Ba, $92.5\times4\times2 = 740$ mol B and $92.5(1 + 12) = 1202.5$ mol O; the nucleating agent contributes 7.5 mol Ti and 15 mol O, for 1217.5 mol O in all.
Convert to mass and divide. Multiplying each count by its atomic weight, $$m_{\text{Ba}} = 12703.03,\; m_{\text{B}} = 7999.40,\; m_{\text{Ti}} = 359.02,\; m_{\text{O}} = 19480.00\ \text{g}$$ These sum to 40 541.45 g, matching the batch total exactly, so no atom has been lost. Expressed as percentages, $$\boxed{31.33\%\ \text{Ba},\; 19.73\%\ \text{B},\; 0.89\%\ \text{Ti},\; 48.05\%\ \text{O}}$$
Oxygen is very nearly half the mass of the material, which is characteristic of every oxide glass and is worth carrying as a sanity check: an elemental analysis of a borate or silicate that does not return roughly 45 to 55 per cent oxygen has an arithmetic error in it. The titanium content, at well under one per cent by weight, is equally typical — a nucleating agent has only to seed crystallisation, not to build the network, so it is added in small amounts. On the reheat, the TiO2 phase-separates into a dense population of nuclei on which the barium-borate crystals then grow, converting the parent glass into a fine-grained glass-ceramic with far better strength and thermal-shock resistance.
Composition of the glass-ceramic in weight percent