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When barium metal is burned in air, an oxide is formed with 81.1% by mass of barium. Calculate the formula of the compound. [Mr of barium=137, oxygen=16]
The answer is simple (BaO2), but I'm looking for a proper working instead of guess & check.
I tried doing this but I failed:
Let compound be Ba(x)O(y)
137x / (137x + 16y) = 0.811
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Edited by secretliker 02 Jul `08, 12:53AM
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Moderator'A' & 'O' Levels Chemistry & Biology Ex-MOE Teacher Tutor
611 posts since May '05-
Whenever the question says "81.1% by mass", let the sample mass of compound be 100g. Then, sample mass of Ba in compound is 81.1g, and sample mass of oxygen in compound is 100 - 81.1 = 18.9g.
Next, work out number of moles of each element in the compound, using the formula "No. of moles = Sample mass / Molar mass". Simplify the mole ratio to simplest ratio (ie. divide all numerical values by the smallest value), and you've got your empirical formula.
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