A sample of an unknown substance was found to contain 0.24 g of carbon, 0.040 g of hydrogen and 0.32 g of oxygen. The empirical formula of the unknown was d) CH20. a) C2h4O2. b) CH3O. C2H20.c)
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Explanation:
carbon- .24/12
hydrogen- .040/1
oxygen - .32/16
Answered by
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Answer:
Explanation:Mass of the given sample compound = 0.24g
Mass of boron in the given sample compound = 0.096g
Mass of oxygen in the given sample compound = 0.144g
% composition of compound = % of boron and % of oxygen
Therefore % of boron = (mass of boron)/(mass of the sample compound) x 100
= 40%
Therefore % of oxygen = (mass of oxygen) / (mass of the sample compound) x 100
= 60%
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