Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.
1 answer:
<em>Acetic acid, HC2H3O2</em> First, calculate for the molar mass of acetic acid as shown below. M = 1 + 2(12) + 3(1) + 2(16) = 60 g Then, calculating for the percentages of each element.<em> Hydrogen:</em> P1 = ((4)(1)/60)(100%) = <em>6.67%</em> <em> Carbon:</em> P2 = ((2)(12)/60)(100%) = <em>40%</em> <em>Oxygen</em> P3 =((2)(16) / 60)(100%) = <em>53.33%</em> <em>Glucose, C6H12O6</em> The molar mass of glucose is as calculated below, 6(12) + 12(1) + 6(16) = 180 The percentages of the elements are as follow, <em> Hydrogen:</em> P1 = (12/180)(100%) = <em>6.67%</em> <em>Carbon:</em> P2 = ((6)(12) / 180)(100%) = <em>40%</em> <em>Oxygen:</em> P3 = ((6)(16) / 180)(100%) = <em>53.33%</em> b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.
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