Empirical formula:
Step 1
Information already provided
The mass percent composition:
40.9 % C
4.58 % H
54.5 % O
Information needed: from the periodic table
For C) 1 mol = 12.01 g
For H) 1 mol = 1.008 g
For O) 1 mol = 15.99 g
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Step 2
A sample of 100 g is assumed, so:
40.9 % C => 40.9 g C
4.58 % H => 4.58 g H
54.5 % O => 54.5 g O
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Step 3
Convert mass into moles:
40.9 g C x (1 mol/12.01 g) = 3.40 moles C
4.58 g H x (1 mol/1.008 g) = 4.54 moles H
54.5 g O x (1 mol/15.99 g) = 3.40 moles O
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Step 4
All moles calculated in step 3 need to be divided by the smallest one.
3.40 moles C/3.40 moles = 1
4.54 moles H/3.40 moles = 1.33
3.40 moles O/3.40 moles = 1
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Step 5
Integer numbers are needed, so let's multiply by 3 all of them in step 4
Therefore,
For C) 3
For H) 3.99 = 4 approx.
For O) 3
All these numbers calculated will be the subindexes in ascorbic acid
Answer:
Empirical formula: C3H4O3