Question

A 3.87 gram sample of ascorbic acid (Vitamin C) yield 5.80 grams of CO₂ and 1.58 grams of H₂O on combustion. What is the empirical formula of this compound? (This compound contains only C, H, and O.)

Answer 1

In one combustion analysis, 3.87 g of ascorbic acid yields 5.8 g of CO2 & 1.58 grams of water
find the empirical formula

convert grams of CO2 & H2O intoo gramms of C & H by using molar masses:
5.8 g CO2 (12.01 g/mol C) / (44.01 g/mol CO2) = 1.5827766 grams of carbon
&
1.58 g H2O (2.016 g H2) / (18.01 g/mol H2O) = 0.1768617 grams of Hydrogen

find grams of oxygen
3.87 g of ascorbic acid - 1.5827766 g C - 0.1768617 g of H = 2.11036 grams of Oxygen

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now find moles using molar masses
(1.5827766 grams of carbon) ( 1 mol C / 12.01 g/mol = 0.131788 mol C
(0.1768617 grams of Hydrogen) ( 1 mole H / 1.008 g ) = 0.175458 mol H
(2.11036 grams of Oxygen) (1 mol O / 16 grams) = 0.131897 mol O

dividing by the smallest, in order to ratio the moles
0.131788 mol C / 0.131788 = 1 mol C
0.175458 mol H / 0.131788 = 1.33136 mol H
0.131897 mol O / 0.131788 = 1 mol O

triple the moles to get whole numbers
& your empirical formula would be
C₃H₄O₃

(which has a fomula mass of 88)

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these questions sometimes as for the molecular formula as well
if so ,
they will provide the molar mass
the molar mass of ascorbic acid is 176 g/mol

so .... since we hav an empirical formula of C₃H₄O₃ has a mass 88,
& the molar mass is double that at 176....
so is the molacular formula
which should be
C₆H₈O₆