Answer
Empirical: $C_{3}H_{4}O_{3}$
Molecular: $C_{6}H_{8}O_{6}$
Work Step by Step
16.01mg$CO_{2}(\frac{12gC}{44.01gCO_{2}})$=4.365mgC/10.68mg=40.87%C
$4.37mgH_{2}O(\frac{2gH}{18.02gH_{2}O})$=0.485H/10.68mg=4.54%H
100%-40.87%-4.57%=54.6%O
$\frac{40.87gH}{12g/mol}=3.4molC$
$\frac{4.54gH}{1g/mol}=4.54molH$
$\frac{54.6gO}{16g/mol}=3.4molO$
$\frac{4.54mol}{3.4mol}=1.3x3=4$
$MW=3M_{C}+4M_{H}+3M_{O}$
MW=3(12)+4(1)+3(16)
MW=88g/mol
$\frac{176.1g/mol}{88g/mol}\approx2$
