CHEMISTRY stoichiometry problem, emperical/molecular formulas HELP!?

Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the comopund yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol.


>>>What are the empirical and molecular formulas of the compound?

THANKS SO MUCH!!!

Answer 1

16.01 mg CO2 x (12.011/44.011) = 4.369 mg C
4.37 mg x (2.016/18.016) = 0.489 mg H
10.68 mg - 4.369 - 0.489 = 5.822 mg O
4.369 mg C /12.011 = 0.364 millimoles C
0.489 mg H / 1.008 = 0.485 millimoles H
5.822 mg O / 16.00 = 0.364 millimoles of O

0.364/0.364 = 1.000 C * 3 = 3.00
0.485/0.364 = 1.329 H * 3 = 3.99.
0.364/0.364 = 1.000 O * 3 = 3.00
Empirical formula is C3H4O3 (MW=88.1)
Since MW of compound is 176.1, the molecular formula is-
C6H8O6

Answer 2

C6H8O6 + 5O2 --> 6CO2 + 4H2O

6.064 x 10^-5 mol C6H8O6 + 3.032 x 10^-4 mol O2 --> 3.638 x 10^-4 mol CO2 + 2.426 x 10^-4 mol H2O

Answer 3

First, you need a balanced equation for this reaction: 2 SO2 + O2 --> 2 SO3 Since the volume of a gas is directly proportional to moles of a gas at the same temperature and pressure, we can see from the equation that one mole of oxygen is needed for each two moles of sulfur dioxide. Therefore, for 3.00 moles of sulfur dioxide, we would need half as many (1.5 mol) moles of oxygen. Since 4.50 moles of oxygen gas is present, oxygen is in excess, and sulfur dioxide is the limiting reactant that will determine the amount of sulfur trioxide produced. From the equation, two moles of sulfur dioxide will produce two moles of sulfur trioxide. Therefore, 3.00 liters of sulfur dioxide gas will produce 3.00 liters of sulfur trioxide gas at constant conditions.