Question

A compound has ony C,H and O. After combustion of a 10.68g sample, 16.01g of CO2 and 4.37g of H2O. If the molar mass is 176.1 what are the Empirical Formulas and molecular formulas?

Answer 1

@thushananth01

Answer 2

@umerlodhi

Answer 3

i found this on yahoo...We don't need the percent composition because the masses from the combustion data will allow us to compute the moles of C, H and O. Grams, milligrams... it won't matter.

CxHyOz + O2(g) --> CO2(g) + H2O(l)
10.68g ................... 16.01g ....4.37g

16.01g CO2 x (1 mol CO2 / 44.0g CO2) x (1 mol C / 1 mol CO2) = 0.364 mol C = 4.37g C
4.37g H2O x (1 mol H2O / 18.0g H2O) x (2 mol H / 1 mol H2O = 0.486 mol H = 0.486 g H

10.68g - 4.37g - 0.486g = 5.82 g O
5.82g O x (1 mol O / 16.0g O) = 0.364 mol O

0.364 mol C / 0.364 = 1 mol C x 3 = 3 mol C
0.486 mol H / 0.364 = 1.33 mol H x 3 = 4 mol C
0.364 mol O / 0.364 = 1 mol O x 3 = 3 mol O

empirical formula = C3H4O3
empirical molar mass = 88.0 g/mol

176.1 g/mol / 88 g/mol = 2 ..... twice as many of each atom

molecular formula = C6H8O6

Answer 4

hope that helps i feel so loved that you tagged me lol

Answer 5

@mrcoolguy

Answer 6

Thank you so much! @umerlodhi

Answer 7

np