When a 0.952 g sample of an organic compound containing C, H,and O is burned completely in oxygen, 1.35 g of CO2 and0.826 g of H2O are produced. What is the empiricalformula of the compound?

Answer 1

Convert 1.35g CO2 to g C.
Convert 0.826 H2O to g H
Add gC+gH and subtract from 0.952 to obtain g O.

Now convert g each to moles. moles = grams/molar mass
Find the ratio of each element to one another with the lowest number being 1.00. The easy way to do that is to divide the smallest number by itself, then divide the other numbers by the same small number.
Post your work if you run into trouble.

Answer 2

C in CO2 is 12/ 44 % = 27.27%
in 1.35 g of CO2 there is (27.27% of 1.35) g of C = 0.3681g or 0.3681 /12 moles of C = 0.031 moles
% C in the compound = .3681 / .952 = 38.67%

H in H2O = 2 / 18 % = 11.111 %
in .826 g of H2O we have 11.11 % of .826 = 0.09177 g of H or 0.09177 / 1 moles of H = 0.09177 moles
% H in the compound = .09177 / .952 = 9.64%

% of O in the compound = 100 - ( 38.67 + 9.64) % = 51.69 %
mass of O = 51.69 % of .925 = 0.4781 g of O or 0.0.4781 / 16 moles of O = 0.02988 moles

ratio of moles of C:H:O = 0.031: 0.09177: 0.02988
divide by the smallest number and we get 1.03 : 3.07: 1

or 1:3:1

empirical formula is CH3O ..