What is its simplest formula?

Question

A 2.20 sample of a compound gave 5.63 g CO[2] and 2.30 g H[2]O on combustion in air. The compound is known to contain only C, H, O.

Answer 1

you have .128 mols co2, and .128 mol of h2o...so both are equal, and therefore have equal coefficients

balanced equation will give you....

2CHO -------> CO2 + H2O

Answer 2

C[x]H[y]O[z] + (x+1/2y-z) O[2] = xCO[2] + y/2 H[2]O

5.63 g CO[2] -> 5.63/(12+16×2) = 0.128 mol
2.30 g H[2]O -> 2.30/(1×2+16) = 0.128 mol

2.20 g (C[x]H[y]O[z]) --- 0.128 mol C
? g --------------------- 1.000 mol C
17.18 ~ 17.20 g (C[x]H[y]O[z])

Trying the numbers (for a whole number):
If x = 5; y = 10; z = 1
5×12 + 10×1 + 1×16 = 86 g/mol = 17.20×5

The compound is C[5]H[10]O
e.g. penthanol, or ethyl-propyl ether, or methyl-buthyl ether (IUPAC: ethoxypropane, methoxybuthane)

Answer 3

5.63/44.0 mol CO2 = 0.128 mol CO2, from 0.128 mol C

2.30/18.0 mol H2O = 0.128 mol H2O, from 0.256 mol H (TWO H in each H2O)

So C:H is 1:2.

0.128 mol of the CH2 fragment is 0.128 x 14.0, or 1.79 g. So 0.41 g is O

0.41/16.0 = 0.025

0.128:0.025 = 5:1, near enough.

So you have C5H10O

Principles: convert g to moles.Convert moles CO2 to C (1:1); convert molesH2O to H (1:2). What isn't C or H is O, so convert left over mass to moles O. Finally, as in all these probelms, divide through by the smallest.