An unknown compound has the formula #C_xH_yO_z#. You burn 0.1523 g of the compound and isolate 0.3718 g of #CO_2# and 0.1522 g of #H_2O#. What is the empirical formula of the compound?

First, work out the mass of each element that was present in the original compound. Carbon is always present as #CO_2# in the ratio (12.011 g / 44.0098 g), and hydrogen is always present as #H_2O# in the ratio (2.0158 g / 18.0152 g).

So you need to work out the mass of carbon in 0.3718 g of #CO_2# and the mass of hydrogen in 0.1522 g of water..

Carbon: 0.3718 x (12.011 / 44.0098) = 0.10147 g
Hydrogen: 0.1522 x (2.0158 / 18.0152) = 0.01703 g

You can determine the mass of oxygen by difference. 0.1523 - 0.10147 - 0.01703 = 0.0338 g
.
Next, convert each of these to numbers of moles:

Carbon: 0.10147 / 12.011 = 0.00845 mol
Hydrogen: 0.01703 / 1.0079 = 0.01689 mol
Oxygen: 0.0338 / 15.994 = 0.002113 mol
Remember that you're working out moles of hydrogen atoms here, not moles of molecular hydrogen gas. So use the atomic weight of hydrogen 1.0079.
.
Next, divide each number of moles by the lowest value, so that you get as close as possible to whole numbers.

Carbon: 0.00845 / 0.002113 = 4
Hydrogen: 0.001689 / 0.002113 = 8
Oxygen: 0.002113/ 0.002113 = 1

So the emprical formula is #C_4H_8O#

The molar mass is 72.1 g/mol - work out the mass of #C_4H_8O# - this is (12 x 4) + (8 x 1) + 16 = 72. Therefore the molecular formula is the same as the empirical..

Balanced equation of cmbustion reaction

#C_xH_yO_z+(x+y/4-z/2)O_2->xCO_2+y/2H_2O#

#"No.of moles of "C_xH_yO_z " reacted"#

#="its mass"/"Its molar mass"=(0.1523g)/(72.1g/"mol")=0.1523/72.1"mol"#

#"No.of moles of "CO_2" formed"#

#="its mass"/"Its molar mass"=(0.3718g)/(44g/"mol")=0.3718/44mol#

#"No.of moles of "H_2O" formed"#

#="its mass"/"Its molar mass"=(0.1522g)/(18g/"mol")=0.1522/18mol#

By the balanced equation 1 mole of the compound produces x moles #CO_2# and #y/2" moles "H_2O#

Hence #0.1523/72.1"mol"# compound will produce #0.1523/72.1*x" mol "CO_2#

#:.0.1523/72.1*x=0.3718/44#

#=>x=0.3718/44*72.1/0.1523~~4#

Again #0.1523/72.1"mol"#
compound will produce #0.1523/72.1*y/2" mol "H_2O#

So
#0.1523/72.1*y/2=0.1522/18#

#=>y=0.1522/18*(72.1*2)/0.1533~~8#

Finally

#"Molar mass of "C_xH_yO_z=72.1#

#=>12x+y+16z=72.1#

#=>12*4+8+16z=72.1#

#z=1#

Hence both empirical and molecular formula of the compound is# C_4H_8O#