Question

Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g
`CO_(2)` and 1.3102 g
`H_(2)O`. Find the empirical formula of the compound.

The answer I got was
`CH_(16)O`, but I'm not sure if I did it right or not. Can someone confirm this for me?

Answer 1

The answer to your question is: C₃H₃O This is my answer.

Step-by-step explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

44 g -------------------- 12 g

3.2007 g ------------ x

x = (3.2007 x 12) / 44

x = 0.8729 g of Carbon

12 g of C -------------- 1 mol

0.8729 g -------------- x

x = (0.8729 x 1) / 12

x = 0.0727 mol of Carbon

For Hydrogen

18 g ---------------------- 1 g

1.3102 g ------------------- x

x = (1.3102 x 1) / 18

x = 0.0727 g of Hydrogen

1 g ------------------------ 1 mol

0.0727g ---------------- x

x = (0.0727 x 1)/1

x = 0.0727 mol of Hydrogen

For oxygen

g of Oxygen = g of sample - g of Carbon - g of hydrogen

g of Oxygen = 1.3109 - 0.8709 - 0.0727

g of Oxygen = 0.3673

16 g of Oxygen ------------- 1 mol of O

0.3673 g --------------------- x

x = (0.3673 x 1)/ 16

x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon 0.0727 / 0.023 = 3.1 ≈ 3

Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3

Oxygen 0.0230 / 0.023 = 1

C₃H₃O

Answer 2

The empirical formula of the compound is
`C_4H_8O_1`

Step-by-step explanation:

Moles of carbon dioxide =
`(3.2007 g)/(44 g/mol)=0.07274 mol`

Moles of carbon in sample =
`1* 0.07274 mol=0.07274 mol`

Moles of water =
`(1.3101 g)/(18 g/mol)=0.07279 mol`

Moles of hydrogen in sample =
`2* 0.07279 mol=0.14558 mol`

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of oxygen

Mass of oxygen =

1.3109 g - (12 g/mol × 0.07274 mol) - (1 g/mol× 0.14558 mol)

Mass of oxygen = 0.29244 g

Moles of oxygen in sample =
`(0.29244 g)/(16 g/mol)=0.01828 mol`

For empirical formula divide the moles of element which are is less amount with all the moles of every element.

Carbon =
`(0.07274 mol)/(0.01828 mol)=3.9\approx 4`

Hydrogen =
`(0.14558 mol)/(0.01828 mol)=7.9\approx 8`

oxygen=
`(0.01828 mol)/(0.01828 mol)=1`

The empirical formula of the compound is
`C_4H_8O_1`