Question

Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g/tex.z-dn.net/?f=CO_%7B2%7D" id="TexFormula1" title="CO_{2}" alt="CO_{2}" align="absmiddle" class="latex-formula"> 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

Answer:

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

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

Answer:

The empirical formula of the compound is $C_4H_8O_1$

Explanation:

Moles of carbon dioxide = $\frac{3.2007 g}{44 g/mol}=0.07274 mol$

Moles of carbon in sample = $1\times 0.07274 mol=0.07274 mol$

Moles of water = $\frac{1.3101 g}{18 g/mol}=0.07279 mol$

Moles of hydrogen in sample = $2\times 0.07279 mol=0.14558 mol$

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 = $\frac{0.29244 g}{16 g/mol}=0.01828 mol$

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

Carbon =$\frac{0.07274 mol}{0.01828 mol}=3.9\approx 4$

Hydrogen =$\frac{0.14558 mol}{0.01828 mol}=7.9\approx 8$

oxygen=$\frac{0.01828 mol}{0.01828 mol}=1$

The empirical formula of the compound is $C_4H_8O_1$