Combustion of a 0.9827-g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of CO2 and 1.070 g of H2O. What is the empirical formula of the compound

The answer to your question is   C₂H₆O

Explanation:

Data

Compound   CxHyOz

mass of the sample = 0.9827

mass of CO₂ = 1.90 g

mass of H₂O = 1.070 g

Process

1.- Find the moles and grams of Carbon

                    44g of CO₂ 12 g of C

                     1.9 g of CO₂  x

                     x = (1.9 x 12) / 44

                     x = 0.52 g

                    12 g of C 1 mol

                    0.52 g     x

                    x = (0.52 x 1) / 12

                    x = 0.043 moles

2.- Find the moles and grams of Hydrogen

                   18 g of H₂O 2 g of H

                   1.07 g of H₂O  x

                   x = (1.07 x 2)/18

                   x = 0.12 g of H

                   1 g of H 1 mol of H

                   0.12 g   x

                   x = (0.12 x 1)/1

                   x = 0.12 moles of H

3.- Calculate the moles and grams of Oxygen

Mass of Oxygen = 0.9827 - 0.12 - 0.52

                            = 0.3427 g

                    16 g of O 1 mol

                     0.3427 g x

                     x = (0.3427 x 1) / 16

                     x = 0.021 moles

4.- Divide by the lowest number of moles

Carbon       0.043 / 0.021 = 2

Hydrogen   0.12 / 0.021 = 5.7 ≈ 6

Oxygen       0.021 / 0.021 = 1

5.- Write the empirical formula

                    C₂H₆O

The empirical formula of the compound is C4H11O2

Explanation:

Step 1: Data given

Mass of the sample = 0.9827 grams

Mass of CO2 produced = 1.900 grams

Mass of H2O produced = 1.070 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of O2 = 32.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 1.900 grams / 44.01 g/mol

Moles CO2 = 0.04317 moles

Step 3: Calculate moles C

In 1 mol CO2 we have 1 mol C

For 0.04317 moles CO2 we have 0.04317 moles C

Step 4: Calculate mass C

Mass C = 0.04317 moles * 12.01 g/mol

Mass C = 0.5185 mass

Step 5: Calculate moles H2O

Moles H2O = 1.070 grams / 18.02 g/mol

Moles H2O = 0.05938 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.05938 moles H2O we have 2*0.05938 = 0.11876 moles H

Step 7: Calculate mass H

Mass H = 0.11876 * 1.01 g/mol

Mass H = 0.1199 grams

Step 8: Calculate mass O

Mass O = 0.9827 grams - 0.5185 grams - 0.1199 grams

Mass O = 0.3443 grams

Step 9: Calculate moles O

Moles O = 0.3443 grams / 16.0 g/mol

Moles O = 0.02152 moles

Step 10: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.04317 moles / 0.02152 moles  = 2

H: 0.11876 moles / 0.02152 moles = 5.5

O: 0.02152 moles / 0.02152 moles = 1

For each 2 C atoms we have 5.5 H atoms and 1 O atom

This means for each 4 C atoms we have 11 H atoms and 2 O atoms

The empirical formula of the compound is C4H11O2

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6