2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g CO 2 and 1.943 g H 2O. What is the empirical formula of the compound?

The empirical formula of the compound is C₃H₆O

We'll begin by calculating the mass of C, H and O in the compound. This can be obtained as follow:

Mass of CO₂ = 4.746 g

Molar mass of CO₂ = 44 g/mol

Molar mass of C = 12 g/mol

Mass of C = 12/44 × 4.746

Mass of H₂O = 1.943 g

Molar mass of H₂O  = 18 g/mol

Molar mass of H₂ = 1 × 2 = 2 g/mol

Mass of H = 2/18 × 1.943

Mass of C = 1.294 g

Mass of H = 0.216 g

Mass of compound = 2.088 g

Mass of O = (Mass of compound ) – (mass of C + mass of H)

Mass of O = 2.088 – (1.294 + 0.216)

C = 1.294 g

H = 0.216 g

O = 0.578 g

Divide by their molar mass

C = 1.294 / 12 = 0.108

H = 0.216 / 1 = 0.216

O = 0.578 / 16 = 0.036

Divide by the smallest

C = 0.108 / 0.036 = 3

H = 0.216 / 0.036 = 6

O = 0.036 / 0.036 = 1

Therefore, the empirical formula of the compound is C₃H₆O

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The empirical formula is C3H6O

Explanation:

Step 1: Data given

Mass of the sample =2.088 grams

The mass contains carbon, hydrogen, and oxygen

Mass of CO2 produced = 4.746 grams

Mass of H2O produced = 1.943 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 4.746 grams/ 44.01 g/mol

Moles CO2 = 0.1078 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.1078 moles CO2 we'll have 0.1078 moles C

Step 4: Calculate mass C

Mass C: moles C * atomic mass C

Mass C: 0.1078 moles * 12.01 g/mol

Mass C= 1.295 grams

Step 5: Calculate moles H2O

Moles H2O = 1.943 grams / 18.02 g/mol

Moles H2O = 0.1078 moles

Step 6: Calculate moles H

For 1 mol H2O we'll have 2 moles H

For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H

Step 7: Calculate mass H

Mass H = 0.2046 moles * 1.01 g/mol

Mass H = 0.218 grams

Step 8: Calculate mass O

Mass O = 2.088 grams - 1.295 grams - 0.218 grams

Mass O = 0.575 grams

Step 9: Calculate moles O

Moles O = 0.575 grams / 16.0 g/mol

Moles O = 0.0359 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.1078 moles / 0.0359 moles = 3

H: 0.2156 moles / 0.0359 moles = 6

O: 0.0359 moles / 0.0359 moles =1

The empirical formula is C3H6O

201cm/s÷5: 18

explanation:

there is no explanation

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.