G An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of a 1.893- gram sample of the compound produced 5.545 grams of carbon dioxide and 1.388 grams of water. The 1.893-gram sample contained 7.002 x 10-3 moles. What is the molecular formula for this compound

C18H22O2

Explanation:

The following were obtained from the question:

Mass of the unknown compound = 1.893g

Mass of CO2 produced = 5.545g

Mass of H2O produced = 1.388g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Now let us obtain the various masses of C, H and O present in the unknown compound.

This is illustrated below:

Mass of C in the unknown compound = 12/44 x 5.545 = 1.512g

Mass of H in the unknown compound = 2/18 x 1.388 = 0.154g

Mass of O in the unknown compound = 1.893 — (1.512+0.154) = 0.227g

Now, let us calculate the empirical formula of the compound.

C = 1.512g

H = 0.154g

O = 0.227g

Divide by their molar Mass

C = 1.512/12 = 0.126

H = 0.154/1 = 0.154

O = 0.227/16 = 0.014

Now, divide by the smallest

C = 0.126/0.014 = 9

H = 0.154/0.014 =11

O = 0.014/0.014 = 1

The empirical formula is C9H11O

from the question,

Number of mole the unknown compound = 7.002x10^-3 mol

Mass of the unknown compound = 1.893g

Molar Mass of the unknown compound = Mass /number of mole = 1.893g/7.002x10^-3 mol = 270.35g/mol

Now, we can easily find the molecular formula of the unknown compound as follows

[C9H11O]n = 270.35

[(12x9) + (11x1) +16 ]n = 270.35

135n = 270.35

Divide both side by the coefficient of n i.e 135

n = 270.35/135 = 2

Therefore, the molecular formula is => [C9H11O]n

=> [C9H11O]2

=> C18H22O2

The molecular formula is C18H22O2

Explanation:

Step 1: Data given

Mass of the sample = 1.893 grams

Mass of CO2 produced = 5.545 grams

Mass of H2O produced = 1.388 grams

The 1.893-gram sample contained 7.002 * 10^-3 moles

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 5.545 grams / 44.01 g/mol

Moles CO2 = 0.1260 moles

In 1 mol CO2 we have 1 mol C

In 0.1260 moles CO2 we have 0.1260 moles C

Step 3: Calculate mass C

Mass C = 0.1260 moles * 12.01 g/mol

Mass C = 1.513 grams

Step 4: Calculate moles and H2O

Moles H2O = 1.388 grams / 18.02 g/mol

Moles H2O = 0.07703 moles

In 1 mol H2O we have 2 moles H

In 0.07703 moles H2O we have 2*0.07703 = 0.15406 moles H

Step 5: Calculate mass H

Mass H = 0.15406 moles * 1.01 g/mol

Mass H = 0.1556 grams

Step 6: Calculate mass O

Mass O = 1.893 - 1.513 - 0.1556

Mass O = 0.2244 grams

Step 7: Calculate moles O

Moles O = 0.2244 grams / 16.0 g/mol

Moles O = 0.01403 moles

Step 8: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.1260 moles/ 0.01403 moles = 9

H: 0.15406 moles / 0.01403 moles = 11

O: 0.01403 moles / 0.01403 moles = 1

The empirical formula is C9H11O

The molecular mass of this formula is 135.2 g/mol

Step 9: Calculate the molar mass of this compound

Molar mass = 1.893 grams / 0.007002 moles

Molar mass = 270.4 g/mol

Step 9: Calculate the molecular formula

We have to multiply the empirical formula by n

n = 270.4 / 135.2  = 2

The molecular formula is 2*(C9H11O) = C18H22O2

1.solid

2.carbon dioxide

3.helium

the volume increases