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Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen,
and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50​

Respuesta :

Eduard22sly

Answer:

C₃H₈O

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 1.031 g

Mass of CO₂ = 2.265 g

Mass of H₂O = 1.236 g

Empirical formula =?

Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:

For carbon, C:

Mass of CO₂ = 2.265 g

Molar mass of CO₂ = 44.01 g/mol

Molar mass of C = 12.01 g/mol

Mass of C =?

Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂

Mass of C = 12.01/44.01 × 2.265

Mass of C = 0.618 g

For hydrogen, H:

Mass of H₂O = 1.236 g

Molar mass of H₂O = 18.02 g/mol

Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol

Mass of H =?

Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O

Mass of H = 2.02/18.02 × 1.236

Mass of H = 0.139 g

For oxygen, O:

Mass of compound = 1.031 g

Mass of C = 0.618 g

Mass of H = 0.139 g

Mass of O =?

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

Mass of O = 1.031 – ( 0.618 + 0.139)

Mass of O = 1.031 – 0.757

Mass of O = 0.274 g

Finally, we shall determine the empirical formula. This can be obtained as follow:

C = 0.618 g

H = 0.139 g

O = 0.274 g

Divide by their molar mass

C = 0.618 / 12.01 = 0.051

H = 0.139 / 1.01 = 0.138

O = 0.274 / 16 = 0.017

Divide by the smallest

C = 0.051 / 0.017 = 3

H = 0.138 / 0.017 = 8

O = 0.017 / 0.017 = 1

Therefore, the empirical formula of the compound is C₃H₈O

mickymike92

Based on the data provided, the empirical formula of the compound is C₃H₈O

What is empirical formula?

The empirical formula of compound is its simplest formula showing the mole ratio of the elements in the compound.

First, the mass of carbon, hydrogen and oxygen in the compound is determined first as follows:

For carbon, C:

Mass of CO₂ = 2.265 g

Molar mass of CO₂ = 44.01 g/mol

Molar mass of C = 12.01 g/mol

Mass of C = 12.01/44.01 × 2.265

Mass of C = 0.618 g

For hydrogen, H:

Mass of H₂O = 1.236 g

Molar mass of H₂O = 18.02 g/mol

Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol

Mass of H = 2.02/18.02 × 1.236

Mass of H = 0.139 g

For oxygen, O:

Mass of compound = 1.031 g

Mass of C = 0.618 g

Mass of H = 0.139 g

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

Mass of O = 1.031 – ( 0.618 + 0.139)

Mass of O = 0.274 g

To determine the empirical formula of a compound, the mole ratio of the elements are determined as follows:

  • Mole ratio = reacting mass/molar mass

Mole ratio of the elements:

Carbon                Hydrogen         Oxygen

0.618 g/12.01    0.139 / 1.01         0.274 / 16

0.051                   0.138              0.017

Divide by the smallest ratio to convert to whole numbers

0.051 / 0.017     0.138 / 0.017  0.017 / 0.017

3           :            8           :               1

Therefore, the empirical formula of the compound is C₃H₈O

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