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Tpy6a [65]
3 years ago
13

A compound contains only carbon, hydrogen, and oxygen. Combustion of 139.1 g of the compound yields 208.6 g of CO2 and 56.93 g o

f H2O. The molar mass of the compound is 176.1 g/mol. *Each part of this problem should be submitted separately to avoid losing your work* 1. Calculate the grams of carbon (C) in 139.1 g of the compound: grams 2. Calculate the grams of hydrogen (H) in 139.1 g of the compound. grams 3. Calculate the grams of oxygen (O) in 139.1 g of the compound. grams
Chemistry
1 answer:
Fofino [41]3 years ago
3 0

Answer:

1. Mass of Carbon is 56.89g

2. Mass of Hydrogen is 6.33g

3. Mass of Oxygen is 75.88

Explanation:

The following were obtained from the question.

Mass of the compound = 139.1g

Mass of CO2 produced = 208.6g

Mass of H2O produced = 56.93

1. Determination of mass of Carbon (C). This is illustrated below:

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

Mass of C = 12/44 x 208.6

Mass of C = 56.89g

2. Determination of the mass of Hydrogen (H). This is illustrated below:

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

Mass of H = 2/18 x 56.93

Mass of H = 6.33g

3. Determination of the mass of oxygen (O).

This is illustrated below:

Mass of the compound = 139.1g

Mass of C = 56.89g

Mass of H = 6.33g

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

Mass of O = 139.1 - (56.89 + 6.33)

Mass of O = 139.1 - 63.22

Mass of O = 75.88

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3. A compound consists of 91.63 grams of carbon, 7.69 grams of hydrogen and 40.81 grams of
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Answer:

Molecular formula = C₁₂H₁₂O₄

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of C = 91.63 g

Mass of H = 7.69 g

Mass pf O = 40.81 g

Molar mass of compound = 220 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 7.69 / 1.01 = 7.61

Number of gram atoms of O = 40.81 / 16 = 2.55

Number of gram atoms of C = 91.63 / 12 = 7.64

Atomic ratio:

            C                      :      H                :         O

           7.64/2.55          :    7.61 /2.55    :       2.55/2.55

               3                     :          3               :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass  = 3×12+ 3×1.01 +16 = 55.03

n = 220 / 55.03

n = 4

Molecular formula = 4 (empirical formula)

Molecular formula = 4 (C₃H₃O)

Molecular formula = C₁₂H₁₂O₄

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3 years ago
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