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Sholpan
11 days ago
10

Classify the following sets of measurements as accurate, precise, both, or neither.

Chemistry
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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ
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Answer: The molecular formula for the specified organic compound is C_{18}H_{20}O_2

Explanation:

The combustion reaction of a hydrocarbon comprising carbon, hydrogen, and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where 'x', 'y', and 'z' denote the subscripts of Carbon, hydrogen, and oxygen respectively.

The information provided includes:

Mass of CO_2=39.61g

Mass of H_2O=9.01g

From our knowledge:

Molar mass of carbon dioxide is 44 g/mol

Molar mass of water is 18 g/mol

For determining the amount of carbon:

In carbon dioxide weighing 44 g, 12 g of carbon is found.

Hence, in 39.61 g of carbon dioxide, \frac{12}{44}\times 39.61=10.80g grams of carbon will be found.

For finding the mass of hydrogen:

In water weighing 18 g, 2 g of hydrogen can be found.

Thus, in 9.01 g of water, \frac{2}{18}\times 9.01=1.00g grams of hydrogen will be present.

Mass of oxygen in the compound is given by (13.42) - (10.80 + 1.00) = 1.62 g

To derive the empirical formula, the following steps must be followed:

  • Step 1: Convert the indicated masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles

  • Step 2: Calculating the ratio of moles of the respective elements.

To find the mole ratio, each mole value is divided by the smallest amount of moles calculated, which is 0.10 moles.

For Carbon = \frac{0.9}{0.10}=9

For Hydrogen = \frac{1}{0.10}=10

For Oxygen = \frac{0.10}{0.10}=1

  • Step 3:Using the mole ratios as subscripts.

The ratio of C: H: O = 9: 10: 1

Therefore, the empirical formula for the mentioned compound is C_9H_{10}O

To ascertain the molecular formula, it is necessary to find the valency, which is multiplied by each element to derive the molecular formula.

The equation to determine the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

Given the data:

Molecular formula mass = 268.34 g/mol

Empirical formula mass = 134 g/mol

Substituting the values into the aforementioned equation yields:

n=\frac{268.34g/mol}{134g/mol}=2

By multiplying this valency with each element's subscripts from the empirical formula, the results are:

C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2

Consequently, the molecular formula for the given organic compound is C_{18}H_{20}O_2.

3 0
2 months ago
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