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Andreyy89
2 years ago
9

What is the empirical formula of a hydrocarbon if complete combustion or 2.900 mg of the hydrocarbon produced 9.803 mg of CO2 an

d 2.006 mg of H2O? Be sure to write C first in the formula.
Chemistry
1 answer:
kow [346]2 years ago
5 0

Answer:

The empirical formula of the hydrocarbon is CH.

Explanation:

The following data were obtained from the question:

Mass of hydrocarbon = 2.9 mg

Mass of CO2 = 9.803 mg

Mass of H2O = 2.006 mg

Next, we shall determine the mass of carbon (C) and hydrogen (H) in the compound since hydrocarbon contains carbon and hydrogen only.

This is illustrated below:

Molar mass of CO2 = 12 + (16x2) = 12 + 32 = 44 g/mol

Mass of CO2 = 9.803 mg

Mass of C in the compound =?

Mass of C in the compound =

12/44 x 9.803

= 2.674 mg

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

Mass of H2O = 2.006 mg

Mass of H in the compound =

2/18 x 2.006

= 0.223 mg

Finally, we shall determine the empirical formula of the hydrocarbon as follow:

Carbon (C) = 2.674 mg

Hydrogen (H) = 0.223 mg

Divide by their molar mass

C = 2.674 /12 = 0.223

H = 0.223 / 1 = 0.223

Divide both side by the the smallest

C = 0.223/0.223 = 1

H = 0.223/0.223 = 1

Therefore, the empirical formula of the hydrocarbon is CH.

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Putting values in equation 1, we get:

\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol

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By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = \frac{1}{1}\times 0.0295=0.108 moles of methyl benzoate

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Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g

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\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%

Hence, the percent yield of the reaction is 32.34 %

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