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Andreyy89
3 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]3 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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1b) Answer is: the number of atoms of gold (Au) is 1.073·10²³.

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M(C₁₂H₂₂O₁₁) = 12 · Ar(C) + 22 · Ar(H) + 11 · Ar(O) · g/mol.

M(C₁₂H₂₂O₁₁) = 12 · 12.01 + 22 · 1.01 + 11 · 16 · g/mol.

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n(O) = 11 · 0.0035 mol.

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N(C) = 0.042 mol · 6.022·10²³ 1/mol.

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