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3 years ago

5

What type of potential energy do fuels contain

2 answers:

NISA [10]3 years ago

6 0

Kinetic energy maybe or something

ale4655 [162]3 years ago

5 0

Answer:

It contains chemical potential energy.

Explanation:

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3 years ago

A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87% H by mass. What is the empirical formula of this

Dmitry [639]

<u>Answer:</u> The empirical formula of the compound becomes $C_5H_{10}O_2$

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Let the mass of the compound be 100 g

Given values:

% of C = 58.8%

% of H = 9.87%

% of O = [100 - 58.8 - 9.87] = 31.33%

Mass of C = 58.8 g

Mass of H = 9.87 g

Mass of O = 31.33 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol$

$\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol$

$\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles

$\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5$

$\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5$

$\text{Mole fraction of O}=\frac{1.96}{1.96}=1$

Converting the mole fraction into whole numbers by multiplying them with 2.

$\text{Mole fraction of C}=2.5\times 2=5$

$\text{Mole fraction of H}=5\times 2=10$

$\text{Mole fraction of O}=1\times 2=2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 5 : 10 : 2

Hence, the empirical formula of the compound becomes $C_5H_{10}O_2$

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