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2 years ago

Calculate the empirical formula of a compound that is 42.9% Carbon and 57.1% Oxygen.

Chemistry

1 answer:

Marrrta [24]2 years ago

4 0

Answer:

CO.

Explanation:

Assuming the given percentages are by mass, we can solve this problem via imagining we have 100 g of the compound, if that were the case we would have:

42.9 g of C

57.1 g of O

Now we convert those masses into moles, using the elements' respective molar masses:

42.9 g of C ÷ 12 g/mol = 3.57 mol C

57.1 g of O ÷ 16 g/mol = 3.58 mol O

As the number of C moles and O moles is roughly the same, the empirical formula for the compound is CO.

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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

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Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

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$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

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