Chemistry, 19.02.2020 21:00 zitterkoph

A 150 g sample of a compound is comprised of 44% C, 9% H and the remainder is O by mass. What is the compound's empirical formula?

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Answer from: ssargeant2559

The empirical formula for the given compound is CH_3O

Explanation:

We are given:

Percentage of C = 44 % of 150 g

Percentage of H = 9 % of 150 g

Percentage of O = 47 % of 150 g

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 66 g

Mass of H = 13.5 g

Mass of O = 70.5 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{66g}{12g/mole}=5.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{13.5g}{1g/mole}=13.5moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{70.5g}{16g/mole}=4.41moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 4.41moles.

For Carbon = $\frac{5.5}{4.41}=1.25\approx 1$

For Hydrogen = $\frac{13.5}{4.41}=3.06\approx 3$

For Oxygen = $\frac{4.41}{4.41}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is CH_3O

Answer from: 501jmw

The empirical formula of the compound has been $\rm \bold{CH_3O}$ .

The empirical formula has been a whole unit formula with the application for the determination of the compound in the ratio of the constituent elements.

The empirical formula has been given by the computation of the moles of elements with respect to another.

Computation for Empirical formula:

The mass of elements in the compound has been calculated as x% of 150 g.

The mass of Carbon has been:

$\rm Mass\;C=\dfrac{44}{100}\;\times\;150\\Mass\;C=66\;g$

The mass of carbon has been 66 grams.

The mass of Hydrogen has been:

$\rm Mass\;H=\dfrac{9}{100}\;\times\;150\\Mass\;H=13.5\;g$

The mass of Hydrogen has been 13.5 grams.

The mass of Oxygen has been:

$\rm Mass\;O=\dfrac{47}{100}\;\times\;150\\Mass\;O=70.5\;g$

The mass of Oxygen has been 70.5 grams.

The moles of an element has been given by:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

The moles of following elements in the compound has been:

$\rm Moles\;C=\dfrac{66}{12} \\Moles\;C=5.5\;mol$

The moles of carbon in the compound has been 5.5 mol.

$\rm Moles\;H=\dfrac{13.5}{1} \\Moles\;H=13.5\;mol$

The moles of hydrogen in the compound has been 13.5 mol.

$\rm Moles\;O=\dfrac{70.5}{16} \\Moles\;O=4.41\;mol$

The moles of oxygen in the compound has been 4.41 mol.

The empirical formula for the compound has been given as the ratio of C : H : O.

C : H : O = 1 : 3 : 1

Thus, the empirical formula of the compound has been $\rm \bold{CH_3O}$ .

For more information about empirical formula, refer to the link:

Answer from: Quest

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Answer from: Quest

that's what i got you can count the number of atoms

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A 150 g sample of a compound is comprised of 44% C, 9% H and the remainder is O by mass. What is the...

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