Question

An oxide of Chromium is found to have the following % composition :
68.4% Cr and 31.6% Oxygen. Determine the empirical formula of the
compound.​

Answer 1

Answer: The empirical formula is $Cr_2O_3$.

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Cr = 68.4 g

Mass of O = 31.6 g

Step 1 : convert given masses into moles.

Moles of Cr = \frac{\text{ given mass of Cr}}{\text{ molar mass of Cr}}= \frac{68.4g}{52g/mole}=1.32moles[/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{31.6g}{16g/mole}=1.98moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cr =$\frac{1.32}{1.32}=1$

For O =$\frac{1.98}{1.32}=1.5$

Converting the ratios into whole number ratios:

The ratio of Cr: O = 2: 3

Hence the empirical formula is $Cr_2O_3$.

Answer 2

Given:

Composition of chromium = 68.4%

Composition of Oxygen = 31.6%

To find:

Empirical formula of the compound

Formula to be used:

$\text {Moles}=\text {Grams} \times \frac{1 \text {mole}}{\text {Atomic mass of the element}}$

Calculation:

Step 1: Convert percentage to gram

68.4% of Cr means 68.4g of Cr

31.6% of O means 31.6g of O

Step 2: Gram to mole conversion

For Cr

$\text {Moles}=\text {Grams} \times \frac{1 \text {mole}}{\text {Atomic mass of the element}}$

$\text {Moles}=68.4 \mathrm{g} \times \frac{1 \mathrm{mole}}{52 \mathrm{g}}$

$\text {Moles}=\frac{68.4}{52}$

$Moles=1.315$

For O

$\text {Moles}=\text {Grams} \times \frac{1 \text {mole}}{\text {Atomic mass of the element}}$

$\text {Moles}=31.6\mathrm{g} \times \frac{1 \mathrm{mole}}{16 \mathrm{g}}$

$\text {Moles}=\frac{31.6}{16}$

$\text {Moles}=1.975$

Step 3: Mole ratio calculation

Divide the obtained values with the least value. The values obtained are 1.315 and 1.975. As 1.315 is least value,

$\text {Mole ratio of } C r=\frac{1.315}{1.315}=1$

$\text {Mole ratio of } O=\frac{1.975}{1.315}=1.5$

Step 4: Conversion of obtained ratios into whole number

Cr = 2 ( 1 multiplied by 2)

O = 1.5 = 3 (1.5 multiplied by 2 to make it as a whole number)

Step 5: Determination of Empirical formula

From the obtained value the empirical formula of the given compound is determined as $Cr_2O_3$.

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