Question

A 170.00g sample of an unidentified compound contains 29.84g sodium,67.49g chromium,and 72.67g oxygen.find the emperical formula.

Answer 1

Answer : The empirical formula for the given compound is $Na_2Cr_2O_7$

Explanation : Given,

Mass of Na = 29.84 g

Mass of Cr = 67.49 g

Mass of O = 72.67 g

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

Step 1: Converting the given masses into moles.

Moles of Na =$\frac{\text{Given mass of Na}}{\text{Molar mass of Na}}=\frac{29.84g}{23g/mole}=1.297moles$

Moles of Cr = $\frac{\text{Given mass of Cr}}{\text{Molar mass of Cr}}=\frac{67.49g}{52g/mole}=1.297moles$

Moles of O = $\frac{\text{Given mass of O}}{\text{Molar mass of O}}=\frac{72.67g}{16g/mole}=4.542moles$

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 1.297 moles.

For Na = $\frac{1.297}{1.297}=1$

For Cr = $\frac{1.297}{1.297}=1$

For O = $\frac{4.542}{1.297}=3.5$

The ratio of Na : Cr : O = 1 : 1 : 3.5

To make the ratio in a whole number we are multiplying the ratio by 2, we get:

The ratio of Na : Cr : O = 2 : 2 : 7

Step 3: Taking the mole ratio as their subscripts.

Empirical formula for the given compound is $Na_2Cr_2O_7$

Hence, the empirical formula for the given compound is $Na_2Cr_2O_7$

Answer 2

The empirical formula of the given sample is $Na_2Ch_2O_7$

Given the following data:

• Total mass of sample = 170.00 g

• Mass of sodium = 29.84 g

• Mass of chromium = 67.49 g

• Mass of oxygen = 72.67 g

Based on science:

• Molar mass of oxygen = 16 g/mol.

• Molar mass of chromium = 52 g/mol.

• Molar mass of sodium = 23 g/mol.

To determine the empirical formula of the given sample:

First of all, we would determine the number of moles for each chemical element in the sample.

For sodium:

$Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{29.84}{23}$

Number of moles = 1.2974 moles

For chromium:

$Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{67.49}{52}$

Number of moles = 1.2979 moles

For oxygen:

$Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{72.67}{16}$

Number of moles = 4.5419 moles

Next, we would divide by the smallest number of moles:

• Sodium = $\frac{1.2974}{1.2974}$ = 1

• Chromium = $\frac{1.2979}{1.2974}$ = 1

• Oxygen = $\frac{4.5419}{1.2974}$ = 3.5

Lastly, we would multiply by 2 to make the ratio a simple whole number:

$(1:1:3.5) \times 2 = 2:2:7$

Empirical formula = $Na_2Ch_2O_7$

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