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Calculate the Empirical formula of a compound that is 32% Cobalt, 15% Nitrogen, and 53% Oxygen Question 2 options: CoNO3 CoNO2 CoNO Co(NO3)2

User Randrumree
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The empirical formula for the compound that is 32% Cobalt, 15% Nitrogen, and 53% Oxygen is CoNO3, with a ratio of 1:1:3 for Cobalt, Nitrogen, and Oxygen, respectively.

To calculate the empirical formula of a compound, we need to determine the ratio of the elements present in the compound.

Given that the compound is 32% Cobalt, 15% Nitrogen, and 53% Oxygen, we can assume that we have 100 grams of the compound.

1. Convert the percentages to grams:

- Cobalt: 32% of 100 grams = 32 grams

- Nitrogen: 15% of 100 grams = 15 grams

- Oxygen: 53% of 100 grams = 53 grams

2. Convert the grams to moles:

- Cobalt: 32 grams / molar mass of Cobalt (Co) = X moles

- Nitrogen: 15 grams / molar mass of Nitrogen (N) = Y moles

- Oxygen: 53 grams / molar mass of Oxygen (O) = Z moles

3. Determine the ratio of the elements:

- Divide the moles of each element by the smallest number of moles obtained. This will give us the simplest whole-number ratio.

Assuming the molar masses are:

- Co: 58.93 g/mol

- N: 14.01 g/mol

- O: 16.00 g/mol

- Co: 32 g / 58.93 g/mol = 0.544 moles

- N: 15 g / 14.01 g/mol = 1.07 moles

- O: 53 g / 16.00 g/mol = 3.31 moles

4. Round the moles to the nearest whole number:

- Co: 0.544 ≈ 1

- N: 1.07 ≈ 1

- O: 3.31 ≈ 3

5. Write the empirical formula using the whole number ratio:

The empirical formula for the compound is CoNO3, with a ratio of 1:1:3 for Cobalt, Nitrogen, and Oxygen, respectively.

User Leigh  McCulloch
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