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The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is?
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The ratio of percentage of carbon and hydrogen in an organic compound ...
Explanation:
To find the empirical formula of the compound, we need to determine the simplest ratio of the atoms present in the compound. The given ratio of carbon to hydrogen is 6:1, which means that for every 6 atoms of carbon, there is 1 atom of hydrogen. Similarly, the ratio of carbon to oxygen is 3:4, which means that for every 3 atoms of carbon, there are 4 atoms of oxygen.
Step 1: Finding the common factor
To find the simplest ratio, we need to determine the common factor between the two ratios. To do this, we can multiply the ratio of carbon to hydrogen by 4 and the ratio of carbon to oxygen by 2. This gives us the following ratios:
Carbon to hydrogen = 24:4
Carbon to oxygen = 6:8
Step 2: Simplifying the ratios
Now we need to simplify the ratios to their simplest form. To do this, we can divide both ratios by the smallest number in each ratio, which is 4 for carbon to hydrogen and 6 for carbon to oxygen. This gives us the following ratios:
Carbon to hydrogen = 6:1
Carbon to oxygen = 1:4/3
Step 3: Determining the empirical formula
The simplified ratios give us the simplest ratio of atoms in the compound. We can now use these ratios to determine the empirical formula of the compound. To do this, we assume that the compound contains 100 grams of the compound, which allows us to convert the ratios into grams.
Assuming 100 grams of the compound:
- Carbon = 75 grams (6/7 of 100 grams)
- Hydrogen = 12.5 grams (1/7 of 100 grams)
- Oxygen = 12.5 grams (1/7 of 100 grams)
Next, we need to convert the grams of each element into moles by dividing by their respective atomic masses. The atomic masses of carbon, hydrogen, and oxygen are 12, 1, and 16 respectively.
- Carbon = 6.25 moles (75 grams / 12 atomic mass)
- Hydrogen = 12.5 moles (12.5 grams / 1 atomic mass)
- Oxygen = 0.78125 moles (12.5 grams / 16 atomic mass)
The simplest ratio of atoms in the compound is therefore:
- Carbon = 6.25 / 0.78125 = 8
- Hydrogen = 12.5 / 0.78125 = 16
- Oxygen = 0.78125 / 0.78125 = 1
Therefore, the empirical formula of the compound is C8H16O.
To find the empirical formula of the compound, we need to determine the simplest ratio of the atoms present in the compound. The given ratio of carbon to hydrogen is 6:1, which means that for every 6 atoms of carbon, there is 1 atom of hydrogen. Similarly, the ratio of carbon to oxygen is 3:4, which means that for every 3 atoms of carbon, there are 4 atoms of oxygen.
Step 1: Finding the common factor
To find the simplest ratio, we need to determine the common factor between the two ratios. To do this, we can multiply the ratio of carbon to hydrogen by 4 and the ratio of carbon to oxygen by 2. This gives us the following ratios:
Carbon to hydrogen = 24:4
Carbon to oxygen = 6:8
Step 2: Simplifying the ratios
Now we need to simplify the ratios to their simplest form. To do this, we can divide both ratios by the smallest number in each ratio, which is 4 for carbon to hydrogen and 6 for carbon to oxygen. This gives us the following ratios:
Carbon to hydrogen = 6:1
Carbon to oxygen = 1:4/3
Step 3: Determining the empirical formula
The simplified ratios give us the simplest ratio of atoms in the compound. We can now use these ratios to determine the empirical formula of the compound. To do this, we assume that the compound contains 100 grams of the compound, which allows us to convert the ratios into grams.
Assuming 100 grams of the compound:
- Carbon = 75 grams (6/7 of 100 grams)
- Hydrogen = 12.5 grams (1/7 of 100 grams)
- Oxygen = 12.5 grams (1/7 of 100 grams)
Next, we need to convert the grams of each element into moles by dividing by their respective atomic masses. The atomic masses of carbon, hydrogen, and oxygen are 12, 1, and 16 respectively.
- Carbon = 6.25 moles (75 grams / 12 atomic mass)
- Hydrogen = 12.5 moles (12.5 grams / 1 atomic mass)
- Oxygen = 0.78125 moles (12.5 grams / 16 atomic mass)
The simplest ratio of atoms in the compound is therefore:
- Carbon = 6.25 / 0.78125 = 8
- Hydrogen = 12.5 / 0.78125 = 16
- Oxygen = 0.78125 / 0.78125 = 1
Therefore, the empirical formula of the compound is C8H16O.
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The ratio of percentage of carbon and hydrogen in an organic compound ...
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The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is?
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The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is?.
The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ratio of percentage of carbon and hydrogen in an organic compound is found to be 6 : 1 and that between carbon and oxygen is 3 : 4 . the empirical formula of the compound is?.
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