
64g of an organic compound contains carbon, of hydrogen and the rest is $ 64g $ oxygen what is the empirical formula of the compound?
Answer
539.4k+ views
Hint: In the above-given question we first need to find out the formula of the compound by calculating the number of atoms included of carbon, hydrogen, oxygen. Then we will find out the simplest ratio of the molecular formula to find out the empirical formula.
Complete step by step answer:
The empirical formula is the simplest ratio of the molecular formula of the compound. In an organic compound, there are three main atoms that are carbon, hydrogen, oxygen. We are provided with the total molecular weight of the compound that is $ 64g $ , now the total mass of the carbon in the compound is $ 24g $ , the atomic mass of carbon is $ 12g $ , therefore the total number of atoms f carbon present is;
$ nC = 24/12 $
$ \Rightarrow $ $ nC = 2 $
Now the total mass of the hydrogen in the compound is $ 8g $ , the atomic mass of carbon is $ 1g $ , therefore the total number of atoms of hydrogen present is;
$ nH = 8/1 $
$ \Rightarrow $ $ nH = 8 $
The only remaining element is oxygen, to find out the mass of oxygen we need to subtract the sum of the mass of carbon and hydrogen from the total molecular mass of the given formula;
$ {m_O} = 64 - ({m_C} + {m_H}) $
$ \Rightarrow $ $ {m_O} = 64 - (24 + 8) $
$ \Rightarrow $ $ {m_O} = 64 - 32 $
$ \Rightarrow $ $ {m_O} = 32 $
The total mass of oxygen present in the formula is $ 32g $ , and the atomic mass of oxygen is $ 16g $ , therefore the total number of atoms of oxygen present is;
$ nO = 32/16 $
$ \Rightarrow $ $ nO = 2 $
By the above calculation, we can conclude that the molecular formula of the compound is $ {C_2}{H_8}{O_2} $ .
To find out the empirical formula we need to divide the formula by the simplest factor that is 2, on doing so we get $ C{H_4}O $ .
The empirical formula of the compound is $ C{H_4}O $
Note:
Organic compounds are generally formed by hydrogen, carbon and oxygen. Compounds which are considered to have central atoms as carbon and there is the bonding between two carbon atoms may be single, double or triple bonds.
Complete step by step answer:
The empirical formula is the simplest ratio of the molecular formula of the compound. In an organic compound, there are three main atoms that are carbon, hydrogen, oxygen. We are provided with the total molecular weight of the compound that is $ 64g $ , now the total mass of the carbon in the compound is $ 24g $ , the atomic mass of carbon is $ 12g $ , therefore the total number of atoms f carbon present is;
$ nC = 24/12 $
$ \Rightarrow $ $ nC = 2 $
Now the total mass of the hydrogen in the compound is $ 8g $ , the atomic mass of carbon is $ 1g $ , therefore the total number of atoms of hydrogen present is;
$ nH = 8/1 $
$ \Rightarrow $ $ nH = 8 $
The only remaining element is oxygen, to find out the mass of oxygen we need to subtract the sum of the mass of carbon and hydrogen from the total molecular mass of the given formula;
$ {m_O} = 64 - ({m_C} + {m_H}) $
$ \Rightarrow $ $ {m_O} = 64 - (24 + 8) $
$ \Rightarrow $ $ {m_O} = 64 - 32 $
$ \Rightarrow $ $ {m_O} = 32 $
The total mass of oxygen present in the formula is $ 32g $ , and the atomic mass of oxygen is $ 16g $ , therefore the total number of atoms of oxygen present is;
$ nO = 32/16 $
$ \Rightarrow $ $ nO = 2 $
By the above calculation, we can conclude that the molecular formula of the compound is $ {C_2}{H_8}{O_2} $ .
To find out the empirical formula we need to divide the formula by the simplest factor that is 2, on doing so we get $ C{H_4}O $ .
The empirical formula of the compound is $ C{H_4}O $
Note:
Organic compounds are generally formed by hydrogen, carbon and oxygen. Compounds which are considered to have central atoms as carbon and there is the bonding between two carbon atoms may be single, double or triple bonds.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

