Find the emprical formula of compound that has 40% carbon,6.7% hydrogen and 53.3% oxygen
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Emperical formula :- It represent the simplest whole number ratio of the atoms of all the elements present in the compound by mole.
molecular formula :- It represent the actual number of atoms of all the elements present in the compound by moles.
Element :- It consists same number of atoms
compound :- when two or more atoms of the different elements combined in a fixed ratio then compound is formed
mole :- It is a amount of a substance which has equal number of molecules as there in a 12g of one atom of carbon - 12 isotope
or it is a amount of an substance which contains exactly avogadro number (6.022×10^23) of particles this particles may be atoms or molecules.
molecular mass :- The sum of the atoms of all the elements present in a compound.
Formula Mass :- Some substance do not contains discrete constituents particles in such cases we use formula mass instead of molecules mass
molar mass :- mass of the one mole of the substance.
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assume 100 g of compound then in 100g ,
40g of carbon
6.7g of hydrogen
53.3 g of oxygen
[ make a 3 coloumns write one by one below ]
elements | C | O | H |
mass. | 40g | 6.7 g | 53.3 |
molar mass | 12g/mol | 16g/mol | 1g/mol |
number of moles | 40/12 = 3.33 | 6.7/16 = 0.41875 | 53.5/1 = 53.5 |
simplest ratio |3.33/0.41875=7.952≈8| 0.41875/0.41875 =1 |53.5/0.41875 | 127.76≈128
take approximate values
Emperical formula = C8H128O.
molecular formula :- It represent the actual number of atoms of all the elements present in the compound by moles.
Element :- It consists same number of atoms
compound :- when two or more atoms of the different elements combined in a fixed ratio then compound is formed
mole :- It is a amount of a substance which has equal number of molecules as there in a 12g of one atom of carbon - 12 isotope
or it is a amount of an substance which contains exactly avogadro number (6.022×10^23) of particles this particles may be atoms or molecules.
molecular mass :- The sum of the atoms of all the elements present in a compound.
Formula Mass :- Some substance do not contains discrete constituents particles in such cases we use formula mass instead of molecules mass
molar mass :- mass of the one mole of the substance.
________________________________
assume 100 g of compound then in 100g ,
40g of carbon
6.7g of hydrogen
53.3 g of oxygen
[ make a 3 coloumns write one by one below ]
elements | C | O | H |
mass. | 40g | 6.7 g | 53.3 |
molar mass | 12g/mol | 16g/mol | 1g/mol |
number of moles | 40/12 = 3.33 | 6.7/16 = 0.41875 | 53.5/1 = 53.5 |
simplest ratio |3.33/0.41875=7.952≈8| 0.41875/0.41875 =1 |53.5/0.41875 | 127.76≈128
take approximate values
Emperical formula = C8H128O.
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