What is the total no of atoms present in 25.0 mg of camphor.C10H16O ?
plz show how to solve it. (whole sol.)
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Molar mass of C10H16O = 10(12.011g/mol)+16(1.008g/mol)+1(15.999g...
Molar mass of C10H16O = 152.237g/mol
Find the number of moles of C10H16O in the sample:
n = m/M
n = 25mg / 152.237g/mol
n = 0.025g / 152.237g/mol
n = 1.642176343 x 10^-4 mol
Since we know that 1 mole is equivalent to 6.02214199 x 10^23, we can calculate the number of MOLECULES of C10H16O:
1.642176343 x 10^-4 mol x 6.02214199 x 10^23 molecules/mol = 9.88941911 x 10^19 molecules
However, since the question asks for the total number of ATOMS, we must multiply 9.88941911 x 10^19 by 27, since there are 27 atoms in each C10H16O molecule:
9.88941911 x 10^19 molecules x 27 atoms/molecule = 2.67 x 10^21 atoms
Therefore, there are 2.67 x 10^21 atoms in 25.0mg of C10H16O.
Hope this helps!
Molar mass of C10H16O = 152.237g/mol
Find the number of moles of C10H16O in the sample:
n = m/M
n = 25mg / 152.237g/mol
n = 0.025g / 152.237g/mol
n = 1.642176343 x 10^-4 mol
Since we know that 1 mole is equivalent to 6.02214199 x 10^23, we can calculate the number of MOLECULES of C10H16O:
1.642176343 x 10^-4 mol x 6.02214199 x 10^23 molecules/mol = 9.88941911 x 10^19 molecules
However, since the question asks for the total number of ATOMS, we must multiply 9.88941911 x 10^19 by 27, since there are 27 atoms in each C10H16O molecule:
9.88941911 x 10^19 molecules x 27 atoms/molecule = 2.67 x 10^21 atoms
Therefore, there are 2.67 x 10^21 atoms in 25.0mg of C10H16O.
Hope this helps!
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