Answer to Question #83347 in Organic Chemistry for Toby

Question #83347
How many atoms of O are present in 78.1gram of oxygen
1
Expert's answer
2018-11-27T07:12:11-0500

Answer on Question #83347, Chemistry/Organic Chemistry

How many atoms of O are present in 78.1gram of oxygen

Solution

N= NA*n

n= m/M

Calculations:

n(O2)= m(O2)/M(O2) = 78.1 g/ 32.00 g/mol = 2.44 mol

N(O2) = 6.02*1023 mol-1*2.44 mol = 1.47*1024 molecules

As one molecule of oxygen (O2) consists of two atoms of oxygen then we should multiply number of oxygen molecules by two to find number of oxygen atoms:

N(O) = N(O2)*2 = 1.47*1024 *2 = 2.94*1024 atoms

Answer: 2.94*1024 atoms

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