# Department of Science, Math, & Technology

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How many atoms is the earth?
Posted: Posted September 11th, 2017
Edited September 11th, 2017 by Louis De Pointe du Lac
 You can get this number with a google search but I'm gonna try to figure it out with basic math. First I need to know the unit used to measure the atomic mass/weight of elements on the periodic table and that happens to be daltons. Now the mass of the earth is 5.972 × 10^24 kg One kilogram is 6.0221366516752 × 10^26 daltons (5.972 × 10^24) * (6.0221366516752 × 10^26) = 3.59642 × 10^51 Therefore the earth's mass in daltons is 3.59642 × 10^51 -----------------next step--------------------- Now the earth's chemical composition is: 32.1% iron 30.1% oxygen 15.1% silicon 13.9% magnesium 2.9% sulfur 1.8% nickel 1.5% calcium 1.4% aluminum 1.2% other elements Now if I'm thinking correctly this should mean the amount of atoms on earth is the dalton mass of the earth divided by the percentage of each element divided again by the element's atomic weight. Then added together. ----------------------------------- Number of iron atoms (3.59642 × 10^51) * 32.1/100 = 1.1544508 × 10^51 (1.1544508 × 10^51) / 55.845 = 2.0672411 × 10^49 ------------------------------------- Number of oxygen atoms (3.59642 × 10^51) * 30.1/100 = 1.0825224 × 10^51 (1.0825224 × 10^51) / 15.999 = 6.7661879 × 10^49 -------------------------------------- Number of silicon atoms (3.59642 × 10^51) * 15.1/100 = 5.4305942 × 10^50 (5.4305942 × 10^50) / 28.0855 = 1.9335936 × 10^49 --------------------------------------- Number of magnesium atoms (3.59642 × 10^51) * 13.9/100 = 4.9990238 × 10^50 (4.9990238 × 10^50) / 24.305 = 2.0567882 × 10^49 ---------------------------------------- Number of sulfur atoms (3.59642 × 10^51) * 2.9/100 = 1.0429618 × 10^50 (1.0429618 × 10^50) / 32.065 = 3.2526487 × 10^48 ----------------------------------------- Number of nickel atoms (3.59642 × 10^51) * 1.8/100 = 6.473556 × 10^49 (6.473556 × 10^49) / 58.6934 = 1.1029445 × 10^48 ------------------------------------------ Number of calcium atoms (3.59642 × 10^51) * 1.5/100 = 5.39463 × 10^49 (5.39463 × 10^49) / 40.078 = 1.3460327 × 10^48 ------------------------------------------ Number of aluminum atoms (3.59642 × 10^51) * 1.4/100 = 5.034988 × 10^49 (5.034988 × 10^49) / 26.981539 = 1.8660863 × 10^48 -------------------------------------------- Number of other elements (represented as the mean atomic weight of all remaining elements) (3.59642 × 10^51) * 1.2/100 = 4.315704 × 10^49 (4.315704 × 10^49) / 151.204004191 = 2.854226 × 10^47 ---------------------------------------------- Sum of all atoms (2.0672411 × 10^49) + (6.7661879 × 10^49) + (1.9335936 × 10^49) + (2.0567882 × 10^49) + (3.2526487 × 10^48) + (1.1029445 × 10^48) + (1.3460327 × 10^48) + (1.8660863 × 10^48) + (2.854226 × 10^47) = 1.3609124 × 10^50 According to my math there are 1.3609124 × 10^50 atoms on earth or: 13,609,124,000,000,000,000,000,000,000,000,000,000,000,000,000,000 atoms or Thirteen quindecillion, six hundred nine quattuordecillion, one hundred twenty-four tredecillion atoms -------------------------------------------------------- In fact there are actually one hundred thirty-three quindecillion atoms that make up the earth. So I was off by a rough factor of 10. Clearly I got something wrong along the way but I was pretty close!
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There are 4 Replies

Your calculations look right to me - I verified your number of iron atoms to be the same. Those are a lot of atoms, and for a very tiny, almost invisible, planet in the scheme of the entire universe. Thanks for the great Math work.

Posted September 12th, 2017 by The Fly

You were actually almost exactly right; you just wrote the final number down incorrectly (the long form, that is; the scientific notation is bang-on).

Edited September 12th, 2017 by Black Yoshi

Oh wow you're right I just double checked! I missed a zero.

Hell yea I won!

Posted September 12th, 2017 by Louis De Pointe du Lac
Louis De Pointe du Lac
No love = No future

Posted September 12th, 2017 by Q
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