how many grams are in 5.0*10^26 molecules of Cl2

how many grams are in 5.0*10^26 molecules of Cl2

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(Greetings Alan, It is diatomic Chlorine, not carbon).

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This question originally had an anonymously appended, incorrect answer to the chemistry question, and a vulgar-worded (probably why it was deleted) opinion stating this is a math site and not a chemistry site. My comments below reflect this.

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The chemistry answer by Anonymous is incorrect. 

The correct solution and the method for its derivation are below.

His/her opinion of the site is subjective. There are tutors on this site who know chemistry. Any science question, which relates to applied mathematics is reasonable.

To solve this:

Divide the quantity of the molecules by Avogadro constant (currently: 6.022141E²³/mol). This constant is the number of atoms or molecules per molar mass of an element or compound.

5.0*10E²⁶ molecules (Cl₂)/ 6.022141E²³/mol = 830.3 mols of Cl₂

Look up the Relative atomic mass (aka Standard atomic weight) of Cl (Chlorine). It is 35.45. Pure Chlorine, in the gas state, usually exists in a diatomic form (Cl₂). This condition is indicated by the subscript. Multiply the Relative atomic mass by this subscript value (2): 35.45 *2 = 70.90. (Note that diatomic forms are considered molecules).

The Relative atomic mass of diatomic Chlorine is 70.90. Convert to gram mass (by multiplying by 1g). 70.90 * 1g = 70.90g. This is the gram molecular mass and it equals 1 mol: 70.90g of Cl₂ = 1 mol. Restated: for each mol of Cl₂ there are 70.90 grams of Cl₂.*

Multiply the 830.3 mols of Cl₂ by 70.90g/mol = 58873.2g of Cl₂. (58.8732Kg of Cl₂).

This is slightly less chlorine than required to maintain "chemical pasteurization” of an Olympic-size non-comercial pool.

 ~~D~~

Avogadro's number N_A = 6.0221415.10²³ molecules per mole

One mole of C₁₂ is 12 grams.

This means 6.0221415.10²³ molecules of C₁₂ is 12 grams.

So 5*10²⁶ molecules of C₁₂ is (12*5*10²⁶)/(6.0221415*10²³⁾ grams

$${\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{26}}}\right)}{\left({\mathtt{6.022\: \!141\: \!5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}\right)}} = {\mathtt{9\,963.233\: \!178\: \!762\: \!073\: \!259\: \!2}}$$  grams

But see David's reply below.

Thanks David, I obviously misread it as Carbon. Shows how important it is to use the formatting options to ensure clarity (or perhaps I should just get stronger spectacles!).