Fundamentals E3- 7th Edition

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Michelle Cervantes 1H
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Joined: Fri Sep 28, 2018 12:18 am

Fundamentals E3- 7th Edition

Postby Michelle Cervantes 1H » Wed Oct 03, 2018 10:27 pm

I am still confused as to how the answer gets 3 astatine atoms to balance the pan of gallium. Can someone explain more about how to get this conclusion? Thanks!

Layal Suboh 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

Re: Fundamentals E3- 7th Edition

Postby Layal Suboh 1I » Wed Oct 03, 2018 10:42 pm


First, in order to know how many atoms of Astatine you will need, you must calculate the mass of Gallium. You use the 9 atoms from the picture and avogadro's number to convert to moles, then the molar mass of Gallium to convert from moles to grams:

Mass Gallium= (9 atoms) *(1 mol Ga/6.02*10^23 atoms Ga)*(70 g Ga/1 mol Ga)= 1.05*10^-21 grams of Gallium

Now, you will use this mass (1.05*10^-21 grams) as the starting point to calculate the number of Astatine atoms because the mass of Gallium and Astatine must be equal. After that, you use the molar mass to convert to moles, then avogadro's number to convert to atoms:

# of At atoms= (1.05*10^-21 g At)*(1 mol At/210 g At)*(6.02*10^23 atoms At/1 mol At)= 3 atoms.

Hope this helps!

Sofia Ban
Posts: 80
Joined: Fri Sep 28, 2018 12:25 am

Re: Fundamentals E3- 7th Edition

Postby Sofia Ban » Wed Oct 03, 2018 11:04 pm

In the visual, you see that there are a total of 9 Gallium atoms. Divide 9 by Avogadro's constant; this is the mole amount of Gallium on that balance. Multiply this number by 70 grams/mol (molar mass of Gallium) in order to find the mass of Gallium. Your value should come out to be 1.05x10^-21

Since the question is asking how many Astatine atoms you need to obtain the same mass as the gallium atoms, you will need 1.05x10^-21 g of Astatine. To find the amount of atoms, do the same process you did to find the mass of Gallium, just backwards. Divide 1.05x10^-21g by 210grams/mol (molar mass of Astatine); you will obtain the mole amount of Astatine needed (4.98x10^-24mol). Find the amount of Astatine atoms by creating a math equation, where x is the amount of Astatine atoms. (x/6.022x10^23)=4.98x10^-24mol. You should get an approximate value of 3 for x.

Jessica Castro 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

Re: Fundamentals E3- 7th Edition

Postby Jessica Castro 2H » Wed Oct 03, 2018 11:29 pm

From the book, we know that the mass of a molecule equals the number of atoms multiplied by the number of atoms (m=nM). If we solve for gallium's and astatine's mass using their respective molar masses and Avogadro's number, we compare the two to see how to make them equal (multiply by 3). By getting 3 as the number to make the masses equal, we can assume that 3 is also the number of astatine atoms needed to make the masses equal because the ratio of atoms equals the ratio of the masses.

Another easier way to see it is through the molar masses given (the molar mass of astatine is 3 times the molar mass of gallium).

Also look through section E in the Fundamentals section of the textbook; they explain this concept (how molar mass, number of atoms, and mass) are related.

Hope this helps! :)

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