Fundamentals E.3

Jessica Ramirez 3D
Posts: 5
Joined: Thu Jul 11, 2019 12:17 am

Fundamentals E.3

The question asks word for word
"In a nanotechnology lab you might have the capability to manipulate individual atoms. The atoms on the left are gallium atoms (molar mass 70 g/mol), and those on the right are atoms of astatine (molar mass 210 g/mol). How many astatine atoms would the pan on the right have to contain for the masses on the two pans to be equal?"

I understand that the atoms of astatine are three times as heavier, but how would I go about this problem? There is also a diagram that has 9 gallium atoms and 9 astatine atoms.

Pegah Nasseri 1K
Posts: 100
Joined: Wed Feb 27, 2019 12:15 am

Re: Fundamentals E.3

There are 9 gallium atoms on the left pan. You have to figure out how many astatine atoms need to be placed on the right pan for the masses on the two sides to be equal. Since it is known that astatine is three times as heavier like you mentioned, you would only need 3 astatine atoms to balance 9 gallium atoms.

Kate Osborne 1H
Posts: 102
Joined: Fri Aug 30, 2019 12:16 am

Re: Fundamentals E.3

The way the diagram is set up is kind of confusing as there are 9 astatine atoms shown, but it is not saying that you need to put all nine it is asking how many you need which is not necessarily all nine as the calculations show.

Maia_Jackson_2C
Posts: 101
Joined: Fri Aug 30, 2019 12:17 am

Re: Fundamentals E.3

I started the problem by finding how many grams of gallium there were so I knew how much astatine I needed to balance the scale, which was 3.794 grams. Then I divided that by the molar mass of astatine, and got 1.8066 x 10^24 moles, and then I divided that by Avogadro's constant to get 3 atoms of astatine. This is just how I solved it to check that the initial ratio of 1:3 was correct.

Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

Re: Fundamentals E.3

I think that the problem is as simple as it seems. Astatine is 3x heavier per mole than gallium, and 1 mole of each is the same amount of atoms, ~6.02x10^23, so each individual atom is 3x heavier. You only need to count up the amount of gallium atoms and divide by 3, giving 3 atoms.

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Re: Fundamentals E.3

I multiplied the 9 gallium atoms by the molar mass, 70 g.mol^-1 and made it equal to a variable, "x" representing the number of astatine atoms because it's unknown and multiplied that by the molar mass of astatine, 210 g.mol^-1. From there the variable I received was 3 atoms of astatine. The math would look like this 9 Ga * 70 g.(mol Ga)^-1 = x*210 g.(mol At) because the pan on both sides must be equal.

kendal mccarthy
Posts: 109
Joined: Wed Nov 14, 2018 12:22 am

Re: Fundamentals E.3

I solved it like this:

9 Ga atoms * 1/6.022e23 * 70g Ga/1 mol Ga = 1.04616e-21g Ga

and since you want the number of atoms of As to be equal to the weight of the number of atoms of Ga, you can use the weight of Ga you just calculated to get the number of atoms of As.

1.04616e-21g As * 1 mol As/210 g As * 6.022e23 atoms As/1 mol As = 3 atoms As

405268063
Posts: 102
Joined: Thu Jul 25, 2019 12:17 am

Re: Fundamentals E.3

E3)
The mass of 9 gallium ions = (70 g/mol)(9 atoms) = 630 g
Divide 630 g by the molar mass of astatine = 3 astatine atoms

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