## Fundamentals E: Moles and Molar Mass Question #23

Ben 1B
Posts: 55
Joined: Sat Aug 17, 2019 12:16 am

### Fundamentals E: Moles and Molar Mass Question #23

What are the steps in solving question E.23 from Fundamentals E: Moles and Molar Mass?
Example: Calculate the amount (in moles) of:
(a) Cu2+ ions in 3.00 g of CuBr2
(b) SO3 Molecules in 7.00 x 10^2 mg of SO3
(c) F^-1 ions in 25.2 kg of UF6
(d) H20 in 2.00 g of (Na2CO3)(10H20)

Posts: 113
Joined: Thu Jul 11, 2019 12:16 am

### Re: Fundamentals E: Moles and Molar Mass Question #23

Ben 1B wrote:What are the steps in solving question E.23 from Fundamentals E: Moles and Molar Mass?
Example: Calculate the amount (in moles) of:
(a) Cu2+ ions in 3.00 g of CuBr2
(b) SO3 Molecules in 7.00 x 10^2 mg of SO3
(c) F^-1 ions in 25.2 kg of UF6
(d) H20 in 2.00 g of (Na2CO3)(10H20)

The steps for solving each part of this problem are basically the same.

1) Convert masses of compounds into moles.
2) Convert moles of compound into moles of whatever is being asked for.

The key in this problem is being able to recognize the ratio between the number of moles of each ion/molecule/whatever in the compound and the moles of the compound itself. For example, for every one mole of CuBr2, there is one mole of Cu2+ ions. Hope this helps!

Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

### Re: Fundamentals E: Moles and Molar Mass Question #23

Here is how you would find the amount of moles of (a) Cu2+ ions ins 3.00g of CuBr2:

1. Find the molar mass of CuBr2: 223.5g/mol CuBr2
2. Perform dimensional analysis to convert known units to mol Cu2+.

$(3.00gCuBr_{2})\cdot (\frac{1molCuBr_{2}}{223.5gCuBr_{2}})\cdot (\frac{1molCu^{2+}}{1molCuBr_{2}})=0.0134molCu^{2+}$

The product of the first two portions of the equation above describe the process of converting the amount of grams of CuBr2 to mol of CuBr2. Then, you can multiply that product by the ratio of mol Cu2+ in one mol of CuBr2. Keep in mind that CuBr2, when broken up, is $Cu^{2+}$ and $Br^{-}$.