# chemistry

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Determine the number of bromide ions in solution when 4.23 g of magnesium bromide and 1.21 g of potassium bromide are dissolved in 120mL of water.

Please explain not sure how to even start this problem, thanks!

chemistry - DrBob222, Wednesday, December 8, 2010 at 7:57pm
Convert 4.23 g MgBr2 to moles and multiply that by 2 to obtain moles Br^- from MgBr2. moles = grams/molar mass.

Convert 1.21 g KBr the same way. Add the moles Br^- together to obtain total moles, then remember that 1 mole of Br^- ions will contain 6.022E23 ions.

chemistry - Stacy, Wednesday, December 8, 2010 at 8:58pm
4.23g MgBr2 Moles= grams/molar mass

4.23/184.113 = 0.023 moles

1.21g KBr Moles= grams/molar mass

1.21/119.002 = 0.033

and than what? I add those two #'s together? 0.033 + 0.023 = 0.033

and I don't know what you mean by 1 mole of Br- ions will contain 6.022e23 ions where did you get that info?

Thanks!

• chemistry - ,

4.23g MgBr2 Moles= grams/molar mass

4.23/184.113 = 0.023 moles
Don't round until the end. The 4.23 gives you 3 significant figure and my calculator says 0.02298 which you might round here at 0.0230. (I know we ended up with the same number and if you went through the same reasoning I did, that's great; however,.....
Now, since there are two Brions in 1 molecule of MgBr2, you want to multiply this number of moles Br ion by 2 to find the Br ions contributed by MgBr2.

1.21g KBr Moles= grams/molar mass

1.21/119.002 = 0.033
Look at this step again. I don't get 0.33 but 0.010167

and than what? I add those two #'s together? 0.033 + 0.023 = 0.033

Yes, add 0.0230 + 0.010167 to find total moles bromide ions. Then, since we know that there are Avogadro's number of molecules in a mole, number of atoms in a mole of atoms, number of ions in a mole of ions etc etc, we know that 1 mole of bromide ions will contain 6.022E23 bromide ions. How many moles of bromide ions do you have? # moles you have x 6.02E23 = ??, then I would round to three significant figures.

and I don't know what you mean by 1 mole of Br- ions will contain 6.022e23 ions where did you get that info?
If you haven't seen this number before, memorize it now and remember it forever. If you know the number of moles of "anything", you can multiply vy 6.022E23 and find the actual number of anythings in the sample.

• chemistry - ,

4.23/184.113= 0.0230

0.0230x2= 0.046

1.21/119.002=0.010167

0.046+0.010167=0.056167

0.056167x6.02E23

= 3.38x10E22 -----is this the right answer?

• chemistry - ,

Yes, that looks ok to me.