We are doing a review of what the previous semester went over, but I took the first part of chemistry in summer 2007. Needless to say I am rusty and really don't know where to start with this.Calculate the total number of ions present in 75.2g SrF2.If I remember correctly the F2 is the ionic part. The molar mass of F2 is 38g/mol. Where should I go now?

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UGFull MemberPosts: 822Mole Snacks: +134/-15Gender:  The entire compound SrF2 is an ionic compound, it consists of cations (positive ions) and anions (negative ions) like all ionic compounds. The cation here is Sr2+ and the anion is F-. There are 2 F- ions for every one Sr2+ ion so that the overall charge is neutral. Now to calculate the number of ions present. You need to find the molar mass of the entire compound. Then, using the mass given and molar mass to find the number of moles of SrF2 present. Once you have this, can you calculate the total number of ions present in the compound (given the information I gave at the beginning)? Think if you had one mole of SrF2, how many ions are present altogether?
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hokychikRegular MemberPosts: 9Mole Snacks: +0/-0Gender: Ok, so the molar mass of the molecule is 125.6g/mol. So 75.2g SrF2/125.6g SrF2 is .599 moles SrF2, correct? I don't really understand how to get the number of ions though. From what you are saying (if I am understanding right), there are 3 ions present. Do I need to get the number moles of each ion present (from the calculated total number of moles for the molecule)?
Yes correct there are three ions present, so all you need to do is multiply 0.599 by three and you will have your final answer It still says it is wrong though. I did .599 mol * 3 ions= 1.797 ions and it wants it in the form of (#)*10^(#) so i put 1.80*10^0 (to keep with the significant figures). Did I need to use Avogadro's number in there somewhere? It is one of hints given when I got it wrong.

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Ooo , of course, the total number of ions, not number of moles, silly me. Yes, then you need to multiply the number of moles of ions by Avagadro's number