Question

What is the empirical formula of an ion whose composition by mass is 57.14% sulfur and 42.86% oxygen?

Answer 1

All the ions have a charge of -2.

Answer 2

For problems like this I just get the moles and assume the total is 100g..

57.14/32 and 42.86/16

however I get 1.79 and 2.677...not sure what to do

Answer 3

@Zale101

Answer 4

@aaronq

Answer 5

Assume the total of grams of compound is 100g. So, there would be 57.14g of sulfur and 42.86g of oxygen.

Answer 6

yeah

Answer 7

Convert to moles and those moles will be in the element's subscript.

Answer 8

I get 1.79 and 2.677

Answer 9

Good, now divide the subscript all by the lowest mole.

Answer 10

1.79 is for sulfur, right?

Answer 11

yes

Answer 12

1.79 mols*

Answer 13

like 1.5

Answer 14

for f

Answer 15

I mean by 2. So I get a whole number

Answer 16

If you get a number such as 1.4, then you'd have to multiply it by 2 to all the subscript (sulfur and oxygen) to get a whole number. If it comes out to be 2.9 that's fine because that is closer to 3 (a whole number).

Answer 17

So I'm looking at S2F3^-2

Answer 18

\(\large S_2F_3\)

Answer 19

^^ That would be the empirical formula

Answer 20

the -2 is just the charge since it's a ion

Answer 21

according to the question

Answer 22

S is sulfur and and F is fluorine and both are nonmetals, so they are covalently bonded together (both of them have negative charges and that's why they share electrons).

Answer 23

When dealing with writing chemical formulas for covalent compounds, make sure you don't focus on the charges because there all negatives and there's no opposite signs that makes them want to balance as + & -

Answer 24

balance out*

Answer 25

makes sense?

Answer 26

Wouldn't the charge still have to be 0?

Answer 27

but all charged elements are ions, including negatives but if a negative element is boned with a negative element , there charges will share and they wont play as "ions" no more.

Answer 28

The charge will have to be zero ONLY if there's an ionic or acids compounds.

Answer 29

Like NaCl, are ionic compounds because there's 1 cation (+) and 1 anion (-)
Na (sodium) has 1+ charge
Cl (chlorine) has 1- charge
1-1=0

Answer 30

but we are dealing with S2F3 and those are covalent and not ionic

Answer 31

Yeah, I see, like in my book when I was learning covalent we had to find the oxidation numbers, but the charge was never 0.

Answer 32

I see what you're getting at. Well, with oxidation states, i dont think we usually include them in this question.

Answer 33

I know, it's just that all 4 answer choices have a charge of -2 lol

Answer 34

OOPs it's oxygen not Florine geez, sorry about the mistake

Answer 35

I'm blind lol

Answer 36

Oh my..lol. I was using florine too..I was kind of wondering why the charge would ever be -2..

Answer 37

\(\LARGE S_{1.79} ~:~O_{2.68}\)
\(\LARGE S_{\frac{1.79}{1.79}} ~:~O_{\frac{2.68}{1.79}}\)
times by 2 to the subscripts to get whole numbers
\(\LARGE S_{2} ~:~O_{3}\)

Answer 38

Yes, then you were right about what you said regarding oxidation states...

Answer 39

Why Fluorine Zale???? lol

Answer 40

Which actually leads to another question...o3=-6 charge?

Answer 41

O is diatomic and written as o2...is that -4 or -2?

Answer 42

there's 2 sulfure, but the charge we don't know (x) and we know the oxygen is 2- and it has3 mols. Also, in oxidation state, the first element has a positive charge.
2 S x (3+)- 3 O (2-)=0
+6-6=0
\(O_3\) is -6 which means that \(S_2\) must be +6

Answer 43

2 S *(3+)- 3 O *(2-)=0

Answer 44

Why is the charge -2 then?

Answer 45

I dont have a good explanation for this. Maybe @aaronq can helps us out.

Answer 46

@Abhisar

Apparently S2O3^-2 is thiosulfate. regular s2o3 is disulfur trioxide. so...i guess if they want an ion, the ^-2 will work