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Calculate the ppm of $\ce{Fe^2+(aq)}$ if $\pu{0.0055 g}$ of $\ce{Fe(NO3)2}$ is dissolved in $\pu{2.0 kg}$ of water.

I have tried converting g of iron nitrate into moles, using mole ratio between iron nitrate and iron(II) ions and then converting to grams of iron(II) ions. I then took these grams and divided by grams of solution ($0.0005 + 20000$).

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You are wrong. Assuming that you are doing ppm in terms of mass: There are $0.0055$ grams of $\ce{Fe(NO3)2}$. Its molar mass is $179.8548$ grams/mol. Thus, you have $3.058022\!\times\! 10^{-5}$ moles. Multiplying this by the molar mass of iron gives you $0.00170775$ grams of $\ce{Fe}$. Thus, you have $0.0037922$ grams of $\ce{NO3}$. The concentration of iron is $$\mathrm c = \ce{\frac{Fe}{Water+Fe(NO3)2}} = \frac{0.00170775}{0.0055+2000}= 8.538726518 \times10^{-7}$$ Thus, $\ce{Fe}$ has a $\pu{ppm}$ of $0.85$

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    $\begingroup$ It is not well received here to answer questions that should be closed for being homework. Although admittedly somehow this one wasn’t closed … $\endgroup$ – Jan May 1 '16 at 11:52
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    $\begingroup$ @Jan It showed effort. That is why I answered it. And I suppose that is the reason why it wasn't closed. $\endgroup$ – CoffeeIsLife May 1 '16 at 11:54
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    $\begingroup$ You can use $\ce{} for chemical formulae, it automatically puts the symbols upright for you. For example: $\ce{NO3-}$ --> $\ce{NO3-}$ $\endgroup$ – orthocresol May 1 '16 at 11:54
  • $\begingroup$ not really …$%MathJax comments hurray! =D$ $\endgroup$ – Jan May 1 '16 at 12:00
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    $\begingroup$ You have way too many significant figures and the ppm of iron should be $\frac{Fe}{Water+\ce{Fe(NO3)2}}$ but the mass of the iron nitrate is really insignificant. $\endgroup$ – MaxW May 1 '16 at 15:57

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