# Calculating ppm of an ion in a solution

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 ($$0.0005 + 20000$$).

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 $$0.85$$
• You can use $\ce{} for chemical formulae, it automatically puts the symbols upright for you. For example: $\ce{NO3-}$ -->$\ce{NO3-}$– orthocresol May 1 '16 at 11:54 • not really …$%MathJax comments hurray! =D$– Jan May 1 '16 at 12:00 • 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. – MaxW May 1 '16 at 15:57