If $\pu{2.54 mg}$ of ammonium phosphate is dissolved in enough water to make $\pu{300 mL}$ of the solution, what is the molar concentration of the solution?

The answer is $\pu{5.68e-5 mol L-1}$. However, the answer I got using dimensional analysis was wrong:

$$\frac{\pu{0.254g} \ \ce{(NH4)3PO4}}{1} \times \frac{\pu{1 mol} \ \ce{(NH4)3PO4}}{\pu{144g} \ \ce{(NH4)3PO4}} \times \frac{1}{\pu{300L}} = \frac{\pu{0.00587mol} \ \ce{(NH4)3PO4}}{\mathrm L}$$

Could someone explain how they got this answer?


closed as off-topic by Jan, Jon Custer, Mithoron, bon, Tyberius Nov 19 '17 at 21:54

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  • 1
    $\begingroup$ Please do read this section: How can I format math/chemistry expressions here? $\endgroup$ – andselisk Nov 19 '17 at 4:42
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    $\begingroup$ Dimensional analysis? There's nothing to be analyzed. Most basic math is sufficient to solve this "problem". $\endgroup$ – aventurin Nov 19 '17 at 12:16
  • $\begingroup$ Is there actually ammonium phosphate? It should lose ammonia as phosphate ion is a stronger base than $\ce{NH_3}$. $\endgroup$ – Oscar Lanzi Nov 19 '17 at 13:06
  • $\begingroup$ You missed the decimal point in $\pu{2.54 mg}$, mistaking it for $\pu{254 mg}$. $\endgroup$ – Oscar Lanzi Nov 19 '17 at 13:19
  • $\begingroup$ $2.54\pu{mg}=2.54\dot10^-3\pu{g}$. Also, $300$mL is different from $300$L. Also, the molar mass of ammonium phosphate is 149g. $\endgroup$ – Tyberius Nov 19 '17 at 21:54

First of all, ammonium phosphate formula is $\ce{(NH4)3PO4}$. Second, I would recommend to write down an expression first, and only plug in the numbers at the end. Third, mind the metric prefixes: $\pu{1 mg} \neq \pu{10^{-1} g}$. That's said, and taking into account correct molecular weight of ammonium phosphate $M(\ce{(NH4)3PO4}) = \pu{149.09 g mol-1}$:

$$C = \frac{n}{V} = \frac{m}{MV} = \frac{\pu{2.54e-3 g}}{\pu{149.09 g mol-1} \cdot \pu{0.3 L}} = \pu{5.68e-5 mol L-1}$$

where $C$ – molar concentration, $n$ – amount, $V$ – volume, $m$ – mass.


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