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Chapter 5: Problem 74

A \(110.520 \mathrm{g}\) sample of mineral water is analyzed for its magnesium content. The \(\mathrm{Mg}^{2+}\) in the sample is first precipitated as \(\mathrm{MgNH}_{4} \mathrm{PO}_{4},\) and this precipitate is then converted to \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7},\) which is found to weigh 0.0549 g. Express the quantity of magnesium in the sample in parts per million (that is, in grams of \(\mathrm{Mg}\) per million grams of \(\mathrm{H}_{2} \mathrm{O}\) ).

Short Answer

Expert verified
Using the steps described, calculate the ppm of magnesium using the given amounts. After completion, the exact quantity of magnesium in parts per million in the sample will be obtained.

Step by step solution

01

Calculate the Molar Mass of Compounds

To make the necessary conversions from grams to moles, it's key to know the molar mass of the compounds. Here's how to calculate these values:\n The molar mass of \(Mg_{2} P_{2} O_{7}\) is \(2 * 24.305 (Mg) + 2 * 30.974 (P) + 7 * 16.00 (O) = 222.57 g/mol\), and \nThe molar mass of Mg is \(24.305 g/mol\).
02

Convert \(Mg_{2} P_{2} O_{7}\) to Mg

Since you know the weight of \(Mg_{2} P_{2} O_{7}\) to be 0.0549 g and it contains Magnesium which you need to find the concentration off. Use the molar mass to convert from grams of \(Mg_{2} P_{2} O_{7}\) to grams of Mg.\n \(\frac{(0.0549 g * 1 mol) / 222.57 g/mol * (2 mol Mg)\}{1 mol Mg_{2} P_{2} O_{7}} * 24.305 g/mol Mg\). The molar quantities cancel out leaving the quantity of Mg in the \(Mg_{2} P_{2} O_{7}\) in grams.
03

Calculate the ppm of Mg

Use the mass of the water sample and the mass of magnesium calculated in the previous step to obtain the quantity of Mg in parts per million. Parts per million is defined as the mass of the substance divided by total mass times 1,000,000. The calculation to find ppm of Magnesium is then \(\frac{Mg (g)}{total mass (g)} * 1,000,000\)

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Most popular questions from this chapter

Which of the following reactions are oxidationreduction reactions? (a) \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{CO}_{2}(\mathrm{g})\) (b) \(2 \mathrm{Li}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow 2 \mathrm{LiOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (c) \(4 \mathrm{Ag}(\mathrm{s})+\mathrm{PtCl}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{AgCl}(\mathrm{s})+\mathrm{Pt}(\mathrm{s})\) (d) \(2 \mathrm{HClO}_{4}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow\) \(2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{Ca}\left(\mathrm{ClO}_{4}\right)_{2}(\mathrm{aq})\)

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Predict in each case whether a reaction is likely to occur. If so, write a net ionic equation. (a) $\mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{CuCl}_{2}(\mathrm{aq}) \longrightarrow$ (b) $\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq})+\mathrm{FeCl}_{2}(\mathrm{aq}) \longrightarrow$ (c) $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow$
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