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Chapter 4: Problem 116

The zinc in a \(1.343-g\) sample of a foot powder was precipitated as \(\mathrm{ZnNH}_{4} \mathrm{PO}_{4} .\) Strong heating of the precipitate yielded \(0.4089 \mathrm{~g} \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} .\) Calculate the mass percent of zinc in the sample of foot powder.

Short Answer

Expert verified
The mass percent of zinc in the sample of foot powder is approximately 12.97 %.

Step by step solution

01

Obtain the moles of zinc in Zn2P2O7

Using the molar mass, we will find the moles of Zn in the zinc pyrophosphate. The molar mass of Zn2P2O7 is calculated as follows: Molar mass of Zn2P2O7 = 2 * Molar mass of Zn + 2 * Molar mass of P + 7 * Molar mass of O Molar mass of Zn2P2O7 = 2(65.38 g/mol) + 2(30.97 g/mol) + 7(16.00 g/mol) Molar mass of Zn2P2O7 = 306.70 g/mol Now we can find the moles of Zn2P2O7: moles of Zn2P2O7 = mass of Zn2P2O7 / Molar mass of Zn2P2O7 moles of Zn2P2O7 = 0.4089 g / 306.70 g/mol moles of Zn2P2O7 = 0.001333 mol
02

Calculate the moles of Zn in Zn2P2O7

Since each molecule of Zn2P2O7 contains 2 Zn atoms, the moles of Zn can be calculated using the stoichiometric ratio of the moles of Zn2P2O7 and Zn: moles of Zn = 2 * moles of Zn2P2O7 moles of Zn = 2 * 0.001333 moles of Zn = 0.002666
03

Determine the mass of zinc in the sample

Now we will convert the moles of Zn to grams using the molar mass of Zn (65.38 g/mol): mass of Zn = moles of Zn * Molar mass of Zn mass of Zn = 0.002666 * 65.38 g/mol mass of Zn = 0.1742 g
04

Calculate the mass percent of zinc in the foot powder sample

Finally, we can determine the mass percent of Zn in the foot powder sample using the formula: mass percent of Zn = (mass of Zn in sample / mass of foot powder sample) * 100 mass percent of Zn = (0.1742 g / 1.343 g) * 100 mass percent of Zn = 12.97 % The mass percent of zinc in the sample of foot powder is approximately 12.97 %.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moles of Zn
To find the number of moles of zinc in a substance, we need to use the formula that relates mass, molar mass, and moles. The formula is given by:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
In this particular exercise, we are dealing with zinc in the compound Zn\(_2\)P\(_2\)O\(_7\). Starting by calculating the molar mass of Zn\(_2\)P\(_2\)O\(_7\), we consider the contribution from each element:
  • Zinc (Zn) has a molar mass of 65.38 g/mol, and there are two zinc atoms per formula unit.
  • Phosphorus (P) has a molar mass of 30.97 g/mol, with two phosphorus atoms per formula unit.
  • Oxygen (O) has a molar mass of 16.00 g/mol, contributing seven oxygen atoms per formula unit.
These calculations help us determine that the molar mass of Zn\(_2\)P\(_2\)O\(_7\) is 306.70 g/mol. With a measured mass of 0.4089 g of Zn\(_2\)P\(_2\)O\(_7\), we find the moles of zinc by dividing by this molar mass. From there, we multiply by 2 (due to two zinc atoms per molecule) to find the total moles of zinc.
Zinc Pyrophosphate
Zinc pyrophosphate is the compound Zn\(_2\)P\(_2\)O\(_7\), formed when zinc reacts with phosphate ions.
It plays a crucial role in this type of quantitative chemical analysis because it provides a known stoichiometric relationship between zinc and phosphorus. When zinc in the sample is precipitated as a zinc-phosphate compound, it allows for precise mass measurement.
In many cases, such as this, heating is required to convert the precipitate to Zn\(_2\)P\(_2\)O\(_7\). This conversion helps in making accurate measurements, since Zn\(_2\)P\(_2\)O\(_7\) is a stable compound with well-defined stoichiometry. This stability ensures that when we weigh the precipitate, it accurately reflects the amount of zinc originally present in the sample.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. In this exercise, stoichiometry is used to relate the moles of Zn in the compound Zn\(_2\)P\(_2\)O\(_7\) with the amount of zinc present in the original foot powder sample.
The stoichiometry tells us that for every mole of Zn\(_2\)P\(_2\)O\(_7\) you have two moles of Zn. Hence, we multiply the moles of Zn\(_2\)P\(_2\)O\(_7\) by two to find the moles of Zn.
This type of problem-solving approach is essential in quantitative analytical chemistry because it allows us to precisely determine the amounts of substances based on their chemical formulas. Understanding how to apply stoichiometry effectively is vital for solving complex chemical calculations, such as determining the composition of a sample.
Foot Powder Analysis
Foot powder analysis is a practical application of chemistry in everyday products. It involves analyzing components to determine their concentration or presence in a formulated product.
The foot powder initially featured in this analysis contains zinc, which is quantified by converting it into a recognizable chemical form like zinc pyrophosphate (Zn\(_2\)P\(_2\)O\(_7\)). Such analyses are crucial for ensuring product effectiveness and safety by verifying that the active ingredients meet the required specifications.
  • The primary goal of this analysis is to determine the mass percent of zinc.
  • This involves several steps, including precipitation, conversion to a stable form, weighing, and stoichiometric calculations.
  • The result indicates how much zinc is present compared to the total weight of the product, which is useful for quality control.
Foot powder analysis exemplifies the practical implementation of laboratory techniques in real-world scenarios.

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Most popular questions from this chapter

Consider the reaction between sodium metal and fluorine \(\left(\mathrm{F}_{2}\right)\) gas to form sodium fluoride. Using oxidation states, how many electrons would each sodium atom lose, and how many electrons would each fluorine atom gain? How many sodium atoms are needed to react with one fluorine molecule? Write a balanced equation for this reaction.

A student titrates an unknown amount of potassium hydrogen phthalate \(\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\right.\), often abbreviated \(\mathrm{KHP}\) ) with \(20.46 \mathrm{~mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution. KHP (molar mass \(=204.22 \mathrm{~g} /\) mol) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0\) ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals \(1.0 \mathrm{~g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The \(\mathrm{VO}^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q)\) To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.
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