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Chapter 4: Problem 130

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 1.343 g sample of foot powder is 13.01%.

Step by step solution

01

Finding the molar mass of zinc and Zn2P2O7

First, we need to find the molar mass of zinc (Zn) and zinc pyrophosphate (Zn2P2O7). The molar mass of an element or a compound is the sum of the atomic masses of the constituent atoms multiplied by their respective moles. The molar mass of Zn is 65.38 g/mol, P is 30.97 g/mol, and O is 16.00 g/mol. To find the molar mass of Zn2P2O7, we can use the equation: Molar mass Zn2P2O7 = 2(Molar mass Zn) + 2(Molar mass P) + 7(Molar mass O)
02

Calculate the molar mass of Zn2P2O7

Now let's calculate the molar mass of Zn2P2O7: Molar mass Zn2P2O7 = 2(65.38) + 2(30.97) + 7(16.00) = 130.76 + 61.94 + 112.00 = 304.70 g/mol
03

Calculate the mass of zinc in zinc pyrophosphate

We have the mass of zinc pyrophosphate, and now we have to find out the mass of zinc only: Mass of Zn in Zn2P2O7 = (mass of Zn2P2O7) * (molar mass of Zn) / (molar mass of Zn2P2O7) Mass of Zn = 0.4089 * (130.76) / 304.70
04

Calculate the mass of zinc

Now let's calculate the mass of zinc: Mass of Zn = 0.4089 * (130.76) / 304.70 = 0.1747 g.
05

Calculate the mass percent of zinc in the sample of foot powder

Finally, we can compute the mass percent of zinc in the sample of foot powder. We'll use the equation: Mass percent of Zn = (mass of Zn in the sample) / (mass of the sample) * 100 Mass percent of Zn = (0.1747) / (1.343) * 100
06

Calculate the mass percent of zinc

In the final step, we can calculate the mass percent of zinc: Mass percent of Zn = (0.1747) / (1.343) * 100 = 13.01 % So, the mass percent of zinc in the sample of foot powder is 13.01%.

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Key Concepts

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

Molar Mass Calculation
Calculating molar mass is essential in stoichiometry. It helps determine how much of an element is in a compound. The molar mass is the total mass (in grams) of one mole of a substance. You find it by summing the atomic masses of all the atoms in a molecule. For example, zinc (Zn) has an atomic mass of 65.38 g/mol.

To calculate the molar mass of zinc pyrophosphate (Zn\(_2\)P\(_2\)O\(_7\)), we combine the molar masses of its components:
  • 2 atoms of Zn (2 \(\times\) 65.38 g/mol)
  • 2 atoms of P (2 \(\times\) 30.97 g/mol)
  • 7 atoms of O (7 \(\times\) 16.00 g/mol)
Summing these values gives 304.70 g/mol for Zn\(_2\)P\(_2\)O\(_7\). Understanding this makes further calculations possible.
Chemical Precipitation
In chemistry, precipitation refers to the process of a solid forming in a solution. This occurs when the reaction of two soluble substances results in an insoluble product. It is a crucial technique for separating specific compounds from a mixture. In our example, zinc in the foot powder forms an insoluble compound, Zn\(\text{NH}_4\)PO\(_4\).

Heating the precipitate results in zinc pyrophosphate, Zn\(_2\)P\(_2\)O\(_7\), being formed. This transformation allows us to measure the amount of zinc present by analyzing the product obtained from heating. Chemical precipitation is valuable when purifying or quantifying components within a sample.
Mass Percent Composition
Mass percent composition is a way of expressing a component's concentration within a compound or mixture. It tells us the percentage of a specific element's mass relative to the total mass of the mixture.

To calculate the mass percent of zinc in the foot powder, we divide the mass of zinc by the total mass of the sample and multiply by 100. This formula is: \[ \text{Mass percent} = \left( \frac{\text{mass of Zn}}{\text{mass of sample}} \right) \times 100 \]
  • Using our values: \( \frac{0.1747}{1.343} \times 100 \approx 13.01\% \)
This calculation gives the proportion of zinc relative to the entire sample, informing us about its composition and content.
Zinc Compounds in Chemistry
Zinc compounds are widely studied due to their numerous applications and reactions. In this example, zinc is present within the foot powder and eventually forms part of zinc pyrophosphate. Understanding zinc's role and how it combines with other elements is crucial in chemistry and industry.

Zinc is valuable for its properties:
  • Has a moderately high melting point
  • Acts as a good reducing agent
  • Used in galvanization to protect metals from corrosion
Exploring zinc compounds like Zn\(_2\)P\(_2\)O\(_7\) helps us comprehend zinc's behavior in various chemical processes, such as electroplating and synthesis reactions.

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Most popular questions from this chapter

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

For the following chemical reactions, determine the precipitate produced when the two reactants listed below are mixed together. Indicate “none” if no precipitate will form \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\)__________________(s) \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow\)__________________(s) \(\mathrm{NaCl}(a q)+\mathrm{KNO}_{3}(a q) \longrightarrow\)__________________(s) \(\mathrm{KCl}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow\)__________________(s) \(\mathrm{FeCl}_{3}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\)__________________(s)

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of 0.5894 g were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO\(_{4}\) obtained was 0.5032 g. What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

An average human being has about 5.0 \(\mathrm{L}\) of blood in his or her body. If an average person were to eat 32.0 \(\mathrm{g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

A 1.42 -g sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\) was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.
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