/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Nickel sulfate forms a blue-gree... [FREE SOLUTION] | 91Ó°ÊÓ

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Nickel sulfate forms a blue-green hydrate with the formula \(\mathrm{NiSO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O}(s) .\) If this hydrate is heated to a high enough temperature, \(\mathrm{H}_{2} \mathrm{O}(g)\) can be driven off, leaving the greenish-yellow anhydrous salt \(\mathrm{NiSO}_{4}(s)\). A \(12.060\) -gram sample of the hydrate was heated to \(300^{\circ} \mathrm{C}\). The resulting \(\mathrm{NiSO}_{4}(s)\) had a mass of \(7.101\) grams. Calculate the value of \(n\) in \(\mathrm{NiSO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O}(s)\)

Short Answer

Expert verified
The formula of the hydrate is \( \mathrm{NiSO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O} \).

Step by step solution

01

Determine the Mass of Water Lost

The mass of water driven off during heating is the difference between the original mass of the hydrate and the mass of the anhydrous salt. \[\text{Mass of } \text{H}_2\text{O} = 12.060\, \text{g} - 7.101\, \text{g} = 4.959\, \text{g}\]
02

Calculate Moles of NiSO4

To find the moles of \( \text{NiSO}_4 \), we use its molar mass. The molar mass of \( \text{NiSO}_4 \) is approximately \(154.75\, \text{g/mol}\).\[\text{Moles of NiSO}_4 = \frac{7.101\, \text{g}}{154.75\, \text{g/mol}} \approx 0.04588\, \text{mol}\]
03

Calculate Moles of Water

To find the moles of water lost, use the molar mass of water, which is approximately \(18.015\, \text{g/mol}\).\[\text{Moles of } \text{H}_2\text{O} = \frac{4.959\, \text{g}}{18.015\, \text{g/mol}} \approx 0.2752\, \text{mol}\]
04

Determine the Ratio of Water to NiSO4

The hydrate's formula involves a certain number \((n)\) of water molecules per formula unit of \( \text{NiSO}_4 \). Determine \(n\) by finding the mole ratio of \(\text{H}_2\text{O}\) to \(\text{NiSO}_4\).\[n = \frac{0.2752\, \text{mol H}_2\text{O}}{0.04588\, \text{mol NiSO}_4} \approx 6\]
05

Write the Formula of the Hydrate

Based on the calculated \(n\), the formula of the hydrate is:\[\text{NiSO}_4 \cdot 6\text{H}_2\text{O}\]

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Key Concepts

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

Molar Mass
Molar mass is immensely helpful in chemistry as it allows us to relate the mass of a substance to the amount of it in moles. It refers to the mass of one mole of a given chemical compound or element. When dealing with compounds, such as nickel sulfate \(\text{NiSO}_4\), the molar mass is the sum of the masses of all the atoms in its chemical formula.
For \(\text{NiSO}_4\), we determine its molar mass by adding:
  • The atomic mass of nickel (Ni): approximately 58.69 \(\text{g/mol}\)
  • Sulfur (S): around 32.07 \(\text{g/mol}\)
  • Oxygen (O): since there are four oxygens, 4 \(\times\) 16.00 \(\text{g/mol}\)
Thus, the molar mass of \(\text{NiSO}_4\) is roughly 154.75 \(\text{g/mol}\). This value is critical for further calculations, like converting grams to moles, which helps in deducing compositions and reactions in chemistry.
Stoichiometry
In chemical reactions, stoichiometry is a key tool that helps in examining the quantitative relationships between reactants and products. It allows us to calculate the amount of reactants or products using balanced chemical equations.
With the nickel sulfate hydrate example, stoichiometry enables the calculation of water loss through the heating process. By knowing the molar masses, we can determine how many moles of water and \(\text{NiSO}_4\) are involved. This analysis employs mole ratios, which are derived from balanced chemical equations to inform us about proportions that substances react or are formed in.
Understanding stoichiometry helps you relate masses, moles, and how substances react to one another, ensuring more accurate predictions and explanations of chemical behavior.
Chemical Formula
Chemical formulas provide a wealth of information about a compound, representing not only the elements it contains but also the proportions in which these elements are present. Consider the hydrate \(\text{NiSO}_4\cdot n \text{H}_2\text{O}\). The part '\(n\text{H}_2\text{O}\)' indicates how many water molecules are associated with each formula unit of nickel sulfate.
The chemical formula tells us what kinds of atoms are present and their quantity; nickel (Ni), sulfur (S), oxygen (O), and water (\(\text{H}_2\text{O}\)) make up the hydrate. Here, determining \(n\) involves figuring out the mole ratio, which tells how many water molecules link to the \(\text{NiSO}_4\). Once calculated, the chemical formula helps in predicting the behavior during reactions, especially when heated where water is lost.
Anhydrous Salt
Anhydrous salts are compounds from which water has been removed, like after heating a hydrate. The term 'anhydrous' means 'without water.' In the example with \(\text{NiSO}_4\), once you heat the hydrate \(\text{NiSO}_4\cdot n \text{H}_2\text{O}\) to drive the water out, the remaining substance is the anhydrous salt, \(\text{NiSO}_4\).
The anhydrous form is often visually different from its hydrated counterpart and can have different chemical properties. Knowing the characteristics of the anhydrous salt is important in understanding how it will behave or react under various conditions, both in industrial applications as well as in lab settings.

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Most popular questions from this chapter

11-76. Glucose is used as an energy source by the human body. The overall reaction in the body is described by the equation $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Calculate the number of grams of oxygen required to convert \(28.0\) grams of glucose to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Also compute the number of grams of \(\mathrm{CO}_{2}(g)\) produced.

The chemical N,N'-diethyl- \(m\) -toluamide, commonly known as DEET, was developed by the U.S. Army following World War II as an insect repellent for use in jungle warfare. DEET works by blocking insect receptors used to locate hosts. The chemical formula of DEET is \(\mathrm{C}_{12} \mathrm{H}_{17} \mathrm{NO} .\) Determine the mass percentage of each of the elements in DEET to four significant figures and show that the sum of these mass percentages totals one hundred percent.

II-92. (*) A 30.450-milligram sample of a chemical known to contain only carbon, hydrogen, oxygen, and sulfur is put into a combustion analysis apparatus, yielding \(54.246\) milligrams of carbon dioxide and \(22.206\) milligrams of water. In another experiment, \(23.725\) milligrams of the compound is reacted with excess oxygen to produce \(10.255\) milligrams of sulfur dioxide. What is the empirical formula of the compound?

Why is it not possible for a reaction to have a percentage yield of the desired product greater than \(100 \%\) (what law is this in violation of)? What is the most likely explanation for a percentage yield measured in the laboratory that is in excess of \(100 \%\) ?

II - 49. Potassium nitrate is widely used as a fertilizer because it provides two essential elements, potassium and nitrogen. It is made by mixing potassium chloride and nitric acid in the presence of oxygen according to the equation $$ \begin{array}{r} 4 \mathrm{KCl}(a q)+4 \mathrm{HNO}_{3}(a q)+\mathrm{O}_{2}(g) \rightarrow \\ 4 \mathrm{KNO}_{3}(a q)+2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ How many kilograms of potassium nitrate will be produced from a solution containing \(50.0\) kilograms of potassium chloride and one containing \(50.0 \mathrm{ki}-\) lograms of nitric acid? An important by-product is chlorine. How many kilograms of chlorine gas will be produced?

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