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Nickel reacts with sulfur to form a sulfide. If \(2.986 \mathrm{~g}\) of nickel reacts with enough sulfur to form \(5.433 \mathrm{~g}\) of nickel sulfide, what is the simplest formula of the sulfide? Name the sulfide.

Short Answer

Expert verified
Answer: The simplest formula of the sulfide is Ni₂S₃, and its name is nickel(II) sulfide.

Step by step solution

01

Find the mass of sulfur reacted

First, we need to find the mass of sulfur in the reaction by subtracting the mass of nickel from the mass of nickel sulfide: Mass of sulfur = Mass of nickel sulfide - Mass of nickel = \(5.433 \mathrm{g} - 2.986 \mathrm{g} = 2.447 \mathrm{g}\)
02

Convert the masses of nickel and sulfur to moles

We convert the mass of nickel and sulfur into moles by dividing each mass by its respective molar mass. The molar mass of nickel (Ni) is 58.69 g/mol, and the molar mass of sulfur (S) is 32.07 g/mol. Moles of nickel = \(\frac{2.986 \mathrm{g}}{58.69 \mathrm{g/mol}} = 0.0509 \mathrm{mol}\) Moles of sulfur = \(\frac{2.447 \mathrm{g}}{32.07 \mathrm{g/mol}} = 0.0763 \mathrm{mol}\)
03

Determine the mole ratio of nickel to sulfur

To find the simplest formula of the sulfide, we need to find the mole ratio of nickel to sulfur. This is done by dividing the moles of each element by the smallest number of moles calculated in the previous step. Ratio_Ni = \(\frac{0.0509 \mathrm{mol}}{0.0509 \mathrm{mol}} = 1\) Ratio_S = \(\frac{0.0763 \mathrm{mol}}{0.0509 \mathrm{mol}} = 1.499\) Since the ratio is very close to 1:1.5, we can assume the ratio should be 1:1.5. However, we need the ratio in whole numbers. Therefore, we can multiply both ratios by 2 to get the simplest whole number ratio. Simplest ratio of Ni to S = \(1 \times 2 : 1.5 \times 2 = 2:3\)
04

Write the simplest formula and name the sulfide

Now that we have the simplest mole ratio of nickel to sulfur as 2:3, we can write the simplest formula of the sulfide using the chemical symbols of nickel (Ni) and sulfur (S). The simplest formula for the nickel sulfide is Ni₂S₃. To name the compound, we use the standard naming conventions for ionic compounds: - The cation (metal) retains its name, which is nickel in this case. - The anion (non-metal) has its ending changed to -ide, which is sulfide for sulfur. Thus, the name of the sulfide is nickel(II) sulfide.

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Key Concepts

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

Stoichiometry
Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It's based on the law of conservation of mass, which states that in a chemical reaction, mass is neither created nor destroyed. To solve stoichiometric problems, you first balance the chemical equation to ensure that the number of atoms for each element is the same on both sides of the reaction.

In the given exercise, stoichiometry comes into play when determining the simple formula of the sulfide formed from nickel and sulfur. The balanced equation allows you to relate the masses of the substances through their molar masses. Based on the calculation of mass difference and mole ratios, we can conclude the simplest recipe for combining reactants to result in a specific product – in this case, nickel sulfide.
Mole Concept
The mole concept is fundamental to understanding chemical reactions at the molecular level. One mole of any substance contains the Avogadro's number (\(6.022 \times 10^{23}\)) of entities, which could be atoms, molecules, or particles. The molar mass, the mass of one mole of a substance, is characteristic of each element and compound and is expressed in grams per mole (g/mol).

In the context of the textbook exercise, converting the mass of nickel and sulfur to moles was an essential step. It allowed us to compare the amounts of nickel and sulfur on a per-atom basis by using their molar masses as a conversion factor. This comparison is what enables us to find the simplest whole number ratio between the nickel and sulfur atoms in the nickel sulfide compound.
Empirical Formula Calculation
The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound. It's an essential piece of information since it represents the basic proportion in which elements combine to form a compound, regardless of the actual number of atoms involved.

In our exercise, after finding the moles of nickel and sulfur, we calculated their mole ratio, which did not give a clear whole number ratio initially. By multiplying the non-whole number ratio by a suitable integer, we arrived at the simplest whole number ratio of 2:3 for nickel to sulfur. This simplest ratio is used to construct the empirical formula, which in this case, is Ni₂S₃. Understanding how to calculate an empirical formula from mass data is crucial in chemistry as it serves as the foundation for more complex topics such as molecular formula determination and quantitative analysis in chemical reactions.

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Most popular questions from this chapter

Tin(IV) chloride is used as an external coating on glass to toughen glass containers. It is prepared by reacting tin with chlorine gas. $$ \mathrm{Sn}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SnCl}_{4}(l) $$ The process requires a \(25.0 \%\) excess of tin and has a yield of \(73.7 \% .\) Calculate the mass of tin and the volume of chlorine gas \(\left(d=2.898 \mathrm{~g} / \mathrm{L}\right.\) at \(\left.25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)\) needed to produce $$ 0.500 \mathrm{~L} \text { of } \mathrm{SnCl}_{4}(d=2.226 \mathrm{~g} / \mathrm{mL}) $$

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Consider the hypothetical reaction $$ 2 \mathrm{R}_{2} \mathrm{X}_{5}+2 \mathrm{Z}_{8} \longrightarrow 5 \mathrm{X}_{2}+4 \mathrm{RZ}_{4} $$ When \(25.00 \mathrm{~g}\) of \(\mathrm{Z}_{8}(\mathrm{MM}=197.4 \mathrm{~g} / \mathrm{mol})\) react with an excess of \(\mathrm{R}_{2} \mathrm{X}_{5}, 21.72 \mathrm{~g}\) of \(\mathrm{X}_{2}\) are produced. (a) How many moles of \(\mathrm{X}_{2}\) are produced? (b) What is the molar mass of \(\mathrm{X}_{2}\) ?

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