Problem 167
Question
Without doing any detailed calculations, arrange the following substances in the increasing order of number of moles: \(20.0 \mathrm{~g} \mathrm{Cl}, 35.0 \mathrm{~g} \mathrm{Br},\) and\(94.0 \mathrm{~g} \mathrm{I}\)
Step-by-Step Solution
Verified Answer
The substances in increasing order of moles are Cl, Br, I.
1Step 1: Understand the relationship between mass and moles
The number of moles of a substance can be calculated by dividing its mass (in grams) by its molar mass (in g/mol). So, more mass means more moles, given the same molar mass. Also, the smaller the molar mass of the substance, the more moles in a given mass.
2Step 2: Approximate molar masses
The molar masses are roughly 35 g/mol for Cl, 80 g/mol for Br, and 127 g/mol for I. You can remember these values from the periodic table.
3Step 3: Compare the ratios
We're not calculating exact number of moles, but we can compare the ratios of mass to molar mass. The smaller the number, the less moles the substance has. So, 20g/35g/mol (for Cl), 35g/80g/mol (for Br) and 94g/127g/mol (for I) would be the approximate ratios for these substances.
4Step 4: Arrange in increasing order
Based on these approximations, Cl has the smallest ratio then it is followed by Br and I. Therefore, the substances in the increasing order of number of moles would be Cl, Br, and then I.
Key Concepts
Understanding Molar MassChemical Moles Calculation SimplifiedStoichiometry: The Chemistry Cookbook
Understanding Molar Mass
Molar mass, one of the foundational concepts in chemistry, is the mass of one mole of a substance (element or compound). It is usually expressed in grams per mole (g/mol) and can be found on the periodic table for each element. For compounds, the molar mass is computed by adding the molar masses of the constituent elements, based on the compound's chemical formula.
As an example, let's consider water (H2O). It consists of 2 hydrogen atoms and 1 oxygen atom. Since the molar mass of hydrogen is approximately 1 g/mol and that of oxygen is about 16 g/mol, the molar mass of water would be \(2 \times 1 g/mol + 16 g/mol = 18 g/mol\). Understanding molar mass is crucial as it links the mass of a sample to the number of particles it contains, serving as a bridge between the macroscopic world we can measure and the microscopic world of atoms and molecules.
As an example, let's consider water (H2O). It consists of 2 hydrogen atoms and 1 oxygen atom. Since the molar mass of hydrogen is approximately 1 g/mol and that of oxygen is about 16 g/mol, the molar mass of water would be \(2 \times 1 g/mol + 16 g/mol = 18 g/mol\). Understanding molar mass is crucial as it links the mass of a sample to the number of particles it contains, serving as a bridge between the macroscopic world we can measure and the microscopic world of atoms and molecules.
Chemical Moles Calculation Simplified
The mole is a central unit in chemistry that quantifies the amount of a substance. One mole contains Avogadro's number of particles (\(6.022 \times 10^{23}\) particles/mole), which could be atoms, molecules, or ions. To calculate the number of chemical moles in a given sample, simply divide the sample's mass by the molar mass of the substance.
The formula for calculating moles is \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). For instance, if you have a 20.0 g sample of chlorine (\text{Cl}), knowing that the molar mass of chlorine is roughly 35.5 g/mol, the number of moles present would be approximately \( \frac{20.0 g}{35.5 g/mol} \).
By mastering this calculation, students transition from understanding the concept of moles to being able to apply this knowledge to practical examples in real-life and experimental scenarios.
The formula for calculating moles is \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). For instance, if you have a 20.0 g sample of chlorine (\text{Cl}), knowing that the molar mass of chlorine is roughly 35.5 g/mol, the number of moles present would be approximately \( \frac{20.0 g}{35.5 g/mol} \).
By mastering this calculation, students transition from understanding the concept of moles to being able to apply this knowledge to practical examples in real-life and experimental scenarios.
Stoichiometry: The Chemistry Cookbook
Stoichiometry can be likened to a recipe in cooking; it's the part of chemistry that tells you how much of each ingredient you need to mix together to make a reaction occur. It entails using balanced chemical equations to calculate the relative quantities of reactants and products.
In practice, once you have calculated the number of moles of each reactant, stoichiometry allows you to determine the proportions needed to react completely without excess. A balanced chemical equation provides the mole ratio of the reactants and products, acting as a conversion factor to predict the outcome of a chemical reaction.
For instance, if a reaction requires one mole of substance A to react with two moles of substance B, stoichiometry would tell you that for every mole of A present, you need twice as much of B to fully carry out the reaction, thus guiding the stoichiometric ‘recipe’ for the desired chemical process.
In practice, once you have calculated the number of moles of each reactant, stoichiometry allows you to determine the proportions needed to react completely without excess. A balanced chemical equation provides the mole ratio of the reactants and products, acting as a conversion factor to predict the outcome of a chemical reaction.
For instance, if a reaction requires one mole of substance A to react with two moles of substance B, stoichiometry would tell you that for every mole of A present, you need twice as much of B to fully carry out the reaction, thus guiding the stoichiometric ‘recipe’ for the desired chemical process.
Other exercises in this chapter
Problem 163
A sample of iron weighing \(15.0 \mathrm{~g}\) was heated with potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) in an evacuated container. The oxygen gener
View solution Problem 164
A sample containing \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaNO}_{3}\) gives the following elemental analysis: Na: 32.08 percent; O:
View solution Problem 168
Without doing any detailed calculations, estimate which element has the highest percent composition by mass in each of the following compounds: (a) \(\mathrm{Hg
View solution Problem 169
Consider the reaction $$6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s)$$ Without doing any detailed calculations, choose on
View solution