Problem 11
Question
In one experiment, the burning of \(0.312 \mathrm{g}\) sulfur produced 0.623 g sulfur dioxide as the sole product of the reaction. In a second experiment, \(0.842 \mathrm{g}\) sulfur dioxide was obtained. What mass of sulfur must have been burned in the second experiment?
Step-by-Step Solution
Verified Answer
The mass of sulfur must have been 0.420 grams in the second experiment
1Step 1: Calculate the mole ratio of sulfur to sulfur dioxide in the first experiment
First, use the given masses and the molar masses of sulfur (S) and sulfur dioxide (SO2) to find the moles of each. The molar mass of S is 32.06 g/mol and the molar mass of SO2 (32.06 g/mol for S plus 32.00 g/mol for two O's) is 64.06 g/mol. So, in the first experiment, \(0.312 \mathrm{g}\) S is \(0.312 \mathrm{g}/32.06 \mathrm{g/mol} = 0.00973 \mathrm{mol}\) S, and \(0.623 \mathrm{g}\) SO2 is \(0.623 \mathrm{g}/64.06 \mathrm{g/mol} = 0.00973 \mathrm{mol}\) SO2. The mole ratio of S to SO2 in the reaction is therefore 1:1.
2Step 2: Use the mole ratio to find how much sulfur is needed in the second experiment
According to the stoichiometry of the reaction, 1 mol of S yields 1 mol of SO2. So, in the second experiment, \(0.842 \mathrm{g}\) SO2 are \(0.842 \mathrm{g}/64.06 \mathrm{g/mol} = 0.0131 \mathrm{mol}\) SO2. Since 1 mol S yields 1 mol SO2, this amount of SO2 requires \(0.0131 \mathrm{mol}\) S for its formation.
3Step 3: Convert that amount of S from moles back to grams
Multiply by the molar mass of S to convert back to grams. The mass of S needed in the second experiment is therefore \(0.0131 \mathrm{mol} \times 32.06 \mathrm{g/mol} = 0.420 \mathrm{g}\) S.
Key Concepts
Molar MassMole RatioChemical Reactions
Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a single molecule of a compound.
For instance, to determine the molar mass of sulfur dioxide (\(\text{SO}_2\)), we add the atomic mass of sulfur (\(32.06\) g/mol) and twice the atomic mass of oxygen (\(2 \times 16.00\) g/mol), which results in \(64.06\) g/mol.
This value is crucial because it allows us to convert between the mass of a substance and the number of moles, providing a bridge between macroscopic and molecular levels. With molar mass, we can ascertain how much of a substance is involved in a reaction based on its weight.
For instance, to determine the molar mass of sulfur dioxide (\(\text{SO}_2\)), we add the atomic mass of sulfur (\(32.06\) g/mol) and twice the atomic mass of oxygen (\(2 \times 16.00\) g/mol), which results in \(64.06\) g/mol.
This value is crucial because it allows us to convert between the mass of a substance and the number of moles, providing a bridge between macroscopic and molecular levels. With molar mass, we can ascertain how much of a substance is involved in a reaction based on its weight.
- Molar mass is unique to each substance and is essential for stoichiometry calculations.
- Knowing the molar mass aids in determining the proportions of elements in chemical compounds.
Mole Ratio
In stoichiometry, the mole ratio is an expression of the relative number of moles of each substance involved in a chemical reaction. It is derived from the coefficients in a balanced chemical equation.
A balanced equation depicts the proportion of reactants and products. For example, if the equation indicates that 1 mole of sulfur \((\text{S})\) reacts to form 1 mole of sulfur dioxide \((\text{SO}_2)\), the mole ratio is \(1:1\).
This ratio is a pivotal factor in determining how much of each reactant is required or how much of each product will be produced in a given chemical reaction. In the provided exercise, the experimentation reveals a\(1:1\) mole ratio between sulfur and sulfur dioxide, which means every mole of sulfur burned corresponds directly to a mole of sulfur dioxide produced.
A balanced equation depicts the proportion of reactants and products. For example, if the equation indicates that 1 mole of sulfur \((\text{S})\) reacts to form 1 mole of sulfur dioxide \((\text{SO}_2)\), the mole ratio is \(1:1\).
This ratio is a pivotal factor in determining how much of each reactant is required or how much of each product will be produced in a given chemical reaction. In the provided exercise, the experimentation reveals a\(1:1\) mole ratio between sulfur and sulfur dioxide, which means every mole of sulfur burned corresponds directly to a mole of sulfur dioxide produced.
- Mole ratios are essential for calculating the amounts of reactants and products in chemical reactions.
- They provide a direct method to relate quantities of different chemical substances to each other.
Chemical Reactions
Chemical reactions are transformative processes where substances known as reactants undergo a change to form new substances called products. This process involves breaking old chemical bonds and forming new ones, often accompanied by observable changes such as color shifts, gas evolution, or energy release.
In our example, when sulfur burns in the presence of oxygen, it forms sulfur dioxide as a product.
Chemical reactions are governed by certain principles such as the conservation of mass and definite proportions. The conservation of mass principle implies that the mass of reactants equals the mass of the products, as seen in the experiment where sulfur reacts to form sulfur dioxide.
In our example, when sulfur burns in the presence of oxygen, it forms sulfur dioxide as a product.
Chemical reactions are governed by certain principles such as the conservation of mass and definite proportions. The conservation of mass principle implies that the mass of reactants equals the mass of the products, as seen in the experiment where sulfur reacts to form sulfur dioxide.
- Reactions can be classified as synthesis, decomposition, single-replacement, or double-replacement.
- Understanding chemical reactions help in predicting the outcomes of mixing substances and in industrial applications like manufacturing.
Other exercises in this chapter
Problem 9
In one experiment, 2.18 g sodium was allowed to react with \(16.12 \mathrm{g}\) chlorine. All the sodium was used up, and 5.54 g sodium chloride (salt) was prod
View solution Problem 10
When \(3.06 \mathrm{g}\) hydrogen was allowed to react with an excess of oxygen, \(27.35 \mathrm{g}\) water was obtained. In a second experiment, a sample of wa
View solution Problem 12
In one experiment, the reaction of \(1.00 \mathrm{g}\) mercury and an excess of sulfur yielded \(1.16 \mathrm{g}\) of a sulfide of mercury as the sole product.
View solution Problem 13
Sulfur forms two compounds with oxygen. In the first compound, \(1.000 \mathrm{g}\) sulfur is combined with \(0.998 \mathrm{g}\) oxygen, and in the second, \(1.
View solution