Chapter 1: Problem 41
Which of the following is wrong? (1) One mole of hydrogen weighs less than one mole of glucose (2) Two moles of hydrogen and one mole of helium weigh the same (3) \(0.1\) mole of oxygen weighs more than one mole of hydrogen (4) One mole of nitrogen and half a mole of oxygen weigh the same
Short Answer
Expert verified
Statements (2) and (4) are incorrect.
Step by step solution
01
Calculate the mass of one mole of hydrogen
Hydrogen (H) has a molar mass of approximately 1 gram per mole. Therefore, one mole of hydrogen weighs 1 gram.
02
Calculate the mass of one mole of glucose
Glucose (\text{C}_6\text{H}_{12}\text{O}_6) has a molar mass calculated as follows: \[ \text{C}_6\text{H}_{12}\text{O}_6 = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ grams} \text{ per mole} \]One mole of glucose weighs 180 grams.
03
Compare the mass of one mole of hydrogen and glucose
One mole of hydrogen (1 gram) weighs less than one mole of glucose (180 grams), so statement (1) is correct.
04
Calculate the mass of two moles of hydrogen
Two moles of hydrogen: \[ 2 \times 1 \text{ gram} = 2 \text{ grams} \]
05
Calculate the mass of one mole of helium
Helium (He) has a molar mass of approximately 4 grams per mole. Therefore, one mole of helium weighs 4 grams.
06
Compare the mass of two moles of hydrogen and one mole of helium
Two moles of hydrogen (2 grams) do not weigh the same as one mole of helium (4 grams). Thus, statement (2) is incorrect.
07
Calculate the mass of 0.1 mole of oxygen
Oxygen (O) has a molar mass of approximately 16 grams per mole. Therefore, \[ 0.1 \times 16 = 1.6 \text{ grams}. \]
08
Compare the mass of 0.1 mole of oxygen and one mole of hydrogen
0.1 mole of oxygen (1.6 grams) weighs more than one mole of hydrogen (1 gram). Hence, statement (3) is correct.
09
Calculate the mass of one mole of nitrogen
Nitrogen (N) has a molar mass of approximately 14 grams per mole. Therefore, one mole of nitrogen weighs 14 grams.
10
Calculate the mass of half a mole of oxygen
Half a mole of oxygen: \[ 0.5 \times 16 = 8 \text{ grams} \]
11
Compare the mass of one mole of nitrogen and half a mole of oxygen
One mole of nitrogen (14 grams) does not weigh the same as half a mole of oxygen (8 grams). Therefore, statement (4) is incorrect.
12
Final Conclusion
Among the given statements, (2) and (4) are incorrect.
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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 a fundamental concept in chemistry, referring to the mass of one mole of a substance (usually in grams per mole). It's a way to connect the microscopic world of atoms and molecules to the macroscopic world we observe. To calculate molar mass, you sum up the atomic masses of all the atoms in a molecule. For instance, in the exercise, we calculated the molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) as follows:
\( 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ grams per mole} \)
This means that one mole of glucose weighs 180 grams. Similarly, the molar masses of hydrogen, helium, and nitrogen were necessary for solving the exercise, showing how versatile this concept is for various chemical calculations.
\( 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ grams per mole} \)
This means that one mole of glucose weighs 180 grams. Similarly, the molar masses of hydrogen, helium, and nitrogen were necessary for solving the exercise, showing how versatile this concept is for various chemical calculations.
Chemical Calculations
Chemical calculations are the bread and butter of solving chemistry problems. They often involve the conversion of masses to moles and vice versa. Understanding how to perform these calculations is crucial. For example, in the exercise, we needed to compare masses of different substances. To do this, we calculated the masses starting from moles:
- One mole of hydrogen: \(1 \text{ gram}\)
- Two moles of hydrogen: \(2 \times 1 = 2 \text{ grams}\)
- One mole of helium: \(4 \text{ grams}\)
- 0.1 moles of oxygen: \(0.1 \times 16 = 1.6 \text{ grams}\)
- One mole of nitrogen: \(14 \text{ grams}\)
- Half a mole of oxygen: \(0.5 \times 16 = 8 \text{ grams}\)
Comparative Analysis of Moles
Comparative analysis of moles allows you to understand the relative amounts of different substances. This is essential for determining the correctness of chemical statements or solving stoichiometry problems. In the exercise, such comparison was critical:
- We found that one mole of hydrogen (1 gram) weighs less than one mole of glucose (180 grams), making statement (1) correct.
- We compared the masses of two moles of hydrogen (2 grams) and one mole of helium (4 grams) to conclude that statement (2) was incorrect.
- We assessed 0.1 mole of oxygen (1.6 grams) against one mole of hydrogen (1 gram) to verify that statement (3) was correct.
- And we analyzed that one mole of nitrogen (14 grams) does not weigh the same as half a mole of oxygen (8 grams), identifying statement (4) as incorrect.
Stoichiometry
Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Although the exercise was not directly about chemical reactions, understanding stoichiometry might still be beneficial. It helps you understand how much of each substance is involved or produced in a chemical reaction. For instance, knowing the molar masses, like for hydrogen (1g/mole) and glucose (180g/mole), allows you to balance equations and determine yield.
In general, stoichiometric calculations start with converting given information into moles, using the molar ratios from the balanced chemical equation, and finally converting moles back to grams if needed. With a firm grasp of stoichiometry, you can predict outcomes in chemical reactions and ensure that reactions proceed efficiently and accurately.
In general, stoichiometric calculations start with converting given information into moles, using the molar ratios from the balanced chemical equation, and finally converting moles back to grams if needed. With a firm grasp of stoichiometry, you can predict outcomes in chemical reactions and ensure that reactions proceed efficiently and accurately.