Problem 97 - Chemistry: Principles and Reactions

Question

97\. Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required) in the blanks provided. (a) The mass (to three significant figures) of \(6.022 \times 10^{23}\) atoms of Na ______ 23.0 g. (b) Boron has two isotopes, B-10 (10.01 amu) and B-11 (11.01 amu). The abundance of B-10 _______the abundance of B-11. (c) If S-32 were assigned as the standard for expressing relative atomic masses and assigned an atomic mass of \(10.00 \mathrm{amu}\), the atomic mass for \(\mathrm{H}\) would be _____\(1.00 \mathrm{amu} .\) (d) When phosphine gas, \(\mathrm{PH}_{3}\), is burned in oxygen, tetraphosphorus decaoxide and steam are formed. In the balanced equation (using smallest whole-number coefficients) for the reaction, the sum of the coefficients on the reactant side is _______ \(7 .\) (e) The mass (in grams) of one mole of bromine molecules is _______ \(79.90\)

Step-by-Step Solution

Answer

Answer: (a) EQ (b) MI (c) LT (d) GT (e) GT

Step 1: Calculating mass of given Na atoms

To calculate the mass of \(6.022 \times 10^{23}\) atoms of Na, multiply the number of atoms by the molar mass of Na (approximately 23.0 g/mol) and divide by Avogadro's constant (\(6.022 \times 10^{23}\)/mol). $$ \text{Mass} = \frac{ (\text{Number of atoms}) \times (\text{Molar mass of Na}) }{(\text{Avogadro's constant})} $$ $$ \text{Mass} = \frac{(6.022 \times 10^{23}) \times 23.0 \text{g/mol}}{6.022 \times 10^{23} \text{atoms/mol}} = 23.0 \text{ g} $$

Step 2: Comparing Mass

Compare the calculated mass (\(23.0 \text{ g}\)) with the provided mass (\(23.0 \text{ g}\)). In this case, they are equal, so the answer is EQ. #---PART(b)---@#

Step 3: Explaining Answer

In this question, there is no given information about the abundance of B-10 and B-11 isotopes, so we cannot compare them. Therefore, the answer is MI (more information required). #---PART(c)---@#

Step 4: Redefining H Atomic Mass

The atomic mass of sulfur (S-32) is given as 10.00 amu. In reality, the atomic mass of S-32 is approximately 32 amu, and the atomic mass of hydrogen (H) is about 1 amu. Applying the same proportion we have: $$ \frac{\text{New H Mass}}{1 \text{amu}} = \frac{10.00 \text{amu}}{32 \text{amu}} $$

Step 5: Calculating New H Atomic Mass

Calculate the new atomic mass for H: $$ \text{New H Mass} = \frac{1 \text{amu} \times 10.00 \text{amu}}{32 \text{amu}} \approx 0.31 \text{amu} $$

Step 6: Comparing New H Mass

Compare the calculated new atomic mass for H (\(0.31 \text{amu}\)) with the given mass (\(1.00 \text{amu}\)). In this case, the calculated mass is less than the given mass, so the answer is LT. #---PART(d)---@#

Step 7: Balancing Chemical Equation

First, balance the chemical equation for the reaction. The balanced equation is: $$ 4 \text{PH}_3 + 5 \text{O}_2 \rightarrow \text{P}_4\text{O}_{10} + 6 \text{H}_2\text{O} $$

Step 8: Adding Reactant Coefficients

Add the coefficients on the reactant side: $$ 4(\text{PH}_3) + 5(\text{O}_2) = 4 + 5 = 9 $$

Step 9: Comparing Coefficients Sum

Compare the sum of the coefficients on the reactant side (\(9\)) with the number \(7\). In this case, the sum is greater than the given number, so the answer is GT. #---PART(e)---@#

Step 10: Bromine Molecular Mass

One mole of bromine molecules, \(\text{Br}_2\), has a molar mass of approximately \(79.90 \text{ g/mol} \times 2 = 159.80 \text{ g/mol}\).

Step 11: Comparing Bromine Mass

Compare the mass of one mole of bromine molecules (\(159.80 \text{ g/mol}\)) with the given mass (\(79.90 \text{ g/mol}\)). In this case, the molar mass of bromine molecules is greater than the given mass, so the answer is GT.

Key Concepts

Atomic Mass Unit

The atomic mass unit (amu) is a fundamental concept in chemistry, serving as the universal unit of mass used to express the weights of atoms and molecules. It is defined as one twelfth of the mass of a carbon-12 atom in its ground state and is approximately equal to 1.66053906660 x 10^-24 grams. Understanding amu is critical when calculating molecular weights and converting between grams and moles. Each element has an average atomic mass listed on the periodic table, which reflects the weighted average of all its isotopes, based on their relative abundance and isotopic mass. This brings us to our textbook exercise, where the atomic mass of hydrogen is recalculated based on a hypothetical standard, S-32, showing us how relative masses are compared using the amu.

Chemical Reaction Balancing

Balancing chemical equations is a staple in chemistry education, ensuring that the principle of conservation of mass is upheld. Each element must have the same number of atoms on both the reactant and product sides of the equation. This means coefficients must be added to the chemical formulas to balance the equation appropriately. In the exercise provided, we're given a reaction involving phosphine gas burning in oxygen to form tetraphosphorus decaoxide and steam. By carefully placing coefficients, we ensure that the number of atoms for each element is balanced, as demonstrated in the step by step solution. This clear balancing act allows us to predictably engage with chemical reactions and understand how different substances interact.

Avogadro's Constant

Avogadro's constant, denoted as 6.022 x 10²³ per mole, is yet another cornerstone of chemistry. It represents the number of particles, usually atoms or molecules, found in one mole of a substance, linking the macroscopic world that we can observe to the microscopic world of atoms and molecules. This number is instrumental when relating a substance's molar mass to its mass in grams, as seen in the textbook exercise where we calculated the mass of a known number of sodium atoms. Understanding Avogadro's constant allows students to seamlessly transition between the number of particles and the tangible mass, facilitating quantitative analysis in chemical reactions and stoichiometric calculations.

Isotopic Abundance

Isotopic abundance reflects the percentage of a particular isotope present in a natural sample of an element. It is a crucial concept for accurately determining an element's average atomic mass, as different isotopes of the same element will have different masses due to varying numbers of neutrons. Isotopes are atoms of the same element that differ only in their neutron count, resulting in identical chemical properties but different physical properties. In our exercise, we encounter boron, which has two isotopes, B-10 and B-11. Without knowing the precise isotopic abundance of these isotopes, the average atomic weight of boron cannot be determined. Thus, the question regarding which isotope is more abundant requires more information, emphasizing the importance this concept plays in the accurate representation of elements' atomic masses.