Chemistry Answer Note #5

Question 1 In the Lewis structure of PO43- in which all atoms obey the octet rule, a.) the formal charge on P is … ; b.) and the formal charge on each O is … . Answer According to the octet rule in Lewis structure, all atoms in a molecule should have eight electrons in their outermost energy level. In the case of PO43- ion, we have four oxygen atoms surrounding a central phosphorous atom. To determine the formal charge on each atom in the PO43- ion, we need to first draw the Lewis structure and then calculate the formal charge for each atom. To draw the Lewis structure of PO43- ion, we first determine the total number of valence electrons. Phosphorous belongs to group 5A and has 5 valence electrons, while each oxygen atom has 6 valence electrons, giving a total of (5+4x6) 29 valence electrons. As the ion has a -3 charge, we add three more electrons (due to 3 extra negative charges) to the total, giving 32 valence electrons in total. Now, we distribute these electrons around the atoms, starting with the outermost electrons and taking into consideration the octet rule. This results in the following Lewis structure for PO43-: :O:; (-) <-- Negative charge on each oxygen atom ---(-)P:; - - - - (-)O ---:O:; (-) According to the octet rule, each oxygen atom should have eight electrons around it, while the central phosphorous atom should have ten. We can see that in this structure, each oxygen atom has six valence electrons (two alone and four shared with phosphorous), while the phosphorous atom has a total of ten electrons (five alone and five shared with oxygen atoms). To calculate the formal charge on each atom, we use the formula FC = V - N - (S/2), where V is the number of valence electrons, N is the number of lone pair electrons, and S is the number of shared electrons. Applying this formula to each atom in PO43-, we get: Formal charge on phosphorous (FCP) = 5 (V) - 0 (N) - (10/2) (S) = 0 Formal charge on each oxygen atom (FCO) = 6 (V) - 2 (N) - (4/2) (S) = -1 Therefore, the formal charge on phosphorous in PO43- is 0, and the formal charge on each oxygen atom is -1. In-depth research on this topic reveals that while the octet rule is a useful guideline for predicting the distribution of electrons in a molecule, there are some exceptions and limitations to this rule. Some molecules, such as those containing elements from period 3 or higher, may have more than eight electrons around the central atom due to the presence of d orbitals in higher energy levels. In some cases, molecules may also have less than eight electrons around the central atom, such as in BCl3 where boron only has six electrons around it. Furthermore, the concept of formal charge is a useful tool for determining the most stable Lewis structure of a molecule, but it is not an accurate reflection of the actual charge distribution within the molecule. Other factors, such as electronegativity and resonance, also play a role in determining the most stable Lewis structure for a molecule. In conclusion, the formal charge on phosphorous in PO43- is 0, while the formal charge on each oxygen atom is -1. While the octet rule is a helpful guide, it is important to consider other factors when determining the most stable Lewis structure of a molecule. Question 2 How many grams of silver chromate will precipitate when 291 mL of 0.250 M silver nitrate reacts with potassium chromate in the following reaction? 2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3 Answer To answer this question, we must first understand the reaction taking place between silver nitrate and potassium chromate. According to the given equation, the two compounds will react to form silver chromate and potassium nitrate. In order to determine the amount of silver chromate precipitated, we must use the principles of stoichiometry and the given values of concentration and volume of silver nitrate. Step 1: Balanced Chemical Equation 2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3 Step 2: Molar Ratio From the balanced equation, we can see that for every 2 moles of silver nitrate, 1 mole of silver chromate is formed. This means that there is a 2:1 molar ratio between silver nitrate and silver chromate. Step 3: Converting Volume to Moles Using the given concentration of silver nitrate (0.250 M) and the volume of 291mL, we can convert the volume to moles using the formula: Moles = Concentration (M) x Volume (L). Moles of AgNO3 = 0.250 M x 0.291 L = 0.07275 moles Step 4: Determining Moles of Silver Chromate Since there is a 2:1 molar ratio between silver nitrate and silver chromate, the number of moles of silver chromate formed will be half of the moles of silver nitrate. Moles of Ag2CrO4 = 0.07275 moles/2 = 0.036375 moles Step 5: Converting Moles to Grams Now, we can use the moles of silver chromate to calculate the mass using its molar mass. The molar mass of silver chromate (Ag2CrO4) is 331.74 g/mol. Mass of Ag2CrO4 = 0.036375 moles x 331.74 g/mol = 12.07 grams Therefore, when 291mL of 0.250 M silver nitrate reacts with potassium chromate, 12.07 grams of silver chromate will precipitate. Question 3 Question 6 P i 50 mL of 1N Ca(OH)2 solution is neutralized by 1 g of the acid having molar mass = 40 g mol¹. The basicity of the acid is Answer To determine the basicity of an acid, we must first understand the concept of basicity. Basicity refers to the number of replaceable hydrogen ions (H+) in an acid molecule. This can also be thought of as the number of moles of a monoprotic acid that can donate one mole of hydrogen ions. In this case, we will be using the term as the number of moles of the acid that can neutralize one mole of a base, in this case, Ca(OH)2. To start with the problem, let's first identify the known values we have: - Volume of base (Ca(OH)2) = 50 mL - Concentration of base (Ca(OH)2) = 1N - Mass of acid = 1 g - Molar mass of acid = 40 g mol¹ Next, we can use the formula for normality (N) to calculate the number of moles of Ca(OH)2: N = moles of solute/ liters of solution Since the concentration is given in 1N (equivalent to 1 mole/L), we can conclude that there are 1 mole of Ca(OH)2 in 1 liter. Since we only have 50 mL, we can convert this to liters by dividing it by 1000: 50 mL/1000 mL = 0.05 L. Therefore, we can say that there are 0.05 moles of Ca(OH)2 in our 50 mL solution. Now, let's determine the number of moles of acid by using the molar mass and mass of the acid given: Number of moles = mass/molar mass Number of moles = 1 g/40 g mol¹ = 0.025 moles From a neutralization reaction, we know that: Number of moles of acid = Number of moles of base Therefore, in this reaction, 0.025 moles of acid are neutralized by 0.05 moles of Ca(OH)2. This means that the basicity of the acid is 0.025 moles or simply 1. Question 4 Maleic acid is an organic compound composed of 41.39% C, 3.47% H and the rest Oxygen. if 0.41 Mole of Maleic acid has mass of 15.0g, what are the empirical formula and molecular formular of Maleic acid? Answer Maleic acid is an organic compound that is commonly used in the production of various chemicals, such as fumaric acid and maleic anhydride. The chemical formula of maleic acid is C4H4O4, with a molar mass of 116.07 g/mol. In this question, we are given that maleic acid is composed of 41.39% C, 3.47% H and the rest oxygen, and that 0.41 mole of maleic acid has a mass of 15.0g. Using this information, we can determine the empirical formula and molecular formula of maleic acid. Empirical Formula: The empirical formula of a compound represents the simplest whole number ratio of the elements present in a molecule. To determine the empirical formula of maleic acid, we first need to find the number of moles of each element present in 0.41 mole of maleic acid. Number of moles of C = 41.39% of 0.41 mole = 0.17 mole Number of moles of H = 3.47% of 0.41 mole = 0.01 mole Number of moles of O = 100% - (41.39% + 3.47%) = 55.14% of 0.41 mole = 0.23 mole Next, we need to find the smallest whole number ratio of these moles. To do this, we divide the number of moles of each element by the smallest value, which in this case is 0.01 mole. This gives us the following ratio: C: 0.17 mole/0.01 mole = 17 H: 0.01 mole/0.01 mole = 1 O: 0.23 mole/0.01 mole = 23 Therefore, the empirical formula of maleic acid is CH2O. Molecular Formula: The molecular formula of a compound represents the actual number of atoms of each element in a molecule. To determine the molecular formula of maleic acid, we first need to calculate the molar mass of the empirical formula, which is CH2O. Molar mass of CH2O = (12.01 x 1) + (1.01 x 2) + (16.00 x 1) = 30.03 g/mol Next, we divide the molar mass of maleic acid (116.07 g/mol) by the molar mass of the empirical formula (30.03 g/mol). This gives us a value of approximately 3.87. We round this value to the nearest whole number, which is 4. This means that the molecular formula of maleic acid is four times the empirical formula, giving us C4H4O4. In conclusion, the empirical formula of maleic acid is CH2O and the molecular formula is C4H4O4. Question 5 chlorine has two naturally occuring isotopes. in a sample of chlorine,75.77% of the atoms are CL-35, with a mass of 34.97 amu. another 24.23% of the atoms are CL-37, with a mass of 36.97 amu. what is the atomic mass of chlorine? Answer The atomic mass of an element is defined as the average mass of the naturally occurring isotopes in a particular sample. In the case of chlorine, there are two naturally occurring isotopes - CL-35 and CL-37. To determine the atomic mass of chlorine, we need to take into account the relative abundance and mass of these two isotopes. As stated in the question, 75.77% of the atoms in a sample of chlorine are CL-35, while the remaining 24.23% are CL-37. This means that CL-35 is the predominant isotope of chlorine. To calculate the atomic mass, we can use the following formula: Atomic mass = (Relative abundance of isotope A * Mass of isotope A) + (Relative abundance of isotope B * Mass of isotope B) Substituting the values given in the question, we get: Atomic mass of chlorine = (0.7577 * 34.97 amu) + (0.2423 * 36.97 amu) = 26.5 amu + 8.95 amu = 35.45 amu Therefore, the atomic mass of chlorine is 35.45 amu.