Small Group Questions
1.
Can the Law of Constant Composition be applied to solutions?2.
What is the purpose of calculating a reactant’s gram equivalent weight?Explanations to Practice Questions
1. D
Ionic compounds are composed of atoms held together by ionic bonds. Ionic bonds associate charged particles with disparate electronegativities; for example, sodium (Na+
) with chloride (Cl-). In ionic bonds, electrons are not really “shared” but are rather donated from the less electronegative atom to the more electronegative atom. As a result, ionic compounds are not formed from true molecules, as are covalent compounds. (A) and (B) both describe covalent compounds; their smallest unit is a molecule, which is typically described in terms of molecular weight and moles. In contrast, ionic compounds are measured using “formula weights” and are made of three-dimensional arrays of their charged particles, as indicated in (D). (C) is incorrect because ionic compounds do not share electrons equally; equal sharing occurs in covalent bonds.2. A
Of the compounds listed, only (A) and (C) are ionic compounds, which are measured in “formula weights.” The other options, (B) and (D), are covalent compounds and thus are measured in “molecular weight.” This clues us in to the fact that we don’t really even need to examine choices (B) and (D). (A) consists of potassium (39.0983 g) plus chloride (35.453 g), which has a total weight of 74.551 grams. (C) is made up of 2 lithiums (2 × 6.941 g) and 2 chlorides (2 × 35.4527 g), whose sum exceeds 75 grams.
3. B
First, it is helpful to know the molecular weight of one mole of H2
SO4, which is found by adding the molecular weight of the atoms that constitute the molecule: 2 × (molecular weight of hydrogen) + 1 × (molecular weight of sulfur) + 4 × (molecular weight of oxygen) = 2 × 1.00794 g + 32.065 g + 4 × 15.9994 g = 98.078 g. Next, you must understand what the term4. C
The definition of an empirical formula is a formula that represents a molecule with the simplest ratio, in whole numbers, of the atoms/elements comprising the compounds. In this case, given the empirical formula CH, any molecule with carbon and hydrogen atoms in a 1:1 ratio would be accurately represented by this empirical formula. Benzene, C6
H6 (A), is thus correct, as is ethyne, C2H2 (B). (D) has eight carbon atoms and eight hydrogen atoms, making it correct as well. (C) is incorrect, and thus the right answer, because it has three carbon atoms while having four hydrogens. Both its molecular and empirical formulas would be C3H4, because this formula represents the smallest whole number ratio of its constituent elements.5. A
The percent composition of any given element within a molecule is equal to the molecular mass of that element in the molecule, divided by the formula or molecular weight of the compound, times 100%. In this case, acetone, C3
H6O, has a total molecular weight of (12.0107 g × 3 + 1.00794 g × 6 + 15.994 g × 1) = 58.074 g/mol, of which 12.0107 g × 3 = 36.0321 g/mol is from carbon. Thus, the percent composition of carbon is 63.132%. With this calculation serving as an example, you can calculate the percent composition for ethanol (C2H6O; MW = 41.023 g/mol) to be 58.556%; for C3H8 (MW = 44.096 g/mol) to be 81.713%; and for methanol (CH4O; MW = 32.036 g/mol) to be 37.491%. These calculations make it clear that although both acetone (A) and ethanol (B) have percent compositions of carbon close to 63%, acetone is closer (within 1%).6. B
This reaction is a classic example of a neutralization reaction, in which an acid and a base react to form water and a new aqueous compound. Although this reaction may also appear to fit the criteria for a double-displacement reaction, in which two molecules essentially “exchange” with each other, neutralization is a more specific description of the process. A single-displacement reaction is typically a redox (reduction/oxidation) reaction in which one element is replaced in the molecules; thus (A) and (D) are incorrect.
7. B